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11.2
Series
In this section, we will learn about:
Various types of series.
INFINITE SEQUENCES AND SERIES
SERIES
If we try to add the terms of an infinite sequence
we get an expression of the form
a1 + a2 + a3 + ··· + an + ∙·∙
1{ }n na
Series 1
INFINITE SERIES
This is called an infinite series (or just a series).
It is denoted, for short, by the symbol
1
orn nn
a a
However, does it make sense to talk about the sum
of infinitely many terms?
INFINITE SERIES
It would be impossible to find a finite sum for the
series
1 + 2 + 3 + 4 + 5 + ∙∙∙ + n + ···
If we start adding the terms, we get the cumulative sums 1, 3, 6, 10, 15, 21, . . .
After the nth term, we get n(n + 1)/2, which becomes very large as n increases.
INFINITE SERIES
However, if we start to add the terms of the series
we get:
1 1 1 1 1 1 1
2 4 8 16 32 64 2n
3 7 15 31 6312 4 8 16 32 64 1 1/ 2, , , , , , , ,n
INFINITE SERIES
The table shows that, as we add more and more
terms, these partial sums become closer and closer
to 1.
In fact, by adding sufficiently many terms of the series, we
can make the partial sums as
close as we like to 1.
INFINITE SERIES
So, it seems reasonable to say that the sum of
this infinite series is 1 and to write:
1
1 1 1 1 1 11
2 2 4 8 16 2n nn
INFINITE SERIES
We use a similar idea to determine whether or not
a general series (Series 1) has a sum.
INFINITE SERIES
We consider the partial sums
s1 = a1
s2 = a1 + a2
s3 = a1 + a2 + a3
s3 = a1 + a2 + a3 + a4
In general,1 2 3
1
n
n n ii
s a a a a a
INFINITE SERIES
These partial sums form a new sequence {sn},
which may or may not have a limit.
If exists (as a finite number), then, as
in the preceding example, we call it the sum of
the infinite series an.
INFINITE SERIES
lim nn
s s
Given a series
let sn denote its nth partial sum:
1 2 31
ii
a a a a
1 21
n
n i ni
s a a a a
SUM OF INFINITE SERIES Definition 2
If the sequence {sn} is convergent and
exists as a real number, then the series ai is called
convergent and we write:
The number s is called the sum of the series.
Otherwise, the series is called divergent.
1 21
orn ii
a a a s a s
SUM OF INFINITE SERIES Definition 2
lim nn
s s
Thus, the sum of a series is the limit of the
sequence of partial sums.
So, when we write , we mean that, by adding
sufficiently many terms of the series, we can get as close as we like to the number s.
1n
n
a s
SUM OF INFINITE SERIES
Notice that:
1 1
limn
n in
n i
a a
SUM OF INFINITE SERIES
Compare with the improper integral
To find this integral, we integrate from 1 to t and then let t → .
For a series, we sum from 1 to n and then let n → .
1 1( ) lim ( )
t
tf x dx f x dx
SUM OF INFINITE SERIES VS. IMPROPER INTEGRALS
An important example of an infinite series is the
geometric series
2 3 1
1
1
0
n
n
n
a ar ar ar ar
ar a
GEOMETRIC SERIES Example 1
Each term is obtained from the preceding one by
multiplying it by the common ratio r.
We have already considered the special case where a = ½ and r = ½ earlier in the section.
GEOMETRIC SERIES Example 1
If r = 1, then
sn = a + a + ∙∙∙ + a = na →
Since does not exist, the geometric series diverges
in this case.
GEOMETRIC SERIES Example 1
lim nns
If r 1, we have
sn = a + ar + ar2 + ∙∙∙ + ar n–1
and
rsn = ar + ar2 + ∙∙∙ +ar n–1 + ar n
GEOMETRIC SERIES Example 1
Subtracting these equations, we get:
sn – rsn = a – ar n
and(1 )
1
n
n
a rs
r
GEOMETRIC SERIES E. g. 1—Equation 3
If –1 < r < 1, we know from Result 9 in Section
11.1 that r n → 0 as n → .
So,
Thus, when |r | < 1, the series is convergent and its sum is a/(1 – r).
(1 )lim lim
1 1
n
nn n
a r as
r r
GEOMETRIC SERIES Example 1
If r –1 or r > 1, the sequence {r n} is divergent by
Result 9 in Section 11.1
So, by Equation 3, does not exist.
Hence, the series diverges in those cases.
GEOMETRIC SERIES Example 1
lim nn
s
The figure provides a
geometric demonstration
of the result in Example 1.
