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Supplemental Notes for Calculus IIEllen Ziliak and Alexander Hulpke
Fall 2011
Alexander HulpkeDepartment of MathematicsColorado State University1874 Campus DeliveryFort Collins, CO, 80523
© 2009 by the authors. �is work is licensed under the Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/us/ or send a letter to Creative Commons, 171 Second Street, Suite 300, SanFrancisco, California, 94105, USA.
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Convergence of Improper Integrals
In this chapter we will look at whether or not improper integrals converge or diverge.
DEFINITION 1An improper integral converges if when it is evaluated the result is a �nite value. An improper integral di-verges if when it is evaluated the result is an in�nite or unde�ned value.
We begin with an example. Suppose a scientist is observing a particle that ismoving in a force �eld. He hasbeen able to determine from several measurements he made that the particle’s acceleration is proportional toits velocity, resulting in a di�erential equation
dvdt
= −2 ⋅ v
where v is the velocity of the particle. He also measured the initial velocity to be 100m/s. Solving this initialvalue problem1 we get v(t) = 100e−2t .
�e scientists now asks whether the particle will move arbitrarily far. To answer this question we need todetermine whether or not ∫ ∞
0100e−2tdt converges. In this case the value of the integral is not di�cult to
calculate using the methods previously described in this section:
limb→∞ ∫ b
0100e−2tdt = lim
b→∞−50e−2t∣
b
0
= limb→∞
−50e−2b + 50
= 50
So the total distance traveled by the particle at the end of time is 50m, which is �nite.
1 �is is the equation for exponential growth:
dvdt
= −2 ⋅ v ⇒ ∫ dvv
= ∫ −2dt
A�er evaluating this integral we get ln ∣v∣ = −2t + C which we can then solve for v(t) to get v(t) = Ae−2t .�e initial condition forv(0) then gives v(t) = 100e−2t .
1
Now suppose instead of such a simple function for the velocity we got instead that the velocity of the
particle is given by v(t) = sin2(t)t2
. If we want to know, whether such a particle travels a �nite distance, we
would have to determine whether ∫ ∞0
sin2(t)t2
dt converges or diverges.We cannot just evaluate the integral,so we must discover some other techniques to answer this question.
Direct Comparison
Our goal is to determine if ∫ ∞a
f (x)dx converges without having to compute the value of the integral. Oneway to do this is to compare with an integral we know something about. Let us �rst consider an examplewhere the limits of integration are �nite to develop intuition about how we compare two functions.
EXAMPLE 1If f (x) = x2 + 2 and g(x) = x2, what can we say about ∫ 1−1 f (x)dx compared to ∫ 1−1 g(x)dx?
Solution First we look at the graphs of f (x) and g(x) below: We know geometrically ∫ 1−1 f (x)dx can
0. 5
1
1. 5
2
2. 5
3
-1 -0.8 -0.6 -0.4 -0.2 0 .2 0. 4 0 .6 0. 8 1x
f(x)=x2+2
f(x)=x2
0
1
2
3
4y
-1 -0.8 -0.6 -0.4 -0. 2 0 .2 0. 4 0 .6 0. 8 1x
be interpreted as the area under the curve of f (x) on [−1, 1] and ∫ 1−1 g(x)dx can be interpreted as the areaunder the curve of g(x) on [−1, 1]. And we notice on the interval from [−1, 1] that f (x) > g(x) (in fact this isalways true but we only care about this interval) the area under g clearly must be smaller than the area underf .
We have seen that if f (x) > g(x) then ∫ 1−1 f (x)dx > ∫ 1−1 g(x)dx.
�is result holds of course for any other interval:
THEOREM 1If on the entire interval [a, b] we have f (x) > g(x) ≥ 0 then ∫ b
af (x)dx > ∫ b
ag(x)dx.
Proof is given by picture on p348
We now want to consider what happens in the limit case:
2
EXAMPLE 2Does ∫ ∞
1
1ex dx converge?
Solution Consider f (x) = 1ex and g(x) = 1
x2 on the interval [1,∞) Let us �rst notice that 1ex is smaller
g(x)=1 ∕ x2
0
0. 2
0. 4
0. 6
0. 8
1
2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0x
f(x)=1 ∕ ex
than 1x2 on the entire interval from [1,∞). We also know that ∫ ∞
1
1x p dx converged (to 1
p−1 ) when p > 1,so the area under 1x2 from [1,∞) is equal to 1
2−1 = 1.�erefore, extrapolating from the �nite case, we would
assume that ∫ ∞1
1ex dx < 1 which would mean that this improper integral converges.
To verify this, let us calculate this integral directly, so we can verify that this guess was correct.
∫ ∞1
1ex dx = lim
b→∞ ∫ b
1e−x dx = lim
b→∞−e−x ∣
b
1
= limb→∞
−e−b + e−1 = 1e= 0.367879 < 1
�is comparison property holds for any other pair of functions. We get the following generalization of�eorem 1:
THEOREM 2Let f(x) and g(x) be continuous on [a,∞) with f (x) > g(x) ≥ 0 on the whole interval.�en if ∫ ∞
af (x)dx
converges, then ∫ ∞a
g(x)dx converges.
With this theorem we can answer the question from the start of the chapter:
EXAMPLE 3 The Particle Revisited
In the example we wanted to know whether or not ∫ ∞1
sin2(t)t2
dt converges.
Solution Since sin2(t) < 1 always, if we compare sin2(t)t2
< 1t2. Since ∫ ∞
1
1t2
dt converges we can con-
clude that ∫ ∞1
sin2(t)t2
dt converges. We therefore know that the distance traveled by the particle is bounded
by ∫ ∞1
1t2
dt = 1 m, however we cannot give a concrete value for this distance.
A similar argument can be used to show that an improper integral diverges.
3
EXAMPLE 4Does ∫ ∞
2
1√x − 1
dx converge?
Solution Consider f (x) = 1√x and g(x) = 1√
x−1 on the interval [2,∞). We see in the graph that
2
4
6
8
10
12
2 4 6 8 1 0x
10
f(x)=1 ∕ √xg(x)=1 ∕ √x-1
1√x−1 >
1√x for x > 1. We also know from previous work that ∫ ∞
1
1√x
dx diverges. since ∫ 211√x
dx is a
�nite value (namely 0.82842112) throwing this part out won’t change the conclusion that the value of theintegral is in�nite, so ∫ ∞
2
1√x
dx diverges.
Since 1√x−1 >
1√x on the entire interval [2,∞), the area under 1√
x−1 on [2,∞) must be larger than the
area under 1√x on [2,∞), so ∫ ∞
2
1√x − 1
dx should also diverge. Again we check the conclusion by actually
calculating the value of this integral:
∫ ∞2
1√x − 1
dx = limb→∞ ∫ b
2
1√x − 1
dx
let u = x−1 then we have dudx = 1 and substituting the limits of integration in we have u(2) = 1 and u(b) = b−1,
yielding:
limb→∞ ∫ b−1
1u−12 du = lim
b→∞2u
12 ∣
b−1
1= lim
b→∞2(b − 1)
12 − 2 =∞
�erefore ∫ ∞2
1√x − 1
dx diverges.
Again we state this observation as a general result:
THEOREM 3Let f(x) and g(x) be continuous on [a,∞) with f (x) < g(x) on the whole interval. �en if ∫ ∞
af (x)dx
diverges, then ∫ ∞a
g(x)dx diverges.
In all of the examples we have seen so far we have had f (x) ≥ g(x) on the entire interval we have con-sidered. In some situations this makes it unnecessaily hard to �nd a function to compare with. We will see inthe next example that one can in fact weaken this condition:
4
EXAMPLE 5
Does ∫ ∞1
sin2(x)x
+ 14
dx converge?
Solution Let f (x) = sin2(x)x + 14 and g(x) = 1
x on the interval [1,∞).�en initially we have f (x) oscil-
f(x)=sin2(x)/x+1∕4
g(x)=1 ∕ x
0. 2
0. 4
0. 6
0. 8
1
5 10 15 20 25x
lating over g(x) from [1, 3.5021575], then a�er that point we have 1x < sin2(x)x + 14 from [3.5021575,∞). Since
∫ 3.50215751
sin2(x)x
+ 14
dx = 1.425137 is �nite and ∫ 3.50215751
1x
dx = 1.2533763 is also �nite, we know thatthe behavior on the interval from 1 to 3.5021575 will not a�ect the convergence or divergence of the improperintegral.We also know that ∫ ∞
1
1x
dx diverges; similarly by splitting up the interval we have
∫ ∞1
1x
dx − ∫ 3.50215751
1x
dx = ∫ ∞3.5021575
1x
dx
letting us conclude that ∫ ∞3.5021575
1x
dx diverges as well.�en since 1x < sin2(x)x + 14 from [3.5021575,∞) we
can conclude that ∫ ∞3.5021575
sin2(x)x
+ 14
dx also diverges.
Comparison now tells us that ∫ ∞3.5021575
sin2(x)x
+ 14
dx diverges, from which we conclude by the same
argument as before that ∫ ∞1
sin2(x)x
+ 14
dx diverges,
Again we state this observation in a theorem which generalizes the theorems stated before in this section.
5
THEOREM 4 Direct Comparison Test for Improper IntegralsLet f(x) and g(x) be piecewise-continuous on [a,∞)
a) If there exists a value b ≥ a such that f (x) ≥ g(x) for all x ≥ b and ∫ ∞a
f (x)dx converges then
∫ ∞a
g(x)dx converges.
b) If there exists a value b ≥ a such that g(x) ≥ f (x) for all x ≥ b and ∫ ∞a
f (x)dx diverges then
∫ ∞a
g(x)dx diverges.
Notice if we happen to choose a f (x) ≥ g(x) and ∫ ∞a
f (x)dx diverges we cannot conclude anything
about ∫ ∞a
g(x)dx. Similarly, if g(x) ≥ f (x) and ∫ ∞a
f (x)dx converges we cannot say anything about
∫ ∞a
g(x)dx.
Limit Comparison Test
Finding a point b in �eorem 4 for which f (x) ≥ g(x) can be hard. It also leaves the case that f and grepeatedly are “oscillating around each other” but “stay close”. For example consider f (x) = 1
xand g(x) =
1 + sin(x)/2x
:
0
0.2
0.4
0.6
0. 8
1
y
10 20 30 40 50x
f(x)=1/x
g(x)=(1+sin(x)/2)/x
One might hope that the functions are “close enough” to let us compare the convergence behavior.�e proper context for describing “close enough” is that of growth, which we studied in section 7.6. We
recall the relevant de�nition:
DEFINITION 2Let f (x) and g(x) be real valued functions which are positive for su�ciently large values of x.
a) f grows faster than g as x tends to in�nity if limx→∞
f (x)g(x) = ∞, or — equivalently — if lim
x→∞g(x)f (x) = 0.
