14 Series
What should be meant by the following expression?
∞
∑
n=1
1
n
2
= 1 +
1
4
+
1
9
+
1
16
+ ···
The ∞-symbol in the summation should lead you to suspect a role for limits. . .
Definition 14.1. Define the n
th
partial sum s
n
of a sequence (a
n
)
∞
n=m
via
s
n
:=
n
∑
k=m
a
k
= a
m
+ a
m+1
+ ···+ a
n
• The (infinite) series
28
∞
∑
n=m
a
n
is the limit lim s
n
of the sequence (s
n
) of partial sums.
• A series converges, s to ±∞ or diverges by oscillation as does the sequence (s
n
).
•
∑
a
n
converges absolutely if
∑
|
a
n
|
converges.
•
∑
a
n
converges conditionally if it converges but not absolutely (
∑
|
a
n
|
diverges to ∞).
To return to our motivating example,
∞
∑
n=1
1
n
2
= lim s
n
where s
n
=
n
∑
k=1
1
k
2
= 1 +
1
4
+ ··· +
1
n
2
We don’t (yet) know whether the series converges or diverges to ∞. . .
Theorem 14.2 (Basic Series Laws). Infinite series behave nicely with respect to addition and scalar
multiplication. For instance:
1. If
∑
a
n
is convergent and k is constant, then
∑
ka
n
= k
∑
a
n
is convergent.
2. If
∑
a
n
and
∑
b
n
are convergent, then
∑
(a
n
±b
n
) =
∑
a
n
±
∑
b
n
are also convergent.
3. If
∑
a
n
= ∞ and k > 0, then
∑
ka
n
= ∞.
4. If
∑
a
n
= ∞ and
∑
b
n
converges, then
∑
(a
n
+ b
n
) = ∞.
Proof. Simply apply the limit/divergence laws to the sequence of partial sums. E.g. for 1,
∑
ka
n
= lim
n→∞
n
∑
j=m
ka
j
finite
sum
= lim
n→∞
k
n
∑
j=m
a
j
limit
laws
= k lim
n→∞
n
∑
j=m
a
j
= k
∑
a
n
The others may be proved similarly.
Note that series do not behave nicely with respect to multiplication (see also Exercise 3):
a
1
b
1
+ a
2
b
2
+ ··· =
∑
a
n
b
n
=
∑
a
n
∑
b
n
=
a
1
+ a
2
+ ···
b
1
+ b
2
+ ···
28
It is common to denote a series
∑
a
n
if the initial term is understood (typically a
0
or a
1
), or is irrelevant to the situation.
Series which may be evaluated exactly
Our major goal is to develop techniques for answering the binary question, “Does
∑
a
n
converge?”
Even when the answer is yes, a precise computation of the limit is usually beyond us. However,
our techniques (the upcoming series tests) will typically rely on comparing
∑
a
n
to a ‘standard’ series
whose properties are completely understood. You have met two such series in elementary calculus.
Definition 14.3 (Geometric series). A sequence (a
n
) is geometric if the ratio of successive terms is
constant: a
n
= ba
n
for some constants a, b. A geometric series is the sum of a geometric sequence.
The computation of the sequence of partial sums should be familiar (for simplicity assume b = 1)
(1 − a)s
n
=
a
m
+ a
m+1
+ ··· + a
n
−
a
m+1
+ a
m+2
+ ··· + a
n
+ a
n+1
= a
m
− a
n+1
from which we quickly conclude:
Theorem 14.4. If a is constant, then
s
n
=
n
∑
k=m
a
k
=
a
m
− a
n+1
1 − a
if a = 1
n + 1 − m if a = 1
=⇒
∞
∑
n=m
a
n
converges to
a
m
1 − a
if
|
a
|
< 1
diverges to ∞ if a ≥ 1
diverges by oscillation if a ≤ −1
In particular,
∑
a
n
converges absolutely if
|
a
|
< 1 and diverges otherwise.
Examples 14.5. 1.
∞
∑
n=−1
2
−
4
5
n
= 2
−
4
5
−1
1 +
4
5
= −
5
2
·
5
9
= −
25
18
2. Consider the series
∑
a
n
=
∞
∑
n=3
2
5
n
+ 2
n
. If this were convergent, then
∑
2
n
=
∑
a
n
−
∑
2
5
n
would converge (Theorem 14.2); a contradiction.
