3 Series
3.14 Infinite Series and the Series Tests
For millennia, certainly since Zeno’s paradoxes of c. 430 BC, mathematicians have been interested in
the meaning and evaluation of infinite sums such as
∞
∑
n=1
1
n
2
= 1 +
1
4
+
1
9
+
1
16
+ ···
The standard approach in modern mathematics is to outsource the definition to that of limits.
Definition 3.1. The n
th
partial sum s
n
of a sequence (a
n
)
∞
n=m
is the finite sum
s
n
:=
n
∑
k=m
a
k
= a
m
+ a
m+1
+ ···+ a
n
• The (infinite) series
23
∞
∑
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 ∞).
We don’t (yet) know whether our motivating example converges, but at least we have a meaning:
∞
∑
n=1
1
n
2
= lim s
n
where s
n
=
n
∑
k=1
1
k
2
= 1 +
1
4
+ ···+
1
n
2
Theorem 3.2 (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.
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
+ ···
23
If the initial term is understood or is irrelevant to the situation, it is common to write
∑
a
n
.
Series which may be evaluated exactly
Given a series
∑
a
n
, our primary goal is to answer a simple question: “Does it converge?” Even when
the answer is yes, a precise computation of the limit will usually be beyond us. We instead develop
techniques (the upcoming series tests) which typically rely on comparing
∑
a
n
to some ‘standard’
series whose properties are completely understood: in particular. . .
Definition 3.3 (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 3.4. Suppose 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 3.5. 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 3.2); a contradiction.
Telescoping series A rarer type of series can be evaluated using the algebra of partial fractions.
Example 3.6. 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)
.
48
The Cauchy Criterion
The starting point for our general series tests uses Cauchy completeness.
Example 3.7. Consider again the series
∑
1
n
2
. We show that the sequence of partial sums (s
n
) is
Cauchy. Suppose ϵ > 0 is 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 most terms cancel analogous to the telescoping series approach. By Cauchy completeness
(Theorem 2.34), (s
n
) converges and we conclude
∑
1
n
2
converges
Computing the value of this series is significantly harder: a sketch argument for why
∑
∞
n=1
1
n
2
=
π
2
6
is in Exercise 10.
Theorem 3.8 (Cauchy criterion for series). A series
∑
a
n
converges precisely when
∀ϵ > 0, ∃N such that m > n > N =⇒
|
s
m
−s
n
|
=
m
∑
k=n+1
a
k
< ϵ
The previous example essentially verified the Cauchy criterion for
∑
1
n
2
.
Proof. Let (s
n
) be the sequence of partial sums. Then
∑
a
n
converges ⇐⇒ ( s
n
) converges
⇐⇒ ( s
n
) is a Cauchy sequence (Theorem 2.34)
⇐⇒
∀ϵ > 0, ∃N such that m > n > N =⇒
|
s
m
−s
n
|
< ϵ
Example 3.9. For contradiction, suppose that the harmonic series
∑
1
n
converges. Take ϵ =
1
2
in the
Cauchy criterion to observe that
∃N such that m > n > N =⇒
m
∑
k=n+1
1
k
<
1
2
However, taking m = 2n (plainly m > n since n > N ≥ 1) results in a contradiction:
1
2
>
m
∑
k=n+1
1
k
=
1
n + 1
+ ···+
1
m
≥
m −n
m
= 1 −
n
m
=
1
2
We conclude that the harmonic series diverges to ∞.
49
The Series Tests
For the remainder of this section we develop tests for the convergence/divergence of an infinite
series: the n
th
-term, comparison, root and ratio tests. The first follows quickly from the Cauchy
criterion.
Theorem 3.10 (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, and that ϵ > 0 is given. Take
m = n + 1 in the Cauchy criterion. Then
∃
˜
N such that m >
˜
N =⇒
|
a
m
|
< ϵ (let
˜
N = N + 1)
Otherwise said, lim a
n
= 0.
Examples 3.11. 1. The series
∑
sin(
nπ
9
) diverges.
2. The n
th
-term test tells us that the geometric series
∑
a
n
diverges whenever
|
a
|
≥ 1. We still need
our earlier analysis (Theorem 3.4) for when
|
a
|
< 1.
3. The converse of the n
th
-term test is false! Example 3.9 provides the canonical example: the
divergent harmonic series
∑
1
n
also satisfies lim
1
n
= 0.
Theorem 3.12 (Comparison test). 1. Let
∑
b
n
be a convergent series of non-negative terms and as-
sume
|
a
n
|
≤ b
n
for all (large) n. Then both
∑
a
n
and
∑
|
a
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 for some fixed M.