We also say that g(x) grows slower than f (x) as x →∞.
6
b) f and g grow at the same rate as x →∞ if limx→∞
f (x)g(x) = L where L is �nite and positive.
For improper integrals to converge, of course we need functions to converge to zero, so we don’t compare“growth” but “decay”. We get this by simply considering the reciprocal function 1/ f (x) and 1/g(x). We getthe following analogous de�nition:
DEFINITION 3Let f (x) and g(x) be real valued functions which are positive for su�ciently large values of x.
a) f decays faster than g as x tends to in�nity if limx→∞
f (x)g(x) = 0, or — equivalently — if lim
x→∞g(x)f (x) =∞.
We also say that g(x) decays slower than f (x) as x →∞.
b) f and g decay at the same rate as x →∞ if limx→∞
g(x)f (x) = L where L is �nite and positive.
Similar to the case of growth, if f decays faster then g there must be a value b such that f (x) < g(x) forall x ≥ b.�is means that we can apply�eorem 4 and make deductions about convergence:
EXAMPLE 6Does ∫ ∞
1
1x3 + 1dx converge?
Solution Let f (x) = 1x3 + 1 and g(x) = 1
x2 + 1 on the interval from [1,∞).We �nd that
limx→∞
g(x)f (x) = lim
x→∞
1x2+11
x3+1= lim
x→∞x3 + 1x2 + 1 =∞
�erefore f (x) decays faster than g(x), which means at some point b, for all x ≥ b we have f (x) ≥ g(x). Wecan see in the graph of the function that this is true.
f (x)=1/(x3+1)
0
0. 2
0. 4
0. 6
0. 8
1
2 4 6 8 10 12 14 16 18 20x
g(x)=1/(x2+1)
Nowwe study if ∫ ∞1
1x2 + 1dx converges.�ere are twoways to do this.�e �rst is to compute the integral
directly.
∫ ∞1
1x2 + 1dx = lim
b→∞ ∫ b
1
1x2 + 1dx = lim
b→∞atan(x)∣
b
1
= limb→∞
atan(b) − atan(1) = π2− π4= π4
7
�erefore the integral converges.(We could also have use the direct comparison test, and compared to 1
x2≥ 1
x2 + 1 so since ∫∞1
1x2
dx
converges we can conclude that ∫ ∞1
1x2 + 1dx converges.)
As 1x3 + 1 decays faster than
1x2 + 1 , we therefore know that ∫
∞1
1x3 + 1dx converges.
Again we get an analogue condition for divergence.If f (x) and g(x) decay at the same rate, convergence of ∫ f (x)dx implies convergence of ∫ g(x) and vice
versa. (�is is not simply a reduction to�eorem 4, but a result that has to be proven separately.) Collectingthe behavior for all three cases, we get the following theorem:
THEOREM 5 Limit Comparison Test for Improper IntegralsLet f (x) and g(x) be piecewise-continuous on [a,∞) and suppose that f (x) ≥ 0 and g(x) ≥ 0 for all x ≥ a
a) If limx→∞
f (x)g(x) = L > 0 then ∫ ∞
af (x)dx and ∫ ∞
ag(x)dx both converge or both diverge.
b) If limx→∞
f (x)g(x) = 0 and ∫ ∞
ag(x)dx converges then ∫ ∞
af (x)dx converges.
c) If limx→∞
f (x)g(x) =∞ and ∫ ∞
ag(x)dx diverges then ∫ ∞
af (x)dx diverges.
One can show that all previous theorems in this chapter, including the direct comparison test, are specialcases of this theorem.�is means that any problemwhich can be done by direct comparison can also be doneby limit comparison. (Still, o�en direct comparison is easier, as no limit needs to be computed.)
Also notice again that if limx→∞
f (x)g(x) = 0 and ∫ ∞
af (x)dx converges, we get no conclusion about ∫ ∞
ag(x)dx
and similarly if limx→∞
f (x)g(x) =∞ and ∫ ∞
af (x)dx diverges, we get no conclusion about ∫ ∞
ag(x)dx.
Let us look at an example for comparison of same decay order:
EXAMPLE 7
Does ∫ ∞2
x3 + x2 + 1x4 − x2
dx converge?
Solution Let f (x) = x3 + x2 + 1x4 − x2
and g(x) = 1xon the interval [2,∞).
Direct comparison is di�cult, so we look at
limx→∞
f (x)g(x) = lim
x→∞
x3+x2+1x4−x21x
= limx→∞
x(x3 + x2 + 1)x4 − x2
= limx→∞
x4 + x3 + xx4 − x2
= 1
8
�erefore we can conclude that these two functions decay at the same rate. Since ∫ ∞2
1x
dx diverges we
thus know that ∫ ∞2
x3 + x2 + 1x4 − x2
dx also diverges.
In this example we could calculate the integral directly, using partial fractions2, giving (of course) thesame result, albeit with far more e�ort.
Finding Comparison Candidates
�e �nal question we want to study in this chapter is how to choose good comparison functions. For this wecan consider growth order of the reciprocal functions. In the previous example, the reciprocal function would
be 1f (x) = x4 − x2
x3 + x2 + 1 . From chapter 7.6, we know that this is of the same growth order as x4−3 = x, which
leads us to a comparison with g(x) = 1x.
Consideration of the growth order will o�en lead to integrals of the form ∫ ∞a
1x p dx as comparison
candidates3. For such integrals we have explicitly determined the criterion that they converge if and only if
2We have
∫ ∞
2
x3 + x2 + 1x4 − x2
dx = limb→∞ ∫ b
2
x3 + x2 + 1x4 − x2
dx = limb→∞ ∫ b
2
x3 + x2 + 1x2(x2 − 1) dx
= limb→∞ ∫ b
2
x3 + x2 + 1x2(x − 1)(x + 1)dx
�e method for solving the integral is partial fractions, we get
x3 + x2 + 1x4 − x2
= Ax2
+ Bx+ C
x − 1 +D
x + 1= A(x2 − 1) + B(x3 − x) + C(x3 + x2) + D(x3 − x2)
x2(x − 1)(x + 1)
= Ax2 − A+ Bx3 − Bx + Cx3 + Cx2 + Dx3 − Dx2
x2(x − 1)(x + 1)Comparing coe�cients for powers of x in the numerator, we get the following equations
B + C + D = 1A+ C − D = 1
B = 0−A = −1
We can then solve for the variables and get A = 1, B = 0,C = 12 ,D = 1
2 . Substituting in the original integral, we get
limb→∞ ∫ b
2
1x2
+ 12(x − 1) +
12(x + 1)dx = lim
b→∞x−1 + 1
2ln∣x − 1∣ + 1
2ln∣x + 1∣∣
b
2
= limb→∞
1b+ 12
ln∣b − 1∣ + 12
ln∣b + 1∣ − 12− 12
ln∣1∣ − 12
ln∣3∣
= 0 +∞+∞− 12− 12
ln(3) =∞
�us the integral diverges, as we found already by the limit comparison test with far less e�ort.3which is the main reason we bothered with stating a theorem for such integrals
9
p > 1.�is not only indicates comparison candidates, but also whether the integral will converge or diverge.If we cannot �nd a function of the same growth order as 1/ f (x), parts b) and c) of�eorem 5 o�en can be
used to �nd functions of strictly faster decay for which the improper integral diverges, respectively functionsof slower decay for which the integral converges.Going back to the example starting this chapter, we �nd for example that for f (t) = 100e−2t the reciprocal
1/ f (t) = e2t/100 is of larger growth than any power of t, thus we could compare to g(t) = 1/t2 to showconvergence of ∫ ∞
0f (t)dt without explicitly calculating the antiderivative.
EXAMPLE 8Determine, by �nding suitable comparisons, whether the following improper integrals converge:
1. ∫ ∞1
3x +
√x
dx
2. ∫ ∞1
ln xx3/2
dx
3. ∫ ∞1
x2
2x dx
4. ∫ ∞1
1(ln x)2 dx
5. ∫ ∞1
x2 + 1x3
√x
dx
6. ∫ ∞1sin( 1
x) dx
Solution
1. If f (x) = 3x +
√x, then 1/ f (x) = 1/3(x +
√x).�is is of the same growth order as x, thus the integral
diverges by limit comparison with ∫ ∞1
1x
dx.
2. If f (x) = ln xx3/2
, then 1/ f (x) = x3/2ln x
. We know that ln x grows slower than any power of x, thus we see
that x3/2−ε for arbitrary small ε, for example x5/4, is a function of smaller growth. We get that
limx→∞
ln xx3/21
x5/4= lim
x→∞ln x ⋅ x5/4
x3/2= lim
x→∞ln xx1/4
= 0
As ∫ ∞1
1x5/4
dx converges, we conclude by limit comparison that ∫ ∞1
ln xx3/2
dx converges.
3. If f (x) = x2
2x , then 1/ f (x) = 2x
x2.�is is of growth larger than any power of x, for example x2.�erefore
limx→∞
x22x1
x2= lim
x→∞x4
2x = 0, and the integral converges by limit comparison with ∫ ∞1
1x2
dx.
10
4. If f (x) = 1(ln x)2 , then 1/ f (x) = (ln x)2. �is grows slower than x, thus the integral diverges by
limit comparison with ∫ ∞1
1x
dx. (Of course we also have that (ln x)2 grows less than x2, but limit
comparison with 1x2is a case in which we cannot draw any conclusion.)
5. If f (x) = x2 + 1x3
√x, then 1/ f (x) = x3
√x
x2 + 1 , which is of the same growth order asx3√
xx2 = x3/2.�e integral
converges by limit comparison with ∫ ∞1
1x3/2
dx.
6. �is one is tricky, as we don’t know the growth order for 1/ sin(1/x).�e 1/x suggests limit comparisonwith 1/x. We get
limx→∞
1x
sin ( 1x )L’Hopital= lim
x→∞−x−2
−x−2 cos ( 1x )= lim
x→∞1
cos ( 1x )= 1
As ∫ ∞1
1x
dx, diverges, so does ∫ ∞1sin( 1
x) dx. (Once we know power series we’ll see another expla-
nation for this behavior: For small values of z, sin(z) behaves like z.�us sin ( 1x )will behave like1x for
large x.)
