Math 104 - Calculus I August 9 (but first, a quick review…)

Math 104 - Calculus I

August 9

(but first, a quick review…)

Series of positive termsConvergence questions for series of positive terms are easiest

to understand conceptually.

Since all the terms a are assumed to be positive, the sequence of partial sums {S } must be an increasing sequence.

So the least upper bound property discussed earlier comes into play -- either the sequence of partial sums has an upper bound or it doesn't.

If the sequence of partial sums is bounded above, then it must converge and so will the series. If not, then the series diverges. That's it.

nn

Tests for convergence of series of positive terms:

The upper bound observations give rise to several "tests" for convergence of series of positive terms. They all are based pretty much on common sense ways to show that the partial sums of the series being tested is bounded are all less than those of a series that is known to converge (or greater than those of a series that is known to diverge). The names of the tests we will discuss are...

Tests...

1. The integral test

2. The comparison test

3. The ratio test

4. The limit comparison test (sometimes called the ratio comparison test)

5. The root test

TODAY

TODAY

The integral test

Since improper integrals of the form

provide us with many examples of telescoping series whose convergence is readily determined, we can use integrals to determine convergence of series:

1dx)x(f

For example, consider the series

From the following picture, it is evident that the nth partial sum of this series is less than

Integral test cont.

1

12

nn

n

x1 dx

1 21

What is the sum?The sum of the terms is equal to the sum of the areas

of the shaded rectangles, and if we start integrating at 1 instead of 0, the

improper integral converges

(question: what is the integral? so what bound to you conclude for the series?).

Since the value of the improper integral (plus 1) provides us with an upper bound for all of the partial sums, the series must converge.

It is an interesting question as to exactly what the sum is. We will answer it next week.

1 2 dxx1

The integral test...

11

1

1

integral the

of divergence knew thealready we-because-series

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new,a us gives This integral). for the checked be

toneedsinfinity at econvergenc(only diverge

bothor converge botheither f integral

improper theand f series the,decreasing

and positive is f(x) function theif that says

.dx

dx)x(

)x(

x

x

Discussion and Connect

-- for which exponents p does the series

converge?

(These are sometimes called p-series, for obvious reasons -- these together with the geometric series give us lots of useful examples of series whose convergence or divergence we know).

1

1

nn p

Question…

Error estimates:Using the picture that proves the integral test for

convergent series, we can get an estimate on how far off we are from the limit of the series if we stop adding after N terms for any finite value of N.

If we approximate the convergent series

by the partial sum

then the error we commit is less than the value of the integral

1

)(fn

n

N

nN ns

1

)(f

Ndxx)(f

Take a closer look...

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the--offfar t isn' estimate thislaer, see shall we

(As . sum infitie thefrom 0.2,or ,

thanlessby differs This 1.46.ely approximat is

which,1 sum theexample,For

1

151

5

1

36005269

251

161

91

41

22

nnx

dx

Question

A) Converge

B) Diverge

diverge?or converge series theDoes1

1 2

nnn

Question

A) Converge

B) Diverge

diverge?or converge series theDoes1

1

)arctan(2

nn

n

Connect

For this latter series, find a bound on the error if we use the sum of the first 100 terms to approximate the limit. (answer: it is less than about .015657444)

Exercise

The comparison testThis convergence test is even more common-

sensical than the integral test. It says that if

all the terms of the series are less than

the corresponding terms of the series

and if converges, then

converges also.

1nna

1nnb

1nna

1nnb

Reverse

This test can also be used in reversed -- if

the b series diverges and the a’s are bigger

than the corresponding b’s, then

diverges also.

1nna

Examples:

diverges.

converges.

1)sin(

12

1

nnn

n

nnn

Question

diverge?or converge series theDoes5

21

kk

A) Converge

B) Diverge

Question

diverge?or converge series theDoes1

21

2

nnn

A) Converge

B) Diverge

Convergence Tests...