Telescoping series A rarer type of series can be evaluated using the algebra of partial fractions.
Example 14.6. To compute
∞
∑
n=1
1
n(n+1)
, first observe that
s
n
=
n
∑
k=1
1
k(k + 1)
=
n
∑
k=1
1
k
−
1
k + 1
=
1
1
−
1
2
+
1
2
−
1
3
+ ··· +
1
n
−
1
n + 1
= 1 −
1
n + 1
It follows that
∞
∑
n=1
1
n(n + 1)
= lim
1 −
1
n + 1
= 1
Similar arguments can be made for other series such as
∑
1
n(n+2)
.
51
The Cauchy Criterion
The starting point for general series convergence uses Cauchy completeness.
Example 14.7. Consider again the series
∑
1
n
2
. We show that the sequence of partial sums (s
n
) is
Cauchy. Let ϵ > 0 be given and let N =
1
ϵ
. Then,
m > n > N =⇒
|
s
m
−s
n
|
=
m
∑
k=n+1
1
k
2
<
m
∑
k=n+1
1
k(k − 1)
=
m
∑
k=n+1
1
k − 1
−
1
k
=
1
n
−
1
m
<
1
N
= ϵ
where we follow the telescoping series approach to cancel most terms. By Cauchy completeness
(Theorem 10.11), (s
n
) converges and we conclude
The series
∑
1
n
2
is convergent
Computing the value of this series rigorously is significantly harder, though a sketch argument for
why
∑
1
n
2
=
π
2
6
is in Exercise 10.
Theorem 14.8 (Cauchy Criterion for Series). A series
∑
a
n
converges if and only if
∀ϵ > 0, ∃N such that m ≥ n > N =⇒
|
s
m
−s
n−1
|
=
m
∑
k=n
a
k
< ϵ
In the previous example we essentially verified the Cauchy criterion for the series
∑
1
n
2
.
Proof. Let (s
n
) be the sequence of partial sums. Then
∑
a
n
converges ⇐⇒ (s
n
) converges ⇐⇒ (s
n
) is Cauchy (Thm 10.11)
⇐⇒
∀ϵ > 0, ∃N such that m > n > N =⇒
|
s
m
−s
n
|
< ϵ
To finish, simply replace n with n −1.
Example 14.9. Assume, for contradiction, that the harmonic series
∑
1
n
converges. Now take ϵ =
1
2
in
the Cauchy criterion:
∃N such that m ≥ n > N =⇒
m
∑
k=n
1
k
<
1
2
However, taking m = 2(n −1) ≥ n (true since n > N ≥ 1) results in a contradiction:
1
2
>
m
∑
k=n
1
k
=
1
n
+ ··· +
1
m
≥
m − (n −1)
m
= 1 −
n −1
m
=
1
2
We conclude that the harmonic series diverges to ∞.
52
The Series Tests
For the remainder of this section we develop several standard tests for the convergence/divergence
of an infinite series: the divergence, comparison, root and ratio tests. The first of these follow quickly
from the Cauchy criterion.
Theorem 14.10 (Divergence/n
th
-term test). If lim a
n
= 0 then
∑
a
n
is divergent.
Proof. We prove the contrapositive. Suppose
∑
a
n
is convergent, let ϵ > 0 be given and take m = n in
the Cauchy criterion. Then
∃N such that n > N =⇒
|
a
n
|
< ϵ
Otherwise said, lim a
n
= 0.
Examples 14.11. 1. The series
∑
sin(
nπ
9
) diverges.
2. The divergence test tells us that the geometric series
∑
a
n
diverges whenever
|
a
|
≥ 1. We still
need our earlier analysis for when
|
a
|
< 1.
3. The converse of the n
th
-term test is false! For the canonical example, consider the divergent har-
monic series
∑
1
n
(Example 14.9, even though lim
1
n
= 0.
Theorem 14.12 (Comparison test). 1. Let
∑
b
n
be a convergent series of non-negative terms and
assume
|
a
n
|
≤ b
n
for all (large n). Then both
∑
b
n
and
∑
|
b
n
|
are convergent.