We close this chapter with the remark that we considered only criteria for convergence tests for the caseof ∫ ∞
af (x)dx. Obviously similar tests are possible for the other types of improper integrals.
EXERCISES
Determine whether the following improper integralsconverge or diverge. You may use direct comparisonor limit comparison (or explicit computation of anantiderivative)
1. ∫ ∞1
1 + cos(x)x2
dx
2. ∫ ∞3
1ln(ln(x))dx
3. ∫ ∞1
2x3x − 1dx
4. ∫ ∞1
x + 2x
x22x dx
5. ∫ ∞2
1(ln(x))2 dx
6. ∫ ∞1
x3
3x dx
7. ∫ ∞1
3x
x3dx
8. ∫ ∞1
−2x√
xdx
9. ∫ ∞1
ln(x)x1.001
dx
10. ∫ ∞1
3x−1 + 13x dx
11. ∫ ∞1
1√x2 + 2
dx
11
2-
Geometric Series
In this chapter we will discuss one of the simplest andmost useful types of in�nite series, the geometric series.�ey are derived from instances where geometric growth occurs, this is growth that is proportional to thecurrent value, i.e the growth rate is constant. (Here we are looking at the discrete situation, in a continuoussituation a di�erential equation for constant growth rate produces exponential functions as solutions.)A simple example of geometric growth is interest accrued annually on a bank account. Each year we take
the amount of money in the account the value of the account and multiply it by the interest rate and add it tothe value to get a new value.�erefore the constant of proportionality is one plus the interest rate.
EXAMPLE 1If you put $100 into a savings account that earned 3% interest accrued annually, how much would you havea�er one year? two years? n years?Solution Let pn be the value of our account at time n, so p0 = 100 since we begin initially with $100.Next at the end of one year we earn 3% interest on this account so p1 = (1.03) ⋅ p0 = (1.03) ⋅ 100, hencep1 is proportional to p0 with the constant of proportionality being 1.03. �is process repeats so that p2 =(1.03) ⋅ p1 = (1.03)2 ⋅ 100. At this point we notice a pattern pn = (1.03)n ⋅ 100.As a second example of geometric growth we consider university scholarships which come from endow-
ments. In this case, an endowment (a �xed sum) is given to the university. A scholarship then is funded fromthe interest of this money (without using up the money, the scholarship therefore exists forever).
EXAMPLE 2 EndowmentsDelighted by the calculus course he took at CSU, an alumnus wants to endow a scholarship for mathematicsstudents that will pay $2000 each year.�e university guarantees an interest rate of 5% per year. How muchmust the alumnus donate to guarantee this scholarship will be available forever?Solution If we let x=the value of the endowment, then a�er one year the endowment is worth 1.05 ⋅ xand we want to give a $2000 scholarship, so we want 1.05x = x + 2000.�erefore the endowment will neverbe touched and the scholarship is funded by the interest earned. Solving this equation we get 0.05x = 2000which gives us that x = 40, 000.�erefore the alumni must give an endowment of $40, 000 to guarantee thescholarship is funded forever.
A geometric series now is a series whose terms show geometric growth, i.e. sn = r ⋅ sn−1 for a constant rindependent of n.�is type of series has applications in several �elds including physics, biology, economicsand �nance. We will look at a few examples of where geometric series occur in our everyday lives including:repeating decimals and calculating dosage of medicine.
13
DEFINITION 1Let r be a ratio and a any nonzero constant. A geometric series is a series of the form
a + ar + ar2 + . . . + arn−1 + . . . =∞∑n=1
arn−1 = a∞∑n=1
rn−1 = a∞∑n=0
rn
As an example of a geometric series we look at the “Race course paradox” of Zeno1. Suppose a runnerwanted to travel a given distance, say one kilometer.�en he must �rst travel the �rst half kilometer, and thenext half kilometer remains. Next the runner must travel half of the next half kilometer – a quarter kilometer– and the next quarter kilometer remains. It might seem (so claimed Zeno) as if the runner never reaches thegoal.
However if we look at the distance traveled a�er k iterations, we get a geometric sum with ratio 12 :
12+ 14+ 18+ 116+⋯ + 1
2k
If the process continues in�nitely many times, we thus get get the following formula for the distance traveled:
12+ 14+ 18+ . . . + 1
2n + . . . =∞∑n=1
12n
We will see that this sums up to 1 (i.e. the full distance is traveled, resolving the paradox).Notice that there is a slight shi� in this formula from the one given in the de�nition of a geometric series
in that we start not with r0 = 1 but with r1 = 12 while in the de�nition this power is zero.�is can be �xed byletting the constant a be 12 ,�is sort of index change arises o�en enough that we will �rst look at how we canmanipulate such in�nite sums.
Reindexing a Series
We begin this discussion with a brief reminder about the Σ notation for sums. Givenk∑n=1
an we call n the
index of summation, an the nth term of the sum, and the upper and lower bounds of summation are k and1 respectively.
Now we consider reindexing our in�nite series, notice in the following examples we are working withgeometric series, however this algorithm will be useful with other types of series that will be discussed later.
1Zeno of Elea, Greek philosopher, ca. 490 BC - ca. 430 BC
14
Most of the in�nite series we have seen so far have a lower bound of summation of one, however their aretimes when we are given an in�nite series where the lower bound of summation will be a value other thanone. We consider �rst a geometric series which does not have a lower bound of one:
−127
+ 181− 1243
+ . . . + (−13)n−1 + . . . =
∞∑n=4
(−13)n−1
�is is still a geometric series, however our de�nition requires the series to have the form∞∑n=1
arn−1, so rein-
dexing is required. To reindex a series we follow the following basic algorithm:
Algorithm 1 Reindexing a series
1 Choose a new letter to be the index of summation (m)
2 Relate m to the original index of summation (n) such that when n=4 we have m=1. So we have theequation m=n-3.
3 Solve for original index of summation (n) and substitute into the nth term of the sum:
n = m + 3⇒∞∑m=1
(−13)m+3−1 =
∞∑m=1
(−13)m+2
Notice if the upper bound of summation were �nite we would have to decrease its value by 3 as well,however since we are summing to in�nity our upper bound of summation will remain in�nite.
4 Solve for an a such that the power on r is m-1.
∞∑m=1
(−13)m+2 =
∞∑m=1
(−13)3 ⋅ (−1
3)m−1
since m-1+3=m+2.
�erefore a = (−13 )3 = −1
27 and r = −13 so we have a geometric series of the form
∞∑m=1
arm−1. Notice that
in this example the ratio was negative, therefore a geometric series can have negative or positive r values. Wenow look at another example of this reindexing algorithm before we develop an equation for the value of ageometric series.
EXAMPLE 3Consider the series
∞∑n=9
42n + 3 , rewrite this series so that the lower bound of summation is n=1.
Solution Let m = n − 8, then n = m + 8 so we have∞∑m=1
42(m + 8) + 3 =
∞∑n=1
42m + 19
Using this algorithm we can always manipulate a given geometric series so that it can be written in the
form∞∑n=1
arn−1 or equivalently∞∑n=0
arn next we consider how to calculate the value of this series.
15
The Formula for the Geometric Series
In this section we will derive a general formula for the value ofk∑n=0
arn. We can apply this in the limit case
k = ∞ to get the value of an in�nite series, for example to show that the series∞∑n=1
12n from Zeno’s paradox
has value 1.
DEFINITION 2�e kth partial sum sk of a Geometric series
∞∑n=0
arn is the (�nite) sum of the �rst k terms:
sk = a + ar + ar2 + . . . + ark−1 =k−1∑n=0
arn
Using this de�nition we can consider the sequence s1, s2, s3, . . . , sk , . . . of partial sums, which converges
to∞∑n=0
arn, since limk→∞
sk = limk→∞
k−1∑n=0
arn =∞∑n=0
arn.�erefore, once we know a formula for the value of sk we
can compute the value of∞∑n=0
arn as the limit.
We begin by consideringsk = a + ar + ar2 + . . . + ark−1
andrsk = ar + ar2 + ar3 + . . . + ark .
Notice that sk and rsk share several terms in common, in fact
sk − rsk = (a + ar + ar2 + . . . + ark−1) − (ar + ar2 + ar3 + . . . + ark) = a − ark
Now if we factor both sides of this equation, we get
sk(1 − r) = a(1 − rk).
As long as r ≠ 1 we can divide both sides by (1 − r) to get
sk =k−1∑
n=0arn= a 1 − rk
1 − r.
(What happens in the case where r=1? We get sk = a + a(1) + a(1)2 + . . . + a(1)k−1 = ka.)
Before we go on to get a formula for the in�nite sum, we see how the formula for sk might be used on itsown:
EXAMPLE 4 The magnitude of repeated growthSuppose we are given a chessboard and place one grain of rice on the �rst square, two grains of rice on thesecond square, four grains of rice on the third square continuing on so that there are 2n−1 grains of rice on thenth square. How many grains of rice are on the chess board? (�ere are 8 ⋅ 8 = 64 squares on a chess board.)
16
Solution Since there are 64 squares on the chess board, we want to compute s64. Our ratio is 2 and a isone, hence we are asked to �nd
s64 = 1 + 2 + 4 + . . . + 263 =1(1 − 264)1 − 2 = 1.84467x1019
A similar calculation underlies the repayment of loans or mortgages:
EXAMPLE 5 Paying back a loanSuppose we take out a loan of L dollars which is paid back periodically (typically monthly). �e periodicpayment is a dollars, the �xed interest rate per period is i. If bk is the loan sum outstanding a�er k timeperiods, we have that
bk+1 = bk ⋅ (1 + i) − a.
Using b0 = L and setting r = 1 + i, in the �rst step we have
b1 = Lr − a
we then continue on to the next step where we get
b2 = b1(r) − a = (Lr − a)(r) − a= Lr2 − (a + ar)
Wemight be starting to notice a pattern, however it becomes obvious in the next step
b3 = b2(r) − a = (Lr2 − (a + ar))(r) − a= Lr3 − (a + ar + ar2)
At this point we can solve this recursion to
bk = L ⋅ rk − (a + ar + ar2 +⋯ + ark−1) = L ⋅ rk −k−1∑n=0
arn Sum formula= L ⋅ rk − a 1 − rk
1 − r.