GEOMETRIC SERIES
If s is the sum of the series,
then, by similar triangles,
So,
s a
a a ar
GEOMETRIC SERIES
1
as
r
GEOMETRIC SERIES
We summarize the results of Example 1 as
follows.
The geometric series
is convergent if |r | < 1.
1 2
1
n
n
ar a ar ar
Result 4
Moreover, the sum of the series is:
If |r | 1, the series is divergent.
1
1
11
n
n
aar r
r
GEOMETRIC SERIES Result 4
Find the sum of the geometric series
The first term is a = 5 and the common ratio is r = –2/3
10 20 403 9 275
GEOMETRIC SERIES Example 2
Since |r | = 2/3 < 1, the series is convergent by
Result 4 and its sum is:
23
53
10 20 40 55
3 9 27 1 ( )
5
3
GEOMETRIC SERIES Example 2
What do we really mean when we say that the
sum of the series in Example 2 is 3?
Of course, we cannot literally add an infinite number of terms, one by one.
GEOMETRIC SERIES
However, according to Definition 2, the total
sum is the limit of the sequence of partial sums.
So, by taking the sum of sufficiently many terms, we can get as close as we like to the number 3.
GEOMETRIC SERIES
The table shows the first ten partial sums sn.The
graph shows how the sequence of partial sums
approaches 3.
GEOMETRIC SERIES
Is the series
convergent or divergent?
2 1
1
2 3n n
n
GEOMETRIC SERIES Example 3
Let us rewrite the nth term of the series in the
form ar n-1:
We recognize this series as a geometric series with a = 4 and r = 4/3.
Since r > 1, the series diverges by Result 4.
12 1 2 ( 1)
11 1 1 1
4 42 3 (2 ) 3 4
3 3
nnn n n n
nn n n n
GEOMETRIC SERIES Example 3
Write the number as a ratio
of integers.
After the first term, we have a geometric series with a = 17/103 and r = 1/102.
3 5 7
17 17 172.3171717 2.3
10 10 10
GEOMETRIC SERIES Example 4
2.317 2.3171717...
Therefore,
3
2
17 1710 10002.317 2.3 2.3
1 991
10 10023 17
10 9901147
495
GEOMETRIC SERIES Example 4
Find the sum of the series
where |x| < 1.
Notice that this series starts with n = 0.
So, the first term is x0 = 1.
With series, we adopt the convention that x0 = 1 even when x = 0.
0
n
n
x
GEOMETRIC SERIES Example 5
Thus,
This is a geometric series with a = 1 and r = x.
2 3 4
0
1n
n
x x x x x
GEOMETRIC SERIES Example 5
Since |r | = |x| < 1, it converges, and Result 4 gives:
0
1
1n
n
xx
GEOMETRIC SERIES E. g. 5—Equation 5
Show that the series
is convergent, and find its sum.
1
1
( 1)n n n
SERIES Example 6
This is not a geometric series.
So, we go back to the definition of a convergent series and compute the partial sums:
1
1
( 1)
1 1 1 1
1 2 2 3 3 4 ( 1)
n
ni
si i
n n
SERIES Example 6
We can simplify this expression if we use the
partial fraction decomposition.
See Section 7.4
1 1 1
( 1) 1i i i i
SERIES Example 6
Thus, we have:
1
1
1
( 1)
1 1
1
1 1 1 1 1 1 11
2 2 3 3 4 1
11
1
n
ni
n
i
si i
i i
n n
n
SERIES Example 6
Thus,
Hence, the given series is convergent and
1lim lim 1 1 0 1
1nn n
sn
1
11
( 1)n n n
SERIES Example 6
The figure illustrates Example 6 by showing the
graphs of the sequence of terms an =1/[n(n + 1)]
and the sequence {sn} of partial sums.
Notice that an → 0 and sn → 1.
SERIES
Show that the harmonic series
is divergent.
1
1 1 1 11
2 3 4n n
HARMONIC SERIES Example 7
For this particular series it’s convenient to consider
the partial sums s2, s4, s8, s16, s32, …and show that
they become large.
1
12 2
1 1 1 1 1 14 2 3 4 2 4 4
2
2
1
1
1 1
1
s
s
s
HARMONIC SERIES Example 7
1 1 1 1 1 1 18 2 3 4 5 6 7 8
1 1 1 1 1 1 1
2 4 4 8 8 8 8
1 1 1
2 2 2
3
2
1
1
1
1
s
HARMONIC SERIES Example 7
Similarly,
1 1 1 1 1 1 116 2 3 4 5 8 9 16
1 1 1 1 1 1 1
2 4 4 8 8 16 16
1 1 1 1
2 2 2 2
4
2
1
1
1
1
s
HARMONIC SERIES Example 7
Similarly,
Similarly, s32 > 1 + 5/2, s64 > 1 + 6/2, and, in
general,
This shows that s2n → as n → , and so {sn} is divergent.