2. If
∑
a
n
= ∞ and a
n
≤ b
n
for all (large) n, then
∑
b
n
= ∞.
Proof. Suppose “large n” means n > M.
1. Let ϵ > 0 be given. Then ∃N ≥ M such that
m ≥ n > N =⇒
m
∑
k=n
a
k
△
≤
m
∑
k=n
|
a
k
|
≤
m
∑
k=n
b
k
< ϵ
2. The n
th
partial sum of
∑
b
n
is
n
∑
k=M
b
k
≥
n
∑
k=M
a
k
→ +∞
Corollary 14.13. 1. Take
|
a
n
|
= b
n
in part 1 to see that
∑
|
a
n
|
converges =⇒
∑
a
n
converges. Thus
absolute convergence implies convergence.
2. If
∑
b
n
is a convergent series of non-negative terms and
|
a
n
|
≤ b
n
for all n, then
∑
a
n
≤
∑
|
a
n
|
≤
∑
b
n
53
Examples 14.14. 1. Since
2n+1
(n+2)3
n
≤ 2 ·3
−n
and the geometric series
∑
2 ·3
−n
converges, we see that
the resulting series converges (absolutely), to some value
∞
∑
n=0
2n + 1
( n + 2)3
n
≤ 2
∞
∑
n=0
3
−n
=
2
1 −
1
3
= 3
2. One can usually find a sensible series to compare with just by thinking about how a
n
behaves
when n is very large. For instance, a
n
=
(n
2
+1)
1/2
(1+
√
n)
4
behaves like
n
n
2
=
1
n
and we see that
a
n
>
n
(1 +
√
n)
4
>
n
(2
√
n)
4
=
1
16n
Comparison with
1
16
∑
1
n
shows that
∑
a
n
diverges to ∞.
3. Since ln n < n =⇒
1
ln n
>
1
n
, we see that
∑
1
ln n
diverges to ∞ by comparison with
∑
1
n
.
4.
∑
sin n
n
2
converges absolutely by comparison to
∑
1
n
2
. Corollary 14.13 gives an estimation for the
value of the series, though it is not accurate! The n
th
partial sums satisfy
∞
∑
n=1
sin n
n
2
≤
∞
∑
n=1
|
sin n
|
n
2
≤
∞
∑
n=1
1
n
2
=
π
2
6
(approximately 1.014 ≤ 1.280 ≤ 1.645)
5. The alternating harmonic series
∞
∑
n=1
(−1)
n+1
n
converges via a sneaky comparison.
Consider the series t =
∞
∑
n=1
1
2n(2n−1)
which converges by comparison with
∑
1
4(n−1)
2
. Its n
th
par-
tial sum is
t
n
=
n
∑
k=1
1
2k(2k −1)
=
n
∑
k=1
1
2k −1
−
1
2k
which is the even partial sum of the alternating harmonic series s
2n
=
∑
2n
m=1
(−1)
m+1
m
.
Take limits of s
2n+1
= s
2n
+
1
2n+1
, we see that lim s
2n+1
= t from which lim s
n
= t. Since the
harmonic series
∑
1
n
diverges (Example 14.9), we conclude that the alternating harmonic series
converges conditionally. We will revisit this discussion in the next section.
6.
∑
n
n+1
n
2
converges by comparison with the geometric series
∑
2
−n
. To see this, note that
n
n + 1
n
=
n + 1
n
1 −
1
n + 1
n+1
−−−→
n→∞
e
−1
<
1
2
from which we see that, for all large n,
n
n + 1
n
<
1
2
=⇒
n
n + 1
n
2
< 2
−n
In fact (compare Exercise 10.10),
n
n+1
n
is monotone-down, whence e
−1
≤
n
n+1
n
≤
1
2
and
0.58198 ≈
1
e −1
=
e
−1
1 −e
−1
=
∞
∑
n=1
e
−n
≤
∞
∑
n=1
n
n + 1
n
2
≤
∞
∑
n=1
2
−n
=
1/2
1 −1/2
= 1
A computer estimate yields
∞
∑
n=1
n
n+1
n
2
≈ 0.8174.