A banknowwould set bm = 0 (wherem is the time a�erwhich the loan should be paid o�, e.g.m = 12⋅30 = 360for a 30 year mortgage) and solve for a to determine the necessary monthly repayment, given the loan sumand interest rate.For example, if we have a mortgage of L = $200, 000, an annual interest rate of 6% (leading to a monthly
rate of i = .06/12 = 0.005, i.e. r = 1.005) and a monthly repayment sum2 of a = $1, 200, we �nd that theoutstanding sum a�er k months is
bk = 200, 000 ⋅ 1.005k − 1, 2001.005k − 1
0.005= 200, 000 ⋅ 1.005k − 240, 000 (1.005k − 1) .
A�er 10 years (120 months) this leaves an outstanding amount of $167, 224.13, a�er 20 years $107, 591.82,roughly half3, a�er 30 years $−903.00 (i.e. the loan is paid o� a�er 30 years less one month4).
2Moralistic remark: Incidentally, initial interest amounts to $1000 per month at the start of the loan. An interest only loan thusdoes not save much and is a very bad deal!
3In general, this means that a loan is paid o� half a�er roughly 23 of its planned life time4�e total cost of the loan then will have been $430, 800, more than double the loan amount. (�ough in�ation means that theactual value will be less.)
17
If the interest rate instead was 7% annually (i = .07/12 = 0.005833 monthly), we get with same repaymentsum a remaining loan amount of $159, 256.18 a�er 30 years, which is not even halfway paid o�. A monthlyrepayment sum of $1330 would be needed5 to have the loan paid o� a�er 30 years.
Now since we saw earlier that the sequence s1, s2, s3, . . . , sk , . . . of partial sums converges to∞∑n=1
arn−1, we
can use the formula just derived to compute a value for the limit of a geometric series.
THEOREM 1�e geometric series a + ar + ar2 + . . . + arn−1 + . . . =
∞∑n=1
arn−1 converges to a1 − r
when ∣r∣ < 1, and diverges
otherwise.
Proof Case 1: If ∣r∣ < 1 then ∣r∣k < 1, and as k gets larger this will cause ∣rk ∣ to get smaller and yieldlim
k→∞rk = 0.�us, by the rules for limits,
limk→∞
sk = limk→∞a(1 − rk)1 − r
= a(1 − 0)1 − r
= a1 − r
.
Case 2:When ∣r∣ ≥ 1 then ∣r∣k ≥ 1.�erefore when k tends to in�nity the numerator a(1 − rk) tends to plus
or minus in�nity, and the denominator 1 − r is a �xed value. So limk→∞
a(1 − rk)1 − r
diverges.
Now that we have a formula for the value of a geometric series let us verify the solution to the Zeno’s
paradox example. We are looking at∞∑n=1
12n . First notice that this sum is not in the correct form, we must
factor ( 12) out of the nth term.We get∞∑n=1
( 12)n =
∞∑n=1
( 12) ⋅ ( 12)n−1 =
∞∑n=0
( 12) ⋅ ( 12)n, hence a = 12 and r = 12 < 1
so this series converges toa1 − r
=121 − 12
=1212= 1.
Next we will look at some interesting examples where we can use this formula.
Examples
In this section we will look at how people may use geometric series in real life applications.In our �rst example we look at how much medication will remain in a persons body a�er taking medica-
tion at regular intervals
EXAMPLE 6 Repeated Drug DosageACSU student comes downwith bacterial laryngitis, the student goes to the Healthcenter to see a doctor whoprescribes 250mg doses of antibiotics that the student must take every six hours over the next several days.
5I.e. a change of one percentage point in the interest rate increased the monthly payment (and thus the total loan cost) by 10%!No wonder people go bonkers about interest rates.
18
Since the body will digest the antibiotics between doses and it is known that 5 percent of the drug remainsin the body at the end of six hours, how can we calculate the quantity of the drug that remains in the studenta�er the ��h dose? At the end of �ve full days on the drug?Solution Let us begin by looking at the quantity of the drug in the body a�er each dose. Let qn representthe quantity of the drug in the students body a�er the nth dose. Initially
q1 = 250
at the next dose we will have 5% of that 250mg remaining and another 250mg is taken, so
q2 = 250(0.05) + 250
At the third dose we have 5% of q2 remaining and another 250mg is taken, so therefore
q3 = 0.05 ⋅ q2 + 250 = 0.05(250(0.05) + 250) + 250= 250(0.05)2 + 250(0.05) + 250.
Now on the fourth dose q4 we have 0.05 ⋅ q3 remaining in the body and another 250mg is taken.
q4 = 0.05 ⋅ q3 + 250 = 0.05(250(0.05)2 + 250(0.05) + 250) + 250= 250(0.05)3 + 250(0.05)2 + 250(0.05) + 250.
Notice at this point that a pattern has developed, we have
qn = sn =n∑k=1250(0.05)n−1
�erefore we can use our formula sn = a(1−rn)1−r to calculate qn = 250(1−0.05
n)1−0.05 . With this information we can
calculate q5 = 250(1−0.055)1−0.05 = 263.1578125mg in the students body a�er 5 doses. If the student continues to
take the antibiotic for 5 full days or 20 doses they will have q20 = 250(1−0.0520)
1−0.05 = 263.1578947mg.
Suppose the student continued to take this drug forever, how much medication would there be in thestudents body at each dose?
�is question is asking us to compute∞∑n=1250(0.05)n−1 which luckily for the student 0.05 < 1 will converge
to a �nite number of 2501−0.05 = 263.1578947mg. Since 0.05
20 = 9.536743x10−27 which is practically zero, weget the same value up to the ten-millionth place if the student took the medication forever as we got whenthe student took the medication for �ve days.
EXAMPLE 7 Repeating DecimalExpress the repeating decimal 0.08 as a fraction of two integers.Solution To express 0.08 as a fraction we �rst notice that 08 is the repeated value, so we want to breakup 0.08 into a sum of the repeated values.�erefore we have 0.08 = 0.08+ 0.0008+ 0.000008+ . . .. Once wehave this written as a sum we can now write each term in the sum as a fraction, hence we have 0.08 = 8
100 ,0.0008 = 8
10,000 =81002 , and 0.000008 =
81003 . Using this information we should recognize a pattern
0.08 = 8100
+ 81002
+ 81003
+ . . . =∞∑n=1
8100n
19
However at this point we notice that we have a geometric series which is not written in the correct form for
us to apply our formula. If we factor out ( 8100) we get∞∑n=1
( 8100
)( 1100
)n−1 which is now in the correct form.
�is gives r = 1100 < 1 so our geometric series converges and since a = 8
100 the series converges to
81001 − 1
100=810099100
= 899
= 0.08
Being able to express a repeating decimal as a fraction is useful when we are calculating with a large numberof values where the precision of the answer is vital since a fraction will not have any round o� error.
Another paradox due to Zeno initially seems even more puzzling, which is the reason we consider it last.Achilles, the fastest runner in ancient Greece, runs 100 times as fast as the Tortoise. But — so the paradoxclaims — if the Tortoise is given an advance, Achilles will never be able to pass the Tortoise.
EXAMPLE 8 Achilles and the TortoiseSuppose that Achilles runs 10m/s and the Tortoise only 0.1m/s, furthermore the Tortoise is given a head startof 100m. A�er 10 seconds, Achilles has reached the place where the Tortoise started. But in this time, theTortoise has run 1m ahead. Achilles will reach this distance in 0.1 seconds. But then the Tortoise has movedanother 0.01m. Achilles will take 0.001 seconds to reach this, and so on. He will never reach the Tortoise.Where is the error in this argument?Solution �e paradox is resolved, if we realize what time period is considered: All events take placewithin
10 + 0.1 + 0.001 +⋯ = 10∞∑n=0
1100n = 10
1 − 1100
= 100099
= 10.10
seconds.�is is exactly the time when Achilles reaches (and overtakes!) the Tortoise.�e paradox arises fromthe implicit (and wrong, as we have calculated!) suggestion, that it describes the events for all time.
EXERCISES
1. What monthly payment is needed to pay a carloan of $ 15, 000 o� a�er 48months if the interest rateis 7%?
2. A standard method for valuing a business (ex-cluding any inventory) is the “discounted cash �ow”method:�e value of the business itself is the presentday value of its future earnings, discounted by an as-sumed interest rate, i.e. the amount of money thatwould have to be invested at an assumed interest rateto get the same income as the business.Calculate the present day value of a business
whose annual earnings are $50, 000, assuming an in-terest rate of 5%.
3. �e goverment – due to an election year inspending mode – enacts a stimulus package that puts$150 billion back into the economy. We assume thatall the people who have extra money to spend wouldspend 80% of it and save 20%. �us, of the extra in-come generated by the package, $150 ⋅ 45 billion = $120billion would be spent again and so become extra in-come to someone else. Assume that these people alsospend 80% of their additional income (which wouldbe $96 = 120 ⋅ 45 = 150(
45)2 billion) and so on.
a) Let C = 150 be the amount of the stimulus package(in billion dollars) and r = 4/5 the spending factor.Write down an in�nite sum, involving C and r, forthe amount of money added to the economy.b) Calculate the value for the in�nite sum in a).
20
3-
Series as Discrete Analogues of ImproperIntegrals
�e geometric series, studied in the previous chapter, are special in that we can completely describe underwhat condition it converges, and also have an explicit formula for its limit. In general �nding the limit ofa series can be very hard, many of the techniques1 needed are far beyond this course. Instead we want tostudy the easier question, whether a given series converges.�is is very similar to what we did with improperintegrals, and indeedwewill see in this chapter that the convergence of series and the convergence of improperintegrals are intimately related: Once we know how to test convergence of improper integrals we can useessentially the same methods to test convergence for series.
The Integral Test
If we are given a series of the form∞∑n=1
an, the only condition for convergence we know is the n-th term test,
which requires that an be non-increasing.�is condition however is not su�cient in general, as for example
the divergent series∞∑n=11nshows: Its terms are decreasing, but the series does not converge.
We nowwant to see, that we can interpret the partial sums of a series as approximation of an integral.�ein�nite sum of the series then becomes an approximation of an improper integral (of type I), and convergenceof the series implies convergence of the improper integral and vice versa.To see why this should hold, we need to go back to the de�nition of integrals which we did using Riemann
sums.
EXAMPLE 1Does
∞∑n=1
1n2converge?
Solution Consider an = 1n2 . �en any partial sum of the form
k∑n=1
1n2can be interpreted as (right end-
1for example using methods from complex analysis, a 400-level course
21
point) Riemann sum2 for 1 + ∫ k
1
1x2
dx (the blue boxes).