1. The integral test

2. The comparison test

3. The ratio test

4. The limit comparison test (sometimes called the ratio comparison test)

5. The root test

The ratio test

The ratio test is a specific form of the comparison test, where the comparison series is a geometric series. We begin with the observation that for geometric series, the ratio of consecutive terms

is a constant (we called it r earlier). n

n

a

a 1

Ratio test (cont.)

• For other series, even if the ratio of consecutive terms is not constant, it might have a limit as n goes to infinity. If this is the case, and the limit is not equal to 1, then the series converges or diverges according to whether the geometric series with the same ratio does. In other words:

The ratio test:

drawn. becan conclusion nothen

exist),not doeslimit theif(or 1r If

1r if -diverges-

1r if -converges-

series then the, lim If1

1

n

nn

n

nar

a

a

Example:

converges. series theso

,01

1 lim

)!1(

! lim lim

have we,!

1For

1

1

nn

n

a

a

n

nnn

n

n

n

For , the ratio is 1 and the ratio

test is inconclusive.

Of course, the integral test applies to these p-series.

Another example:

1

1

npn

Question

diverge?or converge 5

! series theDoes

1

nn

n

A) Converge

B) Diverge

Question

diverge?or converge ln

series theDoes2

kke

k

A) Converge

B) Diverge

Root test• The last test for series with positive terms that we

have to worry about is the root test. This is another comparison with the geometric series. It's like the ratio test, except that it begins with the observation that for geometric series, the nth root of the nth term approaches the ratio r as n goes to infinity (because the nth term is arn and so the nth root of the nth term is a1/nr-- which approaches r since the nth root of any positive number approaches 1 as n goes to infinity.

The root test says...

1.r ifdrawn becan conclusion no and

1r if diverges series the--

1r if converges series the--

then,lim ifthat

1

1

1

nn

nn

nn

a

a

ra n

Example

root test. by the converges

521

n

n

n

n

Question

diverge?or converge 1 series theDoes1

n

nne

A) Converge

B) Diverge

Series whose terms are not all positive

• Now that we have series of positive terms under control, we turn to series whose terms can change sign.

• Since subtraction tends to provide cancellation which should "help" the series converge, we begin with the following observation:

• A series with + and - signs will definitely converge if the corresponding series obtained by replacing all the - signs by + signs converges.

Absolutely convergent series

A series whose series of absolute values converges, which is itself then convergent, is called an absolutely convergent series.

values."

absolute of series" the- be wouldplus tosigns

minus all changingby obtained series then theand

negative, and positive being as series theof

terms theof think tois say this y toAnother wa

1

1

nn

nn

n

a

a

a

Examples...

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see) will we(as convergent is ...1

divergent. is valuesabsolute of series its course)(or and zero) totend

tdoesn' (sincedivergent is ...111111)1(

.convergent absolutley is ...1

61

51

41

31

21

1n

)1(

0n

251

161

91

41

1

)1(

)1(

2

n

nn

nn

n

n

a

Series that are convergent although their series of absolute values diverge (convergent but not absolutely convergent) are called conditionally convergent.

Alternating series

A special case of series whose terms are of both signs that arises surprisingly often is that of alternating series . These are series whose terms alternate in sign. There is a surprisingly simple convergence test that works for many of these:

Alternating series test:

).( termomittedfirst that the valueabsolutein less is and as

sign same thehas sum partial theand series theoflimit ebetween th

difference theMoreover, converges. series then the,0lim if and

...

is that ,decreasing are terms theof) valuesabsolute (the If

series. galternatinan be ...)1(Let

1

210

32100

n

n

nn

nn

n

a

s

a

aaa

aaaaa

Example:The alternating harmonic series clearly

satisfies the conditions of the test and is therefore convergent. The error

estimate tells us that the sum

is less than the limit, and within 1/5. Just to practice using the jargon, the alternating harmonic series, being convergent but not absolutely convergent, is an example of a conditionally convergent series.

...1 41

31

21

127

41

31

211

Classify each of the following...

A) Absolutely convergent

B) Conditionally divergent

C) Divergent

2 ln

)1(

n

n

nn

Classify each of the following...

A) Absolutely convergent

B) Conditionally divergent

C) Divergent

13

sin

k k

k

Classify each of the following...