54
Our final two tests in this section are less powerful, but have the advantage of being easier to use.
Theorem 14.15 (Root test). Let lim sup
|
a
n
|
1/n
= L.
1. If L < 1, then
∑
a
n
converges absolutely.
2. If L > 1, then
∑
a
n
diverges.
If L = 1, then no conclusion can be drawn.
We defer the proof until after seeing some examples.
Corollary 14.16 (Ratio test). Suppose (a
n
) is a sequence of non-zero terms.
1. If lim sup
a
n+1
a
n
< 1, then
∑
a
n
converges absolutely
2. If lim inf
a
n+1
a
n
> 1, then
∑
a
n
diverges
Proof. This follows directly from the root test and Theorem 12.3:
lim inf
a
n+1
a
n
≤ lim inf
|
a
n
|
1/n
≤ lim sup
|
a
n
|
1/n
≤ lim sup
a
n+1
a
n
The versions of these tests familiar from elementary calculus are the special cases when
L = lim
|
a
n
|
1/n
= lim
a
n+1
a
n
Our versions are more general since these limits are not guaranteed to exist.
Examples 14.17. 1. The ratio test is particularly useful for series involving factorials and exponentials.
(a)
∑
n
4
2
n
converges: just observe that lim
a
n+1
a
n
= lim
(n+1)
4
2n
4
=
1
2
< 1.
(b)
∑
n!
2
n
diverges: in this case lim
a
n+1
a
n
= lim
(n+1)!
2n!
= lim
n+1
2
= ∞.
2. Both tests are inconclusive for rational sequences: if a
n
=
b
n
c
n
where b
n
, c
n
are polynomials, then
lim
a
n+1
a
n
= 1 = lim
|
a
n
|
1/n
For example,
∑
n + 5
n
2
⇝ lim
a
n+1
a
n
= lim
( n + 6)n
2
( n + 5)(n + 1)
2
= 1
In fact this example is divergent by comparison with the harmonic series
∑
1
n
.
3. Recall Example 14.14.6: our use of the comparison test was really the root test in disguise
a
n
=
n
n + 1
n
2
=⇒ lim
|
a
n
|
1/n
= lim
n
n + 1
n
= e
−1
< 1 =⇒
∑
a
n
converges
In this case the root test was much easier to apply!
55
4. The ratio test is the weakest test thus far; certainly it does not apply if any of the terms a
n
are
zero! For a more subtle example, consider:
a
n
=
(
2
−n
if n is even
3
−n
if n is odd
First we try applying the ratio test:
a
n+1
a
n
=
(
1
3
2
3
n
if n is even
1
2
3
2
n
if n is odd
=⇒ lim inf
a
n+1
a
n
= 0, lim sup
a
n+1
a
n
= ∞
The ratio test is therefore inconclusive. However
|
a
n
|
1/n
=
(
1
2
if n is even
1
3
if n is odd
=⇒ lim sup
|
a
n
|
1/n
=
1
2
< 1
By the root test, the series
∑
a
n
converges. We need not even have used the root test:
∑
a
n
converges by comparison with
∑
2
−n
!
Proof of the Root Test. 1. Suppose lim sup
|
a
n
|
1/n
= L < 1. Recall that v
N
= sup{
|
a
n
|
1/n
: n > N}
defines a monotone-down sequence converging to L. Choose any ϵ > 0 such that L + ϵ < 1 (say
ϵ =
1−L
2
) to see that
∃N such that v
N
− L < ϵ
But then
n > N =⇒
|
a
n
|
1/n
− L < ϵ =⇒
|
a
n
|
< (L + ϵ)
n
∑
|
a
n
|
therefore converges by comparison with the geometric series
∑
(L + ϵ)
n
.
2. If L > 1 then there exists some subsequence (a
n
k
) such that
|
a
n
k
|
1/n
k
→ L > 1. In particular,
infinitely many terms of this subsequence must be greater than 1. Clearly a
n
does not converge
to zero whence the series diverges by the n
th
term test.