0 2 4 6 8 100
0.5
1.0
x
y
We see thatk∑n=1
1n2
≤ 1 + ∫ k
1
1x2
dx
We now consider the limit k →∞.�e integral on the right will converge (as integral of x−p for p > 1), thusthe sequence of partial sums is bounded from above. On the other hand (we are summing up positive terms)the sequence of partial sums is increasing, thus this sequence (which is the series) must converge.Let us consider the same example once more, but as Riemann sum for le� endpoints (green boxes), ap-
proximating ∫ k+11
1x2
dx:
0 2 4 6 8 100
0.5
1.0
x
y
�e Riemann sum now is an upper approximation, thus we get that
∫ k+11
1x2
dx ≤k∑n=1
1n2
≤ 1 + ∫ k
1
1x2
dx
2As ∫ 10 x−2 diverges, we need to consider the le�most blue box separately
22
We therefore see that we can also consider the series as upper approximation of the integral, by a similarargument as in the previous example we get that convergence of the series ∑∞n=1 1n2 implies convergence ofthe improper integral. ∫ ∞1 1
x2 dx.
�is observation is not speci�c to 1x2 , it holds for any decreasing function. We thus get the following
theorem3:
THEOREM 1 Integral TestLet an be a sequence of positive terms. Suppose that there is a function f de�ned for all positive real numberssuch that
1. an = f (n) for all n,
2. f is piecewise-continuousa,
3. f is decreasing for x > N (where N is some integer, e.g. N = 0 if f is decreasing over the positive realaxis)
�en ∫ ∞1
f (x)dx converges if and only if∞∑n=1
an converges.
aotherwise we cannot guarantee that integrals exist even in the case of �nite limits
Proof Convergence of an in�nite sum or an improper integral is not a�ected by the behavior over a �niteinterval.�us we can consider without loss of generality the case N = 0.�e argument then is essentially thesame as in the discussion for 1x2 .
EXAMPLE 2Does
∞∑n=1
1en converge?
Solution �e function f (x) = 1ex is continuous and decreasing for x > 0. We have seen in the chapter on
improper integrals that ∫ ∞1
1ex dx converges.�us by theorem 1 also
∞∑n=1
1en converges.
EXAMPLE 3 p-series
For which values of p does∞∑n=1
1np converge?
Solution �e function f (x) = 1x p is continuous and decreasing for x ≥ 1. We know that ∫ ∞
1
1x p con-
verges, if and only if p > 1.�us, by theorem 1,∞∑n=1
1np converges if and only if p > 1.
�us (we have seen this previously)∞∑n=1
1n2converges and
∞∑n=11ndiverges.
3In the book this is theorem 9 on p. 773
23
Direct Comparison Test for Series
In the chapter on improper integrals we learned about powerful comparison tests that can be used to deter-mine convergence. If we wanted to test convergence for a series, we therefore could consider the correspond-ing improper integral, and resolve its convergence via a comparison test. Doing so however it turns out to beunnecessarily formal; because theorem 1 has an if and only if condition, we can (by de�ning for a series∑ ana function f (x) with f (x) = an if n ≤ x < n + 1) translate the comparison test for improper integrals to acomparison tests for series. (Compare the statement to theorem 4.)
THEOREM 2 Direct Comparison Test for SeriesLet an and bn be non-increasing sequences of non-negative numbers.
a) If there exists an integer value N such that an ≥ bn for all n ≥ N and∞∑n=1
an converges then∞∑n=1
bn
converges.
b) If there exists an integer value N such that an ≤ bn for all n ≥ N and∞∑n=1
an diverges then∞∑n=1
bn
diverges.
As with improper integrals, this theorem does not reach any conclusion if an ≤ cn and ∑ cn diverges or ifan ≥ dn and∑ dn diverges.
EXAMPLE 4
Does∞∑n=1sin2(n)
n2converge?
Solution We know that sin2(n)n2 ≤ 1
n2 for all values of n. We also know that∞∑n=1
1n2
dx converges. By theo-
rem 2, part a), we can conclude that∞∑n=1sin2(n)
n2dx converges.
EXAMPLE 5Determine if
∞∑n=1
1n3 + 3 converges.
Solution Let us consider∞∑n=1
1n3, which is a p-series for p = 3 > 1 and therefore converges. Since 1
n3+3 ≤1
n3
for all n ≥ 1 the direct comparison test shows∞∑n=1
1n3 + 3 converges.
Limit Comparison Test for Series
With the same argument as used for the ordinary comparison test, the limit comparison test can be adaptedto series. (Again, compare to theorem 5.)
24
THEOREM 3 Limit Comparison Test for SeriesSuppose that an > 0 and bn > 0 for all n ≥ N where N is an integer
a) If limn→∞
anbn
= L > 0, then∑ an and∑ bn both converge or both diverge.
b) If limn→∞
anbn
= 0 and∑ bn converges then∑ an converges.
c) If limn→∞
anbn
=∞ and∑ bn diverges then∑ an diverges.
Again notice that if limn→∞
anbn
= 0 and ∑ bn diverges or limn→∞anbn
=∞ and ∑ bn converges the theoremgives us no conclusion.
EXAMPLE 6Consider
∞∑n=1
1√n + 2
, does this series converge or diverge?
Solution We know that∞∑n=1
1√ndiverges (it is a p-series for p = 12 < 1).
We calculate the limit of the quotient:
limn→∞
1√n+21√n
= limn→∞
√n√
n + 2= lim
n→∞
√n( 1√
n)√n + 2( 1√
n)
= limn→∞
√1√1 + 2n
= 1
So by the limit comparison test we conclude that∞∑n=1
1√n + 2
diverges. (Note that direct comparison will not
work in this example, as 1√n + 2
< 1√n.)
As with improper integrals, candidates for comparison are o�en determined by growth classes.
EXAMPLE 7
Does∞∑n=1
(ln(n))2
n3converge or diverge?
Solution We know that ln(x) grows less than any power of x, thus (ln(x))2 should also grow less thanany power of x.�us we assume that the sum converges.To show this, we just need to pick a suitable exponent xε in relation to ln(x) to make (xε)2x−3 decay
faster than 1x . We pick x18 .�en
(n18 )2
n3= 1
n114
and∞∑n=1
1
n114converges, as 114 > 1.
25
We now use limit comparison:
limn→∞
(ln(n))2n31
n114
= limn→∞
n114 (ln(n))2
n3= lim
n→∞(ln(n))2
n14
To apply L’Hopital’s rule, requiring derivatives, we rewrite this expression in the variable x instead of n:
= limx→∞
(ln(x))2
x14
L’Hopital= limx→∞
2(ln(x))( 1x )14x−34
= limx→∞
8(ln(x))x14
L’Hopital= limx→∞
8( 1x )14x−34= lim
x→∞32
x14= 0.
Since∞∑n=1
1
n114converges, by the limit comparison test we can conclude that
∞∑n=1
(ln(n))2n3
converges.
As in this example (and with improper integrals), comparison to a p-series is o�en useful.Another fruitful comparison candidate is given by geometric series, the method for �nding a suitable
geometric factor is given by the ratio test, which we shall study next:
The Ratio Test
In this section we want to see how the geometric series can be used along with the comparison test to reach
conclusions about the convergence of series. Recall that a geometric series is a series of the form∞∑n=0
arn,
which we have shown will converge if ∣r∣ < 1.Suppose we want to test a given series∑
nan (with nonnegative values an) for convergence by comparing
with a geometric series∑n
arn. For comparison to work, we need that the terms an of our series fall as quickly
as the terms arn of the geometric series.We can ensure this by looking at the ratio of subsequent terms: Assume for the moment that
an+1an
≤ arn+1arn = r
and that ∣r∣ < 1, so we know that∑n
arn converges.
Because of the inequality just assumed, we get — setting b ∶= a0a— that
a1ar1
= a1a0
⋅ a0a⋅ 1
r≤ r ⋅ b ⋅ 1
r= b
a2ar2
= a2a1⋅ a1
ar1⋅ 1
r≤ r ⋅ b ⋅ 1
r= b
⋮an+1arn+1 =
an+1an
⋅ anarn ⋅ 1
r≤ r ⋅ b ⋅ 1
r= b
26
and thereforelim
n→∞anarn ≤ b <∞
By the limit comparison test (�eorem 3) we therefore know that∑n
an also converges.
To do this in a concrete example, of course we would have to determine r. However in the end we don’treally care about the value of r, but whether∑
nan converges.�is is guaranteed if (eventually — think of the
possibility of a positive number N for the comparison test) there is a number r < 1 such that an+1an
≤ r. Butthis is guaranteed, if
limn→∞an+1
an= L < 1
(�ink of the ε-criterion for the limit of the sequence: If the limit L is smaller than 1, we set ε = 1 − L2and can
�nd an N such that an+1an
≤ L + ε < 1 for n ≥ N . We simply set r ∶= L + ε.)
We have thus seen that∑n
an converges, if limn →∞an+1an
< 1.�is is the �rst version of the Ratio Test.
To state this test in general, we note that an analogous argument can be given to show divergence (com-parison with a diverging geometric series), if the ratio has limit > 1.
THEOREM 4 Ratio Test for Series with positive termsLet∑ an be a series with positive terms and suppose that
limn→∞
an+1an
= r
then
(a) the series converges if r < 1
(b) the series diverges if r > 1 or if r is in�nite
(c) the test is inconclusive if r = 1.
To see why we really cannot deduce anything in the case of r = 1, consider two examples:∞∑n=11nwhich
diverges by the p-series test and∞∑n=1
1n2which converges by the p-series test. Let us look at the limits of the
ratios in each of these cases:
limn→∞
1n+11n
= limn→∞
nn + 1 = limn→∞
n( 1n)(n + 1)( 1n)
= limn→∞
11 + 1n
= 1
Similarly
limn→∞
1n2+11
n2= lim
n→∞n2
n2 + 1 = limn→∞n2( 1n2 )
(n2 + 1)( 1n2 )= lim
n→∞1
1 + 1n2
= 1
�erefore we see that in the case where ∣r∣ = 1 the test is inconclusive.
27
EXAMPLE 8
Consider the series∞∑n=1
n2
2n . We want to determine if this series converges or diverges.
Solution Consider the ratio
an+1an
=(n+1)22n+1n22n
= (n + 1)22n
2n+1n2
= (n + 1)22n
2n21n2= (n + 1)2
2n2
Now consider the behavior as n gets large (in other words take the limit as n →∞), we get
limn→∞
(n + 1)22n2
= limn→∞
(n + 1)2( 1n2 )2n2( 1n2 )
= limn→∞
(1 + 1n)2
2= 12< 1.