1. Let ϵ > 0 be given. Since
∑
b
n
converges, ∃N ≥ M such that
m > n > N =⇒
m
∑
k=n+1
a
k
△
≤
m
∑
k=n+1
|
a
k
|
≤
m
∑
k=n+1
b
k
< ϵ
2. The n
th
partial sum (post M) of
∑
b
n
is
n
∑
k=M+1
b
k
≥
n
∑
k=M+1
a
k
→ ∞
Corollary 3.13. 1. (Absolute convergence implies convergence) Take
|
a
n
|
= b
n
in part 1 to see that
∑
|
a
n
|
convergent =⇒
∑
a
n
convergent.
2. (Estimation of series) Suppose
∑
b
n
is a convergent series of non-negative terms and that
|
a
n
|
≤ b
n
for all n. Then
∑
a
n
≤
∑
|
a
n
|
≤
∑
b
n
50
Examples 3.14. 1. Since the geometric series
∑
2
3
n
converges,
2n+1
(n+2)3
n
≤
2
3
n
, we see that
∞
∑
n=0
2n + 1
( n + 2)3
n
≤ 2
∞
∑
n=0
3
−n
=
2
1 −
1
3
= 3
That is, the first series converges (absolutely) to some value ≤ 3.
2. One can sometimes find a sensible comparison series by considering how a
n
behaves for large
n. For instance, when n is large, a
n
=
(n
2
+1)
1/2
(1+
√
n)
4
behaves like
n
n
2
=
1
n
. Indeed, when n ≥ 2,
a
n
>
n
(1 +
√
n)
4
>
n
(2
√
n)
4
=
1
16n
Comparison with the divergent series
1
16
∑
1
n
shows that
∑
a
n
also 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
(Example 3.7). Corollary 3.13 estimates its
value (≤
π
2
6
):
∞
∑
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 s =
∑
∞
n=1
(−1)
n+1
n
converges via a sneaky comparison.
The series t =
∑
∞
n=1
1
2n(2n−1)
converges by comparison with
∑
1
4(n−1)
2
. Its n
th
partial sum
t
n
=
n
∑
k=1
1
2k(2k −1)
=
n
∑
k=1
1
2k −1
−
1
2k
is precisely the even partial sum of the alternating harmonic series s
2n
=
∑
2n
k=1
(−1)
k+ 1
k
.
Plainly lim s
2n
= t. Moreover s
2n+1
= s
2n
+
1
2n+1
=⇒ lim s
2n+1
= t =⇒ lim s
n
= t. Since the
harmonic series
∑
1
n
diverges (Example 3.9), we conclude that the alternating harmonic series
converges conditionally. We’ll revisit this discussion in the next section.
6.
∑
n
n+1
n
2
converges by comparison with
∑
2
−n
. To see this, recall Exercise 2.10.10:
n
n + 1
n
=
n + 1
n
1 −
1
n + 1
n+1
−−−→
n→∞
e
−1
Plainly e
−1
<
1
2
, whence for large n,
n
n + 1
n
≤
1
2
=⇒
n
n + 1
n
2
≤ 2
−n
In fact
n
n+1
n
is monotone-down, whence e
−1
≤
n
n+1
n
≤
1
2
for all n, and so
0.58198 ≈
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.
51
Our last two tests in this section are less powerful but often easier to use.
Theorem 3.15 (Root test). Suppose 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 some examples. By combining with the inequalities of Theorem 2.51
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
we obtain a second familiar test.
Corollary 3.16 (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.
In elementary calculus you likely saw the special cases when
L = lim
|
a
n
|
1/n
= lim
a
n+1
a
n
Our versions are more general since these limits might not exist.
Examples 3.17. 1. The ratio test is particularly useful for series involving factorials and exponentials.
(a)
∑
n
4
2
n
converges, since lim
a
n+1
a
n
= lim
(n+1)
4
2n
4
=
1
2
< 1.
(b)
∑
n!
2
n
diverges, since 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, attempting to apply the ratio test to
∑
n+5
n
2
results in
lim
a
n+1
a
n
= lim
( n + 6)n
2
( n + 5)(n + 1)
2
= 1
This series is divergent by comparison with the harmonic series
∑
1
n
.
3. In Example 3.14.6, 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.
52
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 of its failure, consider
a
n
=
(
2
−n
if n is even
3
−n
if n is odd
=⇒
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, applying the root test it almost trivial!
|
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 =⇒
∑
a
n
converges
We need not even have used the root test:
∑
a
n
plainly converges by comparison with
∑
2
−n
!
For a precise value, note that the sequence of n
th
partial sums converges monotone up to the
sum of two geometric series,
∞
∑
n=0
a
n
=
∞
∑
k=0
2
−2k
+ 3
−2k−1
=
1
1 −1/4
+
1/3
1 −1/9
=
11
8
Proof of the Root Test. 1. Suppose L < 1. Choose any ϵ > 0 such that L + ϵ < 1 (say ϵ =
1−L
2
). Since
v
N
= sup
|
a
n
|
1/n
: n ≥ N
defines a monotone-down sequence converging to L, we 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 and so (a
n
) does not converge
to zero.