Summary The logical flow of the tests in this section is as follows:
(divergence tests) n
th
term
+3
Root
+3
Ratio
(testing both) Definition of
∑
a
n
ks +3
Cauchy Criterion
KS
+3
Comparison
(convergence tests) Root
+3
Ratio
The ratio test is typically the easiest to use, but the least powerful. Every series which converges by
the ratio test can be seen to converge by the root and comparison tests, etc. If a series diverges by the
ratio test, then it in fact diverges by the n
th
term test.
56
Exercises 14. 1. Determine which of the following sequences converge. Justify your answers.
(a)
∑
n−1
n
2
(b)
∑
( −1)
n
(c)
∑
3
n
n
3
(d)
∑
n
3
3
n
(e)
∑
n
2
n!
(f)
∑
1
n
n
(g)
∑
n
2
n
(h)
∑
n!
n
n
(i)
∞
∑
n=2
1
n+(−1)
n
2
(j)
∑
√
n + 1 −
√
n
2. Let
∑
a
n
and
∑
b
n
be convergent series of non-negative terms. Prove that
∑
√
a
n
b
n
converges.
(Hint: start by showing that
√
a
n
b
n
≤ a
n
+ b
n
)
3. (a) If
∑
a
n
converges absolutely, prove that
∑
a
2
n
converges.
(b) More generally, if
∑
|
a
n
|
converges and (b
n
) is a bounded sequence, prove that
∑
a
n
b
n
converges absolutely.
4. Find a series
∑
a
n
which diverges by the root test but for which the ratio test is inconclusive.
5. (Hard) Let (a
n
) be a sequence such that lim inf
|
a
n
|
= 0. Prove that there is a subsequence (a
n
k
)
such that
∑
a
n
k
converges.
(Hint: Try to construct a subsequence which converges to zero faster than
1
k
2
.
6. Prove that the harmonic series
∑
1
n
diverges by comparing with the series
∑
a
n
, where
(a
n
) =
1,
1
2
,
1
4
,
1
4
,
1
8
,
1
8
,
1
8
,
1
8
,
1
16
,
1
16
,
1
16
,
1
16
,
1
16
,
1
16
,
1
16
,
1
16
,
1
32
,
1
32
, . . .
7. Suppose b
n
≤ a
n
for all n and that
∑
b
n
and
∑
a
n
converge. Prove that
∑
b
n
≤
∑
a
n
.
(This also proves part 2 of Corollary 14.13)
8. Use the basic series laws to find the values of
∑
1
(2n)
2
,
∑
1
(2n+1)
2
and
∑
(−1)
n+1
n
2
.
9. The limit comparison test states:
Suppose
∑
a
n
,
∑
b
n
are series of positive terms and that L = lim
a
n
b
n
∈ ( 0, ∞). Then the
series have the same convergence status (both converge or both diverge to ∞).
(a) Use the limit comparison test with b
n
=
1
n
2
to show that the series
∑
1
n
ln
1 +
1
n
converges.
(Hint: Recall that e = lim
1 +
1
n
n
)
(b) Prove the limit comparison test.
(Hint: first show that
L
2
<
a
n
b
n
<
3L
2
for large n)
(c) What can you say about the series
∑
a
n
and
∑
b
n
if L = 0 or L = ∞? Explain.
10. Euler asserted that the sine function, written as an infinite polynomial in the form of a Maclau-
rin series, could also be expressed as an infinite product,
sin x =
∞
∑
n=0
( −1)
n
(2n + 1)!
x
2n+1
= x
1 −
x
2
π
2
1 −
x
2
4π
2
1 −
x
2
9π
2
···
By considering the solutions to sin x = 0, give some weight to Euler’s claim. By comparing
coefficients in these expressions, deduce the fact
∑
1
n
2
=
π
2
6
.
(As we’ve presented it, this argument is non-rigorous!)
57
15 The Integral and Alternating Series Tests
In this section we develop two further standalone series tests, both with narrower applications than
our previous tests.
The first a little out of place given that it requires (improper) integration.
29
Theorem 15.1 (Integral test). Let a
n
= f (n), where f is non-
negative, non-increasing and integrable on [1, ∞). Then
∞
∑
n=1
a
n
converges ⇐⇒
Z
∞
1
f (x) dx converges
in which case
Z
∞
1
f (x) dx ≤
∞
∑
n=1
a
n
≤ a
1
+
Z
∞
1
f (x) dx
The statement is easily modified if the initial term is a
m
.