By the ratio test we therefore conclude that the series∞∑n=1
n2
2n converges.
EXAMPLE 9Determine whether
∞∑n=1
n!en converges or diverges?
Solution We begin by considering the ratio
an+1an
=(n+1)!
en+1n!en
= (n + 1)n!en
ene1n!= (n + 1)
e
Now we look at the behavior as n tends toward in�nity limn→∞
n + 1e
=∞ therefore we can conclude the series∞∑n=1
n!en diverges.
Sometimes the limit in the ratio test requires L’Hopital’s rule for evaluation:
EXAMPLE 10Determine whether
∞∑n=1ln(n)2nn
converges or diverges?
Solution We begin by considering the ratio
an+1an
=ln(n+1)2n+1(n+1)ln(n)2n n
= ln(n + 1)n2n
2n2(n + 1) ln(n) = ln(n + 1)n2(n + 1) ln(n)
Next we take the limit of this function limn→∞
ln(n + 1)n2(n + 1) ln(n) . It has the indeterminant form
∞∞ , therefore we
have to use L’Hopital’s rule.Writing the function in x (to indicate that we now consider it as a continuous, not a discrete function),
we get
28
limx→∞
ln(x + 1)x2(x + 1) ln(x) = lim
x→∞x
x+1 + ln(x + 1)2((1) ln(x) + x+1
x )Again we have the indeterminant form ∞
∞ , so we apply L’Hopital’s rule once more, and get.
= limx→∞
(x+1)(1)−x(1)(x+1)2 + 1
x+12( 1x +
(x)(1)−(x+1)(1)x2 )
= limx→∞
1(x+1)2 +
(x+1)(x+1)2
2( xx2 −
1x2 )
= limx→∞
x+2(x+1)22x−2
x2
= limx→∞
(x + 2)(x2)(x + 1)2(2x − 2) = lim
x→∞x3 + 2x2
2x3 + 2x2 − 2x − 2 =12lim
x→∞x3 + 2x2
x3 + x2 − x − 1At this point we have only polynomials le�. We know that the behaviour of such a quotient is determined
by the leading terms (formally, one applies L’Hopital’s rule multiple times), thus limx→∞
x3 + 2x2x3 + x2 − x − 1 = 1 and
limx→∞
an+1an
= 12< 1.
�erefore we can conclude by the ratio test that the series∞∑n=1ln(n)2nn
converges.
EXAMPLE 11Determine whether
∞∑n=1
n!nn converges or diverges?
Solution We begin by considering the ratio
an+1an
= (n + 1)!nn
(n + 1)n+1n! =(n + 1)n!nn
(n + 1)n(n + 1)n!= nn
(n + 1)n
We note that limn→∞( n
n + 1)n has indeterminant form 1∞. Using the fact that logarithm and exponential
function are continuous, we therefore use that
limx→∞( n
n + 1)n = exp( lim
x→∞ ln((n
n + 1)n))
We now consider the inner limit
limn→∞ ln(
nn + 1)
n = limn→∞ n ln( n
n + 1) = limn→∞ln( n
n+1)1n
.
�is has the indeterminant form 00 which again we will resolve by applying L’Hopital’s rule. (Again, we write
this as a function in x, to indicate that this is using a continuous function.) By L’Hopital’s rule we get
= limx→∞
ln( xx+1)1x
= limx→∞
1x
x+1⋅ (x+1)(1)−x(1)
(x+1)2−1x2
= limx→∞
x+1x
1(x+1)2−1x2
= limx→∞
1x(x+1)−1x2
= limx→∞
−x2
x(x + 1) = limx→∞
−x(x + 1)
= limx→∞
−x( 1x )(x + 1)( 1x )
= limx→∞
−1(1 + 1/x) = −1.
29
We thus know that
limn→∞
an+1an
== exp( limx→∞ ln((
nn + 1)
n)) = exp(−1) = e−1 < 1
and therefore we can conclude that∞∑n=1
n!nn converges.
So far all the terms of the series we considered were positive. Once we get to power series, however alsonegative terms will occur. We therefore want to establish that — by considering absolute values — the ratiotest also applies to these series.
THEOREM 5 Ratio Test for SeriesLet∑ an be a series with positive terms and suppose that
limn→∞ ∣an+1
an∣ = r
then
(a) the series converges if r < 1
(b) the series diverges if r > 1 or if r is in�nite
(c) the test is inconclusive if r = 1.
EXAMPLE 12Consider the series
∞∑n=1
(−2)n
n!does it converge or diverge?
Solution As not all terms are positive, we cannot immediately apply the ratio test.�erefore we begin bylooking at how this series compares to the corresponding series of positive terms
∞∑n=1
∣(−2)n
n!∣ =
∞∑n=12n
n!.
Each term in our original series is smaller than or equal to the terms of this new series, since (−2)n
n!≤ 2
n
n!.
However, the comparison test as given only holds for positive series, so this is not enough to make a conclu-sion.
�erefore, we also consider the corresponding series of negative terms:∞∑n=1
(−1)2n
n!. In this series each
term is smaller than or equal to the terms of our original series, since −2n
n! ≤(−2)n
n! .
30
n
n!n
n!n
n!
�erefore, it seems plausible that if both∞∑n=12n
n!and −
∞∑n=12n
n!converge, we can conclude that
∞∑n=1
(−2)n
n!converges since the value of its terms are squeezed between the two other series terms and partial sums of theoriginal series therefore are bounded in terms of the partial sums of the positive and negative series, both ofwhich are converging.Let us look at this in the context of the ratio test. By taking the absolute value of quotients we treat all
three series “simultaneously”: For∞∑n=1
(−2)n
n!we thus get the ratio
∣an+1an
∣ =
RRRRRRRRRRRRRRRR
(−2)n+1(n+1)!(−2)n
n!
RRRRRRRRRRRRRRRR
=∣−2∣n+1(n+1)!∣−2∣n
n!
=2n+1(n+1)!2nn!
= 2n+1(n + 1)!
n!2n = 2n ⋅ 2 ⋅ n!
(n + 1)n! ⋅ 2n = 2n + 1 .
As limn→∞
2n + 1 = 0 < 1, we conclude by the ratio test that this series converges.
Since∞∑n=1
(−1)2n
n!= (−1)
∞∑n=12n
n!this is just a scalar multiple of
∞∑n=12n
n!and scalar multiples do not a�ect
31
convergence.�erefore, we can conclude∞∑n=1
(−1)2n
n!converges. Hence by the argument above we conclude
∞∑n=1
(−2)n
n!must converge as well. �erefore if we have a series with negative terms, we can determine the
convergence using the ratio test on the absolute value of the series. In fact the proof of this statement is thesame as the proof of the ratio test, but using ∣an∣ instead of an.
EXERCISES
Determine whether the following series converge ordiverge. You may use the integral test, direct compar-ison, limit comparison, or the ratio test.
1.∞∑n=1
110n
2.∞∑n=1
ln(n)n
3.∞∑n=12n
3n
4.∞∑n=12n
n2
5.∞∑n=1
nn3 + 5
6.∞∑n=1
√n
n2 + 1
7.∞∑n=1sin( 1
n)
8.∞∑n=1
1n n√n
9.∞∑n=11 − nn2n
10.∞∑n=1
n3
3n
11.∞∑n=1
n10
n!
32
4-
Working with Power Series
In this chapter we �nally see the reasonwhywewere bothering with series. A power series is a series involvingpowers of x, thus as long as it converges we obtain a series value for every x-value. In other words: the seriesdescribes a function. Such a function might not look as nice as the functions you have seen so far, but in facthas many advantages. By calculating partial sums we can approximate the value of the function. (�is is infact how your calculator evaluates functions such as sine or logarithm. When you press sine there is no littleman in the calculator measuring the length among the circle; instead power series is evaluated.)We have seen in the main textbook that Power series have an (possibly in�nite) interval, centered around
the center of the series (that’s why its called “center”), in which they are converging (absolute value of ratio< 1).�ey diverge outside the closed interval. (In this course we ignore the behavior at the end points.)
Power Series as Polynomials
Power series look a lot like polynomials. We therefore would like to treat them like polynomials — in fact thisis one of the main reasons for using power series. Before we can do so, however we will have to establish thatthis is a valid thing to do, for example we have to show that adding two power series gives the same result asadding the coe�cients o� the power series:
∞∑n=1
an +∞∑n=1
bn =∞∑n=1
(an + bn)
If we only had �nite sums this would be immediately clear, the law of commutativity tells us that the result isthe same.�is is something you are probably using without actually thinking about it because you learned itas soon as you learn the addition of numbers.Surprisingly (though you will see more examples of such behavior in advanced calculus), this is far more
di�cult for in�nite sums.�e problem is basically that the law of commutativity only lets usmove numbers inaddition by �nitely many places, but we need to move them in�nitely many places.�at can cause di�culties,because it potentially allows us to keep a particular summand for “later addition” in perpetuity, thus changingthe value of any �nite approximation (and thus also the value of the sum).�e following example illustratesthis in a particular case.
EXAMPLE 1What value should the in�nite sum
1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 +⋯
33
have?Solution Let us �rst note that it is not clear whether this in�nite series actually converges. What we willbe doing now therefore is also an illustration of the need to prove convergence.What we will be doing looks rather harmless at �rst sight. We will simply change the order of addition,
by introducing pairs of parentheses in the sum.�is changes the order of addition and is equivalent to rear-ranging the terms of this series. In each of the cases we can then ”evaluate” the sum easily:
1 − 1 + 1 − 1 + 1 − 1 +⋯ = (1 − 1) + (1 − 1) + (1 − 1) +⋯ = 0 + 0 + 0 + 0 +⋯ = 0= 1 + (−1 + 1) + (−1 + 1) + (−1 + 1) +⋯ = 1 + 0 + 0 + 0 +⋯ = 1
�is cannot be right (or the world is insane, a possibility which we do not want to consider here). In fact onecould do even worse things. We could group each −1 with the +1 following three positions later and thereforeget the value of 2.
�e root of the problem turns out to be the negative summands in between. If all were positive, we hadthe sum 1+ 1+ 1+⋯which clearly diverges. In such a case we say, that this series is not absolutely convergent.�e example shows, that rearrangement is not a valid operation for such a series.1
Fortunately a theorem from advanced calculus tells us that this obstacle we observed here is the only one.If the series converges absolutely, i.e. it would converge even if all minuses were replaced by pluses, we arepermitted to rearrange.