Therefore, the harmonic series diverges.
HARMONIC SERIES Example 7
21
2n
ns
HARMONIC SERIES
The method used in Example 7 for showing that
the harmonic series diverges is due to the French
scholar Nicole Oresme (1323–1382).
If the series is convergent, then1
nn
a
lim 0n
na
SERIES Theorem 6
Let sn = a1 + a2 + ∙∙∙ + an
Then, an = sn – sn–1
Since an is convergent, the sequence {sn} is convergent.
SERIES Theorem 6 - Proof
Let
Since n – 1 → as n → , we also have:
SERIES Theorem 6 - Proof
lim nn
s s
1lim nn
s s
Therefore,
1
1
lim lim
lim lim
0
n n nn n
n nn n
a s s
s s
s s
SERIES Theorem 6 - Proof
With any series an we associate two sequences:
The sequence {sn} of its partial sums
The sequence {an} of its terms
SERIES Note 1
If an is convergent, then
The limit of the sequence {sn} is s (the sum of the series).
The limit of the sequence {an}, as Theorem 6 asserts, is 0.
SERIES Note 1
The converse of Theorem 6 is not true in general.
If , we cannot conclude that an is
convergent.
SERIES Note 2
lim 0nn
a
Observe that, for the harmonic series 1/n, we
have an = 1/n → 0 as n → ∞.
However, we showed in Example 7 that 1/n is divergent.
SERIES Note 2
If does not exist or if ,
then the series is divergent.
lim nna
lim 0nn
a
1n
n
a
THE TEST FOR DIVERGENCE Test 7
The Test for Divergence follows from Theorem 6.
If the series is not divergent, then it is convergent.
Thus,
TEST FOR DIVERGENCE
lim 0nn
a
Show that the series diverges.
So, the series diverges by the Test for Divergence.
2
21 5 4n
n
n
TEST FOR DIVERGENCE Example 8
2
2 2
1 1lim lim lim 0
5 4 5 4 / 5nn n n
na
n n
If we find that , we know that an
is divergent.
If we find that , we know nothing about
the convergence or divergence of an.
SERIES Note 3
lim 0nna
lim 0nn
a
Remember the warning in Note 2:
If , the series an might converge or
diverge.
SERIES
lim 0nn
a
Note 3
If an and bn are convergent series, then so are the
series can (where c is a constant), (an + bn), and
(an – bn), and
1 1
1 1 1
1 1 1
i.
ii.
iii.
n nn n
n n n nn n n
n n n nn n n
ca c a
a b a b
a b a b
SERIES Theorem 8
These properties of convergent series follow from
the corresponding Limit Laws for Sequences in
Section 11.1
For instance, we prove part ii of Theorem 8 as follows.
SERIES
Let
1 1
1 1
n
n i ni n
n
n i ni n
s a s a
t b t b
THEOREM 8 ii—PROOF
The nth partial sum for the series Σ (an + bn) is:
1
n
n iii
u a b
THEOREM 8 ii—PROOF
Using Equation 10 in Section 5.2, we have:
1
1 1
1 1
lim lim
lim
lim lim
lim lim
n
n i in n
i
n n
i in
i i
n n
i in n
i i
n nn n
u a b
a b
a b
s t s t
THEOREM 8 ii—PROOF
Hence, Σ (an + bn) is convergent, and its sum is:
1
1 1
n nn
n nn n
a b s t
a b
THEOREM 8 ii—PROOF
Find the sum of the series
The series 1/2n is a geometric series with a = ½ and
r = ½. Hence,
1
3 1
( 1) 2nn n n
SERIES Example 9
12
11 2
11
2 1nn
In Example 6, we found that:
So, by Theorem 8, the given series is convergent
and
1 1 1
3 1 1 13
( 1) 2 ( 1) 2
3 1 1
4
n nn n nn n n n
SERIES Example 9
1
11
( 1)n n n
A finite number of terms does not affect the
convergence or divergence of a series.
SERIES Note 4
For instance, suppose that we were able to show
that the series is convergent.
Since
it follows that the entire series is convergent.
34 1n
n
n
SERIES Note 4
3 31 4
1 2 3
1 2 9 28 1n n
n n
n n
31 1n
n
n
Similarly, if it is known that the series
converges, then the full series
is also convergent.
1n
n N
a
1 1 1
N
n n nn n n N
a a a
SERIES Note 4