∑
a
n
thus diverges by the n
th
-term test.
Summary The logical flow of the tests in this section is as follows:
(divergence testing) Comparison n
th
-term
+3
Root
+3
Ratio
Definition of
∑
a
n
ks +3
KS
Cauchy criterion
KS
(convergence testing) Comparison
+3
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, and the Cauchy criterion. If
you find that a series diverges by the ratio test, you could have just used the n
th
-term test!
53
Exercises 3.14. Key concepts: Infinite series, Cauchy criterion, Comparison/root/ratio tests
1. Determine which of the following series 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
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. Suppose 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 3.13)
8. Given
∑
∞
n=1
1
n
2
=
π
2
6
, 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 presented, this argument is non-rigorous!)
54
3.15 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 is a little out of place given that it requires (improper) integration.
24
Theorem 3.18 (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 such a situation,
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 not a
1
.
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 as describing upper and lower Riemann sums:
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 (∗)
Take limits as n → ∞ for the result.
Even for divergent sums, (∗) allows us to estimate the growth of (s
n
). For greater accuracy, we may
evaluate the first few terms explicitly and modify the integral test to estimate the remainder.
An important application of the integral test is to provide a complete description of the convergence
status of p-series: a useful family of series to which others may be compared.
Corollary 3.19 (p-series). Let p > 0. The series
∑
1
n
p
converges if and only if p > 1.
Examples 3.20. 1.
∑
1
n
3
converges (it is a p-series with p > 1). Moreover,
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, particularly the lower bound. For a quick improvement, 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.
24
Which in turn requires limits of functions:
R
∞
1
f (x) dx := lim
b→∞
R
b
1
f (x) dx. While we haven’t rigorously developed
these concepts, the relevant computations should be familiar from elementary calculus.
55
2. In Example 3.9, we used the Cauchy criterion to show that the harmonic series diverges to ∞.
The integral test makes this much easier and allows us to estimate how many terms are required
for the partial sum to s
n
to reach a certain threshold: 10 say. Since
ln(n + 1) =
Z
n+1
1
1
x
dx ≤ s
n
=
n
∑
k=1
1
k
≤ 1 +
Z
n
1
1
x
dx = 1 + ln n
we see that s
n
≈ 10 requires
ln(n + 1) ≤ 10 ≤ 1 + ln n =⇒ e
9
≤ n ≤ e
10
−1 =⇒ 8104 ≤ n ≤ 22025
The harmonic series diverges to infinity, but it does so very slowly.
3. The integral test shows that
∑
∞
n=2
1
n ln n
= ∞. To exceed 10, somewhere between 10
3223
and 10
6631
terms are required. To exceed 100 requires ‘roughly’ 10
6×10
42
terms: 1 followed by 1000 zeros
for each water molecule in Lake Tahoe puts you at least in the right ballpark. . .
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 unique in that it can detect conditional conver-
gence. The canonical example is the alternating harmonic series
(Example 3.14.5). 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. Consider the behavior of the sequence of partial
sums (s
n
), in particular 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)
Each bracketed term is non-negative, so (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).
56
The above discussion depends only on two simple properties of the sequence (a
n
). We’ve therefore
proved a general statement.
Theorem 3.21 (Alternating series test). Suppose (a
n
) is monotone-down and that lim a
n
= 0. Then:
1. The alternating series s =
∑
( −1)
n
a
n
converges.
2. If (s
n
) is the sequence of partial sums, then
|
s −s
n
|
≤ a
n+1
.
Think about where the hypotheses regarding (a
n
) are used in the proof.
It can be shown that the alternating harmonic series converges to ln 2, though the estimates provided
by the alternating series test are very poor: to guarantee accuracy to two decimal place requires us to
sum 100 terms of the series!
Examples 3.22. 1. Since a
n
=
1
n!
converges monotone-down to zero, the alternating series
∑
(−1)
n
n!
converges. By evaluating s
8
and s
9
explicitly, we see that
0.3678791887 . . . ≤
∞
∑
n=0
( −1)
n
n!
≤ 0.3678819444 . . .
yielding an estimate of 0.36788 to 5 decimal places (the exact value is in fact e
−1
). The alternating
series test is only needed for the estimate, since the series 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 demonstrates convergence.
Rearranging Infinite Series
A rearrangement of an infinite series
∑
a
n
is a series that results from changing the order of the terms
of the sequence (a
n
) before computing the partial sums. The new series must still use every term of
the original. Since the new sequence of partials sums is likely different, we shouldn’t assume that the
rearranged series has the same convergence properties as the old.