THEOREM 1 The Rearrangement Theorem for Absolutely Convergent Series
Suppose that∞∑n=1
an converges absolutely, i.e.∞∑n=1
∣an∣ converges as well, and b1, b2, . . . , bn , . . . is any arrange-
ment of the sequence {an}, then∑ bn converges absolutely, and
∞∑n=1
bn =∞∑n=1
an
�is theorem is stated as�eorem 17 in the textbook on p. 790, a sketch of the proof is outlined on pg 794exercise 60. We will be using this theorem in this lecture only to establish�eorems 2-5 below.
For power series we are now in luck: because of the ratio test (whichwe also could dowith absolute values)a power series converges absolutely inside its interval of convergence.We therefore are permitted to rearrangeits terms. In particular, when adding to a power series (centered at the same point) we can collect their termstogether in the same way as we would do for �nite sums. We get the following important theorem:
1Modern Physics has a method called Renormalization, which is essentially about such reordering.
34
THEOREM 2 Term-by-Term Summation for Power Series
Suppose that f (x) =∞∑n=l
cn(x − a)n and g(x) =∞∑n=0
dn(x − a)n are two power series, both centered at the
same point a, and that b is chosen inside the interval of convergence of both f and g. (�us f (b) and g(b)are both absolutely convergent series.)�en
f (b) + g(b) =∞∑n=0
(cn + dn)(b − a)n
We write thatf (x) + g(x) =
∞∑n=0
(cn + dn)(x − a)n
for x inside the interval of convergence for both series.
EXAMPLE 2Let f (x) = ∑
n
xn
nand g(x) = ∑
n
xn
n!. Determine the common interval of convergence for both series and
write for x inside this interval f (x) + g(x) as a single sum.Solution
�e same holds for products. Again, once we are permitted to rearrange terms, we can take the sameprocess as used for multiplying polynomials, and generalize it to the case of power series, as long as long aswe are inside the interval of convergence. �e resulting formula is in fact the same as holds for the generalmultiplication of polynomials. (In the textbook this is theorem 21 on p. 803.)
THEOREM 3 Term-by-Term Multiplication for Power Series
Suppose that f (x) =∞∑n=l
cn(x − a)n and g(x) =∞∑n=0
dn(x − a)n are two power series, both centered at the
same point a, and that b is chosen inside the interval of convergence of both f and g.�en
f (b) ⋅ g(b) =∞∑n=0
⎛⎝
n∑k=0
ck ⋅ dn−k⎞⎠(b − a)n
We write that
f (x) ⋅ g(x) =∞∑n=0
⎛⎝
n∑k=0
ck ⋅ dn−k⎞⎠(x − a)n
for x inside the interval of convergence for both series.
Proof To simplify notation we assume that a = 0. (�e general proof works the same way by replacing x
35
by x − a, but this is more messy to write down.) We multiply out:
(c0x0 + c1x1 + c2x2 +⋯) ⋅ (d0x0 + d1x1 + d2x2 +⋯)= c0x0 (d0x0 + d1x1 + d2x2 +⋯) + c1x1 (d0x0 + d1x1 + d2x2 +⋯)
+c2x2 (d0x0 + d1x1 + d2x2 +⋯) +⋯= c0d0x0x0 + c0d1x0x1 + c0d2x0x2 +⋯ + c1d0x1x0 + c1d1x1x1 +⋯
c2d0x2x0 + c2d1x2x1 +⋯= c0d0x0+0 + c0d1x0+1 + c0d2x0+2 +⋯ + c1d0x1+0 + c1d1x1+1 +⋯
c2d0x2+0 + c2d1x2+1 +⋯
Now we collect the summands according to the power of x. �is is where we need to rearrange the terms,which is permitted because of the absolute convergence.
= (c0d0)´¹¹¹¹¹¸¹¹¹¹¹¹¶
n=0, k=0x0 + (c0d1 + c1d0)
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶n=1, k=0,1
x1 + (c0d2 + c1d1 + c2d0)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
n=2, k=0,1,2x2 +⋯
+ (c0dn + c1dn−1 + c2dn−2 +⋯ + cnd0)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
k=0,1,2,...,nxn +⋯
EXAMPLE 3Let f (x) =∑
n
xn
nand g(x) =∑
n
xn
n!. For x inside the common interval of convergence, write f (x) ⋅ g(x) as
a sum.Solution
Since we are doing calculus, we don’t only want to do arithmetic with power series, but also di�erentiateand integrate. Two theorems from advanced calculus state, that this also can be done, as one would do it forpolynomials:
THEOREM 4 The Term-by-Term Differentiation Theorem
Let f (x) =∞∑n=0
cn(x − a)n.�en f (x) is di�erentiable for x inside the interval of convergence, and we have
thatf ′(x) = d
dxf (x) =
∞∑n=1
ncn(x − a)n−1,
and this derived series converges (absolutely).
Notice that the series now begins with n = 1 because the derivative of the constant term is zero.
36
THEOREM 5 The Term-by-Term Integration Theorem
Let f (x) =∞∑n=0
cn(x − a)n.�en for any x inside the interval of convergence of this series also the series
F(x) =∞∑n=0
cn(x − a)n+1
n + 1
converges (absolutely), and we have that ddx
F(x) = f (x).�erefore f (x) has an antiderivative, and
∫ f (x)dx =∞∑n=0
cn(x − a)n+1
n + 1 + C
�is establishes that inside the interval of convergence, we can treat power series just like polynomials.
A power series as solution to a differential equation
To give you a real world example consider theBessel functions2.�ese functions �rst arose in solvingKepler’sequations for planetary motion. Since that time these functions have been applied in many di�erent physicalsituations — for example the temperature distribution in a circular plate.
EXAMPLE 4 The Bessel functionWe de�ne the Bessel function of order 0 as
J0(x) =∞∑n=0
(−1)nx2n
22n(n!)2
What is its interval of convergence?Solution To �nd the interval of convergence, set an = (−1)nx2n/(22n(n!)2).�en
∣an+1an
∣ =RRRRRRRRRRR
(−1)n+1x2(n+1)
22(n+1)((n + 1)!)2⋅ 22n(n!)2
(−1)nx2nRRRRRRRRRRR= ∣x∣24(n + 1)2 → 0 < 1
�us the series converges for all real x, the Bessel function is de�ned on the whole real axis.If we want to see how the Bessel function looks we can start looking at the approximations given by partial
sums:
s2(x) = 1− x2
4, s3(x) = 1− x2
4+ x4
64, s4(x) = 1− x2
4+ x4
64− x6
2304, s6(x) = 1− x2
4+ x4
64− x6
2304+ x8
147456
as shown below.�e second graph is the full Bessel function J0(x).
As we have absolute convergence within the interval of convergence (and thus may reorder terms!) wecan treat arithmetic with power series in the same way as arithmetic with polynomials (i.e. sum up term-wise or form the product from partial sums), which anyhow looks like the natural thing to do. We also see
2Friedrich Bessel, German astronomer, 1784-1846
37
Degree=2
Degree=4
Deg=6
Deg=8
that we can calculate derivative and anti-derivative of a power series very easily while in general calculatinganti-derivatives is hard.Later we will see that we can express solutions to di�erential equations easily in terms of power series.
EXAMPLE 5 The Bessel function as solution for a differential equationIn continuation of the previous example, we want to see that the Bessel function J0(x) is a solution to thedi�erential equation
x2y′′(x) + x y′(x) + x2y(x) = 0
Solution For this, we calculate the �rst two derivatives (note that we can leave out the n = 0 term in thederivatives):
J0(x) =∞∑n=0
(−1)nx2n
22n(n!)2 , J′0(x) =∞∑n=1
(−1)n ⋅ 2n ⋅ x2n−122n(n!)2 , J′′0 (x) =
∞∑n=1
(−1)n2n(2n − 1)x2n−222n(n!)2
and multiply appropriately (doing an index-shi� to have all three series start at n = 1) to get
x2J0(x) =∞∑n=0
(−1)nx2n+222n(n!)2 =
∞∑n=0
−(−1)(n+1)x2(n+1)
22((n+1)−1)((n + 1) − 1)!)2m = n + 1=
∞∑m=1
−(−1)mx2m
22(m−1)((m − 1)!)2,
as well as:
xJ0′(x) =∞∑n=1
(−1)n ⋅ 2n ⋅ x2n
22n(n!)2 , x2J′′0 (x) =∞∑n=1
(−1)n2n(2n − 1)x2n
22n(n!)2
38
and therefore (collecting according to powers of x2n):
x2J′′0 + xJ′0 + x2J0 =∞∑n=1
(−1)n (− 122(n−1)((n − 1)!)2
+ 2n22n(n!)2 +
2n(2n − 1)22n(n!)2 ) x2n
=∞∑n=1
(−1)n−4n2 + 2n + 2n(2n − 1)22n(n!)2 x2n
=∞∑n=1
(−1)n 022n(n!)2 x2n = 0
�is veri�es that the Bessel function is a solution to the given di�erential equation.
�ere are many other so-called special functions, de�ned as power series, which occur in applications.
Expressing known functions as power series
To be able to work with power series and “ordinary” functions, the next big question then is how to translatebetween “traditional” functions, given by a formula, and power series.We can do some cases using partial fractions and geometric series.
EXAMPLE 6 Power series for rational functionsDetermine a power series for 9
x3 + 3x2 − 4 .Solution In a partial fraction decomposition, we write
9x3 + 3x2 − 4 =
9(x − 1)(x + 2)2 =
1x − 1 −
1x + 2 −
3(x + 2)2
�en we note that (by geometric series):
1x − 1 = −1 − x − x2 − x3 −⋯
1x + 2 = 1
2− 14
x + 18
x2 − 116
x3 +⋯ =∞∑n=0
(−1)n 12n+1 x
n
For the power series for 1(x + 2)2 , remember that ∫
1(x + 2)2 dx = −1
x + 2 , we can thus get a series for1
(x + 2)2 by term-by-term di�erentiation of a series for−1
x + 2 (which we know already):
1(x + 2)2 = d
dx−1
x + 2 =d
dx
∞∑n=0
−(−1)n 12n+1 x
n =∞∑n=1
(−12)n+1nxn−1
= (−12)2 + (−1
2)3 ⋅ 2x + (−1
2)4 ⋅ 3x2 + (−1
2)5 ⋅ 4x3 +⋯
= 14− 14
x + 316
x2 − 18
x3 +⋯
39
Note that one could get this result also by multiplying the series for 1x+2 with itself, using the formula for products:
1(x + 2)2 = ( 1
2⋅ 12) + (− 1
2⋅ 14− 14⋅ 12) x + ( 1
2⋅ 18+ 14⋅ 14+ 18⋅ 12) x2
+(− 12⋅ 116− 14⋅ 18− 18⋅ 14− 116⋅ 12) x3 +⋯
= 14− 14
x + 316
x2 − 18
x3 +⋯
by geometric series and multiplication (remember reindexing!)In summary we get that
9x3 + 3x2 − 4 = −
94− 2716
x2 − 916
x3 +⋯.