0
f (x)
0 1 2 3 4 5 6
x
n
a
1
a
2
a
3
a
4
a
5
Proof. We need only interpret the picture:
Z
n+1
1
f (x) dx ≤
n
∑
k=1
a
k
= s
n
= a
1
+
n
∑
k=2
a
k
≤ a
1
+
Z
n
1
f (x) dx (∗)
Taking limits gives the result.
Even for divergent sums, (∗) allows us to estimate s
n
and analyze its rate of growth. For greater accu-
racy, explicitly evaluate the first few terms and use a modified integral test to estimate the remainder.
The big application of the integral test is a complete description of the convergence status of p-series,
another useful collection of series to which others may be compared.
Corollary 15.2 (p-series). Let p > 0. The series
∑
1
n
p
converges if and only if p > 1.
Examples 15.3. 1.
∑
1
n
3
converges (it is a p-series with p > 1). For a simple estimate, observe that
Z
∞
1
1
x
3
dx = lim
b→∞
−
1
2
x
−2
b
1
=
1
2
=⇒
1
2
≤
∞
∑
n=1
1
n
3
≤
3
2
This is a poor estimate, especially the lower bound. For a quick improvement, we could explic-
itly evaluate the first term and re-run the test starting at n = 2:
1 +
Z
∞
2
1
x
3
dx ≤
∞
∑
n=1
1
n
3
≤ 1 +
1
8
+
Z
∞
2
1
x
3
dx =⇒ 1 +
1
8
≤
∞
∑
n=1
1
n
3
≤ 1 +
1
4
If greater accuracy is required, more terms can be explicitly evaluated.
29
Which in turn requires limits of functions:
R
∞
1
f (x) dx := lim
b→∞
R
b
1
f (x) dx. Even though we haven’t developed these
concepts, the relevant computations should be familiar from elementary calculus.
58
2. In Example 14.9, we used the Cauchy criterion to show that the harmonic series diverges to ∞.
The integral test makes this much easier! The integral test also allows us to estimate how many
terms are required for the partial sum to s
n
to reach a certain threshold, say 10. Since
ln(n + 1) =
Z
n+1
1
1
x
dx ≤
n
∑
k=1
1
k
≤ 1 +
Z
n
1
1
x
dx = 1 + ln n
we see that
s
n
≈ 10 =⇒ ln(n + 1) ≤ 10 ≤ 1 + ln n =⇒ e
9
≤ n ≤ e
10
−1 =⇒ 8104 ≤ n ≤ 22025
Somewhere between 8 and 22 thousand terms are required! The harmonic series diverges to
infinity, but it does so very slowly.
3. The integral test shows that
∑
∞
n=2
1
n ln n
= ∞ and moreover that, to exceed 10, somewhere be-
tween 10
3223
and 10
6631
terms are required!
4. The series
∑
2n+1
√
4n
3
−1
diverges to ∞ by comparison with the p-series
∑
1
√
n
.
Alternating Series and Conditional Convergence
Our final test is the only one capable of detecting conditional
convergence, the canonical example of which is the alternat-
ing harmonic series (recall Example 14.14.5). With an eye on
generalization, we re-index so that the first term is a
0
= 1:
s =
∞
∑
n=0
( −1)
n
n + 1
= 1 −
1
2
+
1
3
−
1
4
+ ···
=
∞
∑
n=0
( −1)
n
a
n
= a
0
− a
1
+ a
2
− a
3
+ ···
0
10 20
s
n
(−1)
n
a
n
s
+
n
s
−
n
The alternating ±-signs give the series its name. For us, however, the behavior of the sequence (s
n
)
of partial sums is more interesting. Consider two subsequences (s
+
n
) = (s
2n
) and (s
−
n
) = (s
2n−1
):
s
+
n
=
2n
∑
k=0
( −1)
k
a
k
= 1 −
1
2
−
1
3
|{z}
a
1
−a
2
−
1
2
−
1
3
|{z}
a
3
−a
4
−··· −
1
2n
−
1
2n + 1
| {z }
a
2n−1
−a
2n
(n ≥ 0)
s
−
n
=
2n−1
∑
k=0
( −1)
k
a
k
=
1 −
1
2
|{z}
a
0
−a
1
+
1
3
−
1
4
|{z}
a
2
−a
3
+ ··· +
1
2n −1
−
1
2n
| {z }
a
2n−2
−a
2n−1
(n ≥ 1)
Since the brackets are non-negative, (s
+
n
) is monotone-down and (s
−
n
) monotone-up. Moreover,
1
2
= s
−
1
≤ s
−
n
≤ s
−
n
+ a
2n
= s
+
n
≤ s
+
0
= 1 (†)
from which both subsequences are bounded and thus convergent. Not only this, but
lim
s
+
n
−s
−
n
= lim a
2n
= 0
shows that the limits of both subsequences are identical (of course both are s).