Example 3.23. 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 of 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 11. This behavior is quite different to
that of finite sums, where the order of summation makes no difference at all.
57
The general situation is summarized in a famous result of Riemann. The first part says that absolutely
convergent series behave just like finite sums. Conditionally convergent series are much stranger.
25
Theorem 3.24 (Riemann rearrangement). 1. If a series converges absolutely, then all rearrange-
ments converge to the same limit.
2. If a series converges conditionally and s ∈ R ∪{±∞} is given, then there exists a rearrangement
which tends to s.
We omit the proofs since they are prohibitively lengthy. Instead we illustrate the rough idea of part 2
via an example.
Example 3.25. We show how to construct a rearrangement of the alternating harmonic series which
converges to s =
√
2 = 1.41421 . . .
First we convince ourselves that the sum of the positive terms
∑
a
+
n
diverges to infinity. The compar-
ison test makes this easy:
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 S
1
= a
+
1
+ a
+
2
+ ··· + a
+
n
1
in order until the partial sum exceeds
s: plainly S
1
= 1 +
1
3
+
1
5
≈ 1.53333 will do here.
2. Add negative terms starting at the beginning of the sequence until the sum is less than s:
S
2
= S
1
+ (a
−
1
+ a
−
2
+ ··· + a
−
m
1
) = 1 +
1
3
+
1
5
−
1
2
= 1.0333 ··· < s
3. Repeat: add positive terms until the sum just exceeds s, then add negative terms, etc.,
S
3
= S
2
+
1
7
+
1
9
+
1
11
+
1
13
= 1.4551 . . . > s, S
4
= S
3
−
1
4
= 1.2051 . . . < s
Continuing the process ad infinitum, we claim that
s =
√
2 = 1 +
1
3
+
1
5
−
1
2
+
1
7
+
1
9
+
1
11
+
1
13
−
1
4
+
1
15
+
1
17
+
1
19
+
1
21
−
1
6
+
1
23
+
1
25
+ ···
To see why, observe:
• Since
∑
a
+
n
= ∞ and
∑
a
−
n
= −∞, at each stage we need 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 final term used at the n
th
stage. The right hand side converges
to zero (n
th
-term test!), whence lim S
n
= s.
25
Riemann’s second result is in fact even stronger. Conditionally convergent series also have rearrangements whose
sequence of partial sums diverges by oscillation to any given lim inf s
n
< lim sup s
n
!
58
Exercises 3.15. Key concepts: Integral test and approximation, Alternating series and approximation
1. Use the integral test to determine whether the series
∑
∞
n=1
1
n
2
+1
converges or diverges.
2. Prove Corollary 3.19 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 3.20.3) Verify the claim that
∑
∞
n=2
1
n ln n
= ∞ and the claim regarding the estimate.
5. (a) Use calculus to show that a
n
=
ln n
n
2
is monotone-down whenever n ≥ 2.
(b) Show that lim a
n
= 0, and that the hypotheses of the integral test are therefore satisfied.
(c) Determine whether the series
∑
∞
n=2
ln n
n
2
converges or diverges.
6. (a) Give an example of a series
∑
a
n
which converges, but for which
∑
a
2
n
diverges.
(Exercise 3.14.3 really requires that
∑
a
n
be absolutely convergent!)
(b) Give an example of a divergent series
∑
b
n
for which
∑
b
2
n
converges.
7. Suppose (a
n
) satisfies the hypotheses of the alternating series test except that lim a
n
= a is
strictly positive. What can you say about the sequences (s
+
n
) and (s
−
n
) and the series
∑
( −1)
n
a
n
?
8. Let a
n
=
1
n
have partial sum s
n
=
∑
n
k=1
a
n
, and define a new sequence (t
n
) by
t
n
= s
n
−ln n = 1 +
1
2
+ ··· +
1
n
−ln n
Prove that (t
n
) is a positive, monotone-down sequence, which therefore converges.
26
(Hint: You’ll need the mean value theorem from elementary calculus)
9. Suppose
∑
a
n
is conditionally convergent and let
∑
a
+
n
be the series obtained by summing, in
order, the positive terms of the sequence (a
n
). Prove that
∑
a
+
n
= ∞.
10. (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 3.25, use a calculator to state the first twelve terms in a rearrangement
of the series in part (a) which converges to 0.
11. Recall the rearrangement of the alternating harmonic series in Example 3.23.
(a) Verify that the subsequence of partial sums (s
3n
) is monotone-up, by checking that
b
n
:=
1
4n −3
+
1
4n −1
−
1
2n
> 0, for all n ∈ N
(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)
26
The limit γ := lim t
n
≈ 0.5772 is the Euler–Mascheroni constant. It appears in several mathematical identities, and yet
very little about it is understood; it is not even known whether γ is irrational!
59