�e general case will have to wait for the next chapters.
EXERCISES
1. Determine a power series for 1(x + 2)(x − 2) .
2. Determine a power series for 1(x − 2)2 from a
power series for its antiderivative.
3. Determine the �rst four terms of a power seriesfor 1
(x − 2)2 , by multiplication. Compare with the re-sult of problem 2.
4. We know that ddxln(1+ x) = 1
1 + x. Determine a
power series for 11 + x
, and use term-wise integrationto obtain a power series for ln(1 + x).
40
5-
Taylor polynomials as functionapproximations
One of the principal uses of Taylor polynomials is to approximate functions. �e key to this is the formulafor the error term. Let us recall the formula:
Let f be a function that is in�nitely o�en di�erentiable and M be a positive constant such that∣ f (n+1)(t)∣ ≤ M for all t between x and a, then the remainder term Rn(x) = f (x) = Tn(x) inTaylor’s�eorem satis�es
∣Rn(x)∣ ≤ M ∣x − a∣n+1(n + 1)! .
If f is well-behaved (essentially any function you will consider in calculus is), then for each �nite intervalthere will be one (possibly very large) value ofM, such that ∣ f (n+1)(t)∣ ≤ M for every n.�e reason for this is,that otherwise higher derivatives of f must have larger and larger values, and therefore oscillate rather wildly.
�e remainder estimate however then is essentially ∣x − a∣n
(n + 1)! . From the study of series we remember that
n! grows faster1 than xn.�us the error becomes smaller and smaller, the larger n gets.Vice versa, for a �xed n we can make the error arbitrary small, by choosing x to be close to the center a.
�is principle is fundamental to applications in Physics and Engineering:
Approximation principle for well-behaved functionsIf f is a well-behaved function, and Tn the Taylor polynomial for f of degree n centered at a, then the functionTn(x) is an approximation for f (x).�e approximation quality gets better, the larger n is and the closer x isto a.For values of x very close to a the linear approximation T1(x) is o�en su�ciently good.
We ilustrate this principle in the case of deciding which function is largest:
1Once n > x the subsequent factors in n! are larger than the factors in xn
41
EXAMPLE 1 Using Power Series to Approximate FunctionsSuppose we wish to compare the values of functions x
1−x , sin(x), and ex − 1 for small (positive) values of x.Which function is largest?Solution We begin expressing each of these functions as power series, centered at a = 0:
x1 − x
= x( 11 − x
) = x ⋅∞∑n=0
xn =∞∑n=0
xn+1 = x + x2 + x3 +⋯
sin(x) =∞∑n=0
(−1)nx2n+1(2n + 1)! = x − x3
3!+ x5
5!−⋯
ex − 1 =∞∑n=0
xn
n!− 1 =
∞∑n=1
xn
n!= x + x2
2!+ x3
3!+⋯
We see that all three functions agree on the Taylor approximations T0(x) and T1(x), thus we need to considerat least T2(x). We get the following approximations:
x1 − x
≃ x + x2
sin(x) ≃ x
ex − 1 ≃ x + x2
2
For positive values of x, we then note that x + x2 > x + x2
2> x, thus for small values of x we have that
x1 − x
> ex − 1 > sin(x).
A look at the graph veri�es this result: �e graph for x ∈ [0..0.01] shows that for even smaller values of x
xx
sin xex
xx
sin xex
xx
sin xex
the graphs are almost identical, this is a consequence of the fact that the linear approximation of all threefunctions is the same. Of course, for larger values of x the behaviour can be quite di�erent, as the third graphshows.
42
EXERCISES
1. By looking at their Taylor series, decide which ofthe following functions is largest, and which smallestfor x near 0:
ex , 1√1 − 2x
, 11 − x
2. Consider the power series
W(x) =∞∑n=1
(−n)n−1n!
xn .
a) Determine the interval (open interval: you do notneed to decide on the behavior at the end points) ofconvergence for W(x). You may use without proof
that limn→∞( n
n + 1)n+1
= 1e
b) Calculate a power series forW ′(x).c) Let I = [0, 0.1]. You are given the information, thatfor the second derivative ∣W(2)(x)∣ ≤ 2 for x ∈ I.Using the formula for the error term of a Taylor poly-nomial, show that ∣W(x) − x∣ ≤ x2 (and thus x−x2 ≤W(x) ≤ x + x2) for x ∈ I.
43
-
Homework Assignments
EXERCISES
8.2.A Evaluate the integrals
∫ sin3 x cos3 xdx
∫ sin2 x cos2 xdx
7.1.A Let f (x) = 13x + 53 . Find the inverse f −1(x)
and identify the domain and range of f −1. As a check,show that f ( f −1(x)) = f −1( f (x)) = x.
7.1.B Let f (x) = (x − 2)3. Show that f is one-to-one on R. Find a formula for d f −1/dx at x = 8 usingtheorem 1.
7.6.A Given that α = arccos(4/9), �nd sin α, tan αand sec α.
7.6.B Simplify the expression tan(arcsin(x/7)).
7.6.C Evaluate the integrals
∫ 1√−x2 − 6x − 8
dx
∫ 4r√1 − r4
dr
7.2.A Di�erentiate:d
dxln( cos x√
x2 + 1(x − 1)5)
7.2.B Evaluate the integrals
∫ (ln x)2x
dx
∫ (3x − 1)9 − 2x + 3x2 dx
7.3.A Evaluate the integrals
∫ et√et − 1dt
∫ t ⋅ 3t2dt
7.4.A Verify that the function e−x(x − 2)−2 is a so-lution to the initial value problem
(x − 2)y′ + x y = 0, y(0) = 14
7.4.B Solve the separable di�erential equation
y′ ⋅√1 − x2 = x y
7.4.C When cane sugar is dissolved in water, it con-verts to invert sugar over time. �e amount f (t) ofunconverted cane sugar at time t decreases, follow-ing the di�erential equation f ′ = −0.2 f . Suppose westart with a solution containing 100g of cane sugar.How much cane sugar remains a�er 5 hours? a�er 10hours?How long will it take, until only 1g of cane sugar re-mains?
45
8.0.A Evaluate the integrals
∫ x − 53 − x
dx
∫ 1x2 − 14x + 50dx
8.1.A Evaluate the integrals
∫ x−9 ⋅ ln xdx
∫ arccos tdt
8.3.A Evaluate the integrals
∫ x2
(1 + x2)3/2dx
∫ x2
(1 − x2)3/2dx
8.4.A Expand the quotient
2x + 4(x − 2)(x2 + 4)
by partial fractions.
8.4.B Evaluate the integral
∫ 1(x − 4)2(x − 1)dx
8.7.A Evaluate the improper integral
∫ ∞2
1(x + 4)3 dx
10.1.A Find a formula for the terms of the followingsequences:
1, 4, 7, 10, 13, . . .12,−24, 68,−2416, . . .
10.1.B Give an N/ε-proof, that the sequence an =1 − 2n1 + 2n converges to −1.
10.7.A Determine the open (i.e. ignoring the behav-ior at the endpoints) interval of convergence for
∞∑
n=10x2n+13n + 1
∞∑n=1
(x − 3)n
7
10.7.B Consider the power series
f (x) = x − x2
2+ x3
3− x4
4+⋯
i) determine the (open, ignoring endpoints) interval,in which f (x) converges.ii) Determine its term-by-teem derivative f ′(x).iii) Determine the value of f ′(x) as a fraction (Hint:Geometric series).iv) Give a formula for the value of f (x), using an an-tiderivative of the function found in iii)
10.7.C Determine a power series for
1(2 + x3)2 .
10.7.D We know that ddxln(1 + x) = 1
1 + x. Deter-
mine a power series for 11 + x
, and use term-wise in-tegration to obtain a power series for ln(1 + x).
10.8.A Find the �rst four terms of a Taylor seriesabout a = 0 for
sin(x)1 − x
10.8.B Determine the Taylor series for f (x) =√4 − x about a = 0.
10.9.A Determine a Power series for the followingfunctions about a = 0:
f (x) = x2 ⋅ ex2
f (x) = cos(√
x)
f (x) = sin2(x) = 1 − cos(2x)2
46
10.9.B Identify the following functions from theirTaylor series about a = 0
1 + x3 + x6
2!+ x9
3!+ x12
4!+⋯
1 − 4x + 42x2 − 43x3 + 44x4 − 45x5 +⋯
x4 − x12
3+ x20
5− x28
7⋯
10.9.C We approximate sin(x) by x − x36 on the in-
terval [−1, 1]. What estimate can you make for the er-ror?Suppose we want to approximate sin(x) on the inter-val [−1, 1] with an error of at most 10−5, what degreeTaylor polynomial do we need to choose?
10.9.D By looking at their Taylor series, decidewhich of the following functions is largest, and whichsmallest for x near 0:
ex , 1√1 − 2x
, 11 − x
10.10.A Find a power series solving the di�erentialequation
y′ − 2y = 4xunder the initial condition y(0) = 1. Identify the so-lution as a function.
10.10.B Find a power series solving the di�erentialequation
y′′ + y = 0under the initial conditions y(0) = 2, y′(0) = 1. Iden-tify the solution as a function.
11.2.A Calculate the arc length of ln(cos(x)) from0 to π
4 .
11.3.A Convert the equation of the curve
r = 12 − cos θ
into cartesian coordinates
11.3.B Convert the equation of the curve
x y = 1
into polar coordinates
11.4.A Find the points of intersection of the polarcurves
r = 2 + sin θ , r = 2 − sin θ
11.5.A Sketch the curves r1 = 2 + sin 2θ and r2 =2 + cos 2θ.Determine the area inside the curve r1 = 2 + sin 2θbut outside the curve r2 = 2 + cos 2θ.
A.7.A Determine all complex ��h roots of 32.
A.7.B Calculate 12+i√3
A.7.C Determine all complex roots of the equationx4 + 3 ∗ x2 + 9 = 0
47