59
The above discussion depended only on two simple properties of the sequence (a
n
); we’ve therefore
proved a general statement.
Theorem 15.4 (Alternating series test). Let (a
n
) be monotone-down with lim a
n
= 0.
1. The series
∑
( −1)
n
a
n
converges.
2. If (s
n
) is the sequence of partial sums converging to s =
∑
( −1)
n
a
n
, then
|
s − s
n
|
≤ a
n+1
.
Think about where our assumptions on (a
n
) are used in the proof. It can, in fact, be shown that
the alternating harmonic series converges to ln 2, although the estimates provided by the alternating
series test make this a terrible method of approximation. Even summing the first 100 terms only
results in 2 decimal places of accuracy!
Examples 15.5. 1. Since a
n
=
1
n!
converges monotone-down to zero, the alternating series
∑
(−1)
n+1
n!
converges. By taking the first 9 and 10 terms of this series, we see that
0.9010498898 . . . ≤
∞
∑
n=1
( −1)
n+1
n!
≤ 0.90105016538 . . .
which at least yields the estimate 0.9015 to 5 decimal places. The alternating series test is not
required for this example, since it in fact converges absolutely.
2. The series
∑
∞
n=2
sin
π
2
n
ln n
can be viewed as an alternating series since every even term is zero. Writ-
ing m = 2n + 1, we obtain
∞
∑
n=2
sin
π
2
n
ln n
=
∞
∑
m=1
sin(πm +
π
2
)
ln( 2m + 1)
=
∞
∑
m=1
( −1)
m
ln( 2m + 1)
Since
1
ln(2m+1)
decreases to zero, the alternating series test shows convergence.
Rearranging Infinite Series
A rearrangement of an infinite series
∑
a
n
arises when we change the order of the terms of the sequence
(a
n
) before computing the sequence of partial sums. We still have to use every term a
n
in the new
series. Since the resulting sequence of partials sums is likely completely different, we shouldn’t
assume that the new series has the same convergence properties as the old.
Example 15.6. We rearrange the alternating harmonic series by summing two positive terms before
each negative term:
1 +
1
3
−
1
2
+
1
5
+
1
7
−
1
4
+
1
9
+
1
11
−
1
6
+ ··· +
1
4n −3
+
1
4n −1
−
1
2n
+ ···
Every term in the original sequence is used here, so this is a genuine rearrangement. It is perhaps
surprising to discover that the new series converges, though its limit is not the same as the original
alternating harmonic series! We leave the details to Exercise 9.
60
This behavior is quite different to that of finite sums, where the order of summation makes no differ-
ence at all. The situation can be summarized in a famous result of Riemann.
Theorem 15.7 (Riemann rearrangement). 1. If
∑
a
n
is conditionally convergent and s ∈ R ∪ {±∞},
then there exists a rearrangement of the series which tends
30
to s.
2. If
∑
a
n
converges absolutely, then all rearrangements converge to the same limit.
The second part says that absolutely convergent series behave just like finite sums! We omit the
proofs since they are lengthy and require nasty notation. Instead we illustrate the rough idea of part
1 via an example.
Example 15.8. We show how to construct a rearrangement of the alternating harmonic series which
converges to s =
√
2 −1 = 0.41421 . . .