A) Absolutely convergent

B) Conditionally divergent

C) Divergent

12 5

cos

n n

nn

Power seriesLast week's project was to try and sum series

using your calculator or computer. The answers correct to ten decimal places are:

Sum((-1)^n/(2*n+1),n=0..infinity) = evalf(sum((-1)^n/(2*n+1),n=0..

infinity));

Sum(1/factorial(n),n=0..infinity)=evalf(sum(1/

factorial(n),n=0..infinity));

0

12)1( 7853981635.

nn

n

0

!1 718281828.2

nn

Power series (cont.)Sum(1/n^2,n=1..infinity)=evalf(sum(1/n^2,n=1..infinity));

Sum((-1)^(n+1)/n,n=1..infinity)=evalf(sum((-1)^

(n+1)/n,n=1..infinity));

We can recognize these numbers as

1

1 6644934068.12

nn

1

)1( 6931471806.)1(

nn

n

).2(ln and , , 64

2 e

Two directions:

1. Given a number, come up with a series that has the number as its sum, so we can use it to get approximations.

2. Develop an extensive vocabulary of "known" series, so we can recognize "familiar" series more often.

Geometric series revisited

1). (provided -1

...

:series geometric the

friend, oldour begin with We

432

0

rr

a

ararararaarn

n

r as a variableChanging our point of view for a minute (or a

week, or a lifetime), let's think of r as a variable. We change its name to x to emphasize the point:

So the series defines a function (at least for certain values of x).

)1(for 1

...)(f0

32

xx

aaxaxaxaaxx

n

n

Watch out...We can identify the geometric series when we see it,

we can calculate the function it represents and go back and forth between function values and specific series.

We must be careful, though, to avoid substituting values of x that are not allowed, lest we get nonsensical statements like

!!1...168421...22221 432

Power seriesIf you look at the geometric series as a function, it

looks rather like a polynomial, but of infinite degree.

Polynomials are important in mathematics for many

reasons among which are:

1. Simplicity -- they are easy to express, to add, subtract, multiply, and occasionally divide

2. Closure -- they stay polynomials when they are added, subtracted and multiplied.

3. Calculus -- they stay polynomials when they are differentiated or integrated

Infinite polynomials

So, we'll think of power series as "infinite polynomials", and write

0

33

2210 ...

n

nn xaxaxaaxa

Three (or 4) questions arise...

1. Given a function (other than ), can it be expressed

as a power series? If so, how? 2. For what values of x is a power series representation valid?

(This is a two part question -- if we start with a function f(x) and form "its" power series, then

(a) For which values of x does the series converge?

(b) For which values of x does the series converge to f(x) ?

[There's also the question of "how fast".]

x

ax

1)(f

continued

3. Given a series, can we tell what function it came from?

4. What is all this good for?

As it turns out, the questions in order of difficulty, are 1, 2(a), 2(b) and 3. So we start with question 1:

The power series of a function of f(x)

Suppose the function f(x) has the power series...)(f 4

43

32

210 xaxaxaxaax

Q. How can we calculate the coefficients a from a knowledge of f(x)?

A. One at a time -- differentiate and plug in x=0!

i

Take note...

(0)/2f have should weso

...,201262)(f

:derivativeAnother

(0)f

have should weso ,...432)(f

write toreasonable seemsIt :(first?) Second

f(0) that have weso

...000)0(f :zeroth)(or First

2

35

2432

1

34

2321

0

03

32

210

a

xaxaxaax

a

xaxaxaax

a

aaaaa

Continuing in this way...

n!by divided

zero,at evaluated f of derivativenth

:generalIn

etc.... (0)/24,f (0)/6,f 43

na

aa

ExampleSuppose we know, for the function f, that f(0)=1 and

f ' = f.

Then f '' = f ', f ''' = f '' etc... So f '(0) = f ''(0) = f '''(0) = ... = 1.

From the properties of f we know on the one hand that So we get that...

)!previously did wesums theof oneget to1(Set x

!...

!4!3!21

0

432

n

nx

n

xxxxxe

xex )(f

Good night…

See you Wednesday!