First we convince ourselves that the sum of the positive terms
∑
a
+
n
diverges to infinity. In this case
the comparison test comes to our rescue:
1
2n −1
>
1
2n
=⇒
∑
a
+
n
=
∑
1
2n −1
>
1
2
∑
1
n
= ∞
The negative terms also diverge
∑
a
−
n
= −∞. Construction of the rearrangement is inductive.
1. Sum just enough positive terms until the partial sum exceeds s: plainly S
1
= 1 will do.
2. Sum negative terms starting at the beginning of the sequence until the sum is less than s:
S
2
= 1 −
1
2
−
1
4
= 0.25 < s
3. Repeat: add positive terms until the sum just exceeds s, then add negative terms, etc.,
S
3
= S
2
+
1
3
= 0.583 . . . > s, S
4
= S
3
−
1
6
−
1
8
= 0.291 . . . < s
Continuing the process ad infinitum, we claim that
s = 1 −
1
2
−
1
4
+
1
3
−
1
6
−
1
8
+
1
5
−
1
10
+
1
7
−
1
12
−
1
14
+
1
9
−
1
16
−
1
18
+
1
11
−
1
20
+
1
13
−···
To see why, observe:
• Since
∑
a
+
n
= ∞ and
∑
a
−
n
= −∞, at each stage we only add/subtract finitely many terms.
• All terms of the original sequence (a
n
) are eventually used since we add the positive (negative)
terms in order. E.g., a
495
=
1
495
appears, at the latest, during the 495
th
positive-addition phase.
•
|
S
n
−s
|
≤
|
a
m
n
|
, where a
m
n
is the last term used at the n
th
stage. This plainly converges to zero,
whence lim S
n
= s.
30
Riemann’s result is in fact even stronger. Rearrangements also exist which diverge by oscillation between any given
lim inf and lim sup!
61
Exercises 15. 1. Use the integral test to determine whether the series
∞
∑
n=1
1
n
2
+1
converges or diverges.
2. Prove Corollary 15.2 regarding the convergence/divergence of p-series.
3. Let s
n
=
n
∑
k=1
1
√
k
. Estimate how many terms are required before s
n
≥ 100.
4. (Example 15.3.3) Verify the claim that
∞
∑
n=2
1
n ln n
= ∞. If you want a challenge, verify the estimate
claim also.
5. (a) Give an example of a series
∑
a
n
which converges, but for which
∑
a
2
n
diverges.
(Exercise 14.3 really requires that
∑
a
n
be absolutely convergent!)
(b) Give an example of a divergent series
∑
b
n
for which
∑
b
2
n
converges.
6. Suppose (a
n
) satisfies the hypotheses of the alternating series test except that lim a
n
= a > 0.
What can you say about the sequences (s
+
n
) and (s
−
n
) and the series
∑
( −1)
n
a
n
?
7. We know that the harmonic series has a growth rate comparable to ln n. Let a
n
=
1
n
and define
a new sequence (t
n
) by
t
n
= s
n
−ln n = 1 +
1
2
+ ··· +
1
n
−ln n
where s
n
=
n
∑
k=1
a
n
is the n
th
partial sum. Prove that (t
n
) is a positive, monotone-down sequence,
which therefore converges.
31
(Hint: you’ll need the mean value theorem from elementary calculus)
8. (a) Show that the series
∞
∑
n=1
(−1)
n
n
n
2
+1
is conditionally convergent to some real number s.
(b) How many terms are required for the partial sum s
n
to approximate s to within 0.01.
(c) Following Example 15.8, use a calculator to state the first 15 terms in a rearrangement of
the series in part (a) which converges to 0.
9. In Example 15.6 we rearranged the terms of the alternating harmonic series by taking two pos-
itive terms before each negative term.
(a) Verify, for each n ∈ N, that
b
n
:=
1
4n −3
+
1
4n −1
−
1
2n
> 0
whence the subsequence of partial sums (s
3n
) is monotone-up.
(b) Use the comparison test to show that
∑
b
n
converges.
(c) Prove that the rearranged series converges, to some value s >
5
6
.
(Thus s > ln 2 ≈ 0.69, the limit of the original alternating harmonic series)
31
The limit γ := lim t
n
≈ 0.5772 is the Euler–Mascheroni constant. It appears in many mathematical identities, and yet
very little about it is known; it is not even known whether γ is irrational!
62
