10.4.1 Convergence Because Of Cancellation
So far, the tests for convergence have been applied to non negative terms only.
Sometimes, a series converges, not because the terms of the series get small fast enough,
but because of cancellation taking place between positive and negative terms. A
discussion of this involves some simple algebra.

Let

and

be sequences and let

∑n
An ≡ ak,A −1 ≡ A0 ≡ 0.
k=1

Then if p < q

q q q q
∑ a b = ∑ b (A − A ) = ∑ b A − ∑ b A
n=p nn n=p n n n− 1 n=p n n n=p n n−1

∑q q−∑ 1 q∑−1
= bnAn − bn+1An = bqAq − bpAp −1 + An (bn − bn+1) (10.6)
n=p n=p− 1 n=p
(10.6)

This formula is called the partial summation formula. It is just like integration by
parts.

Theorem 10.4.1 (Dirichlet’s test) Suppose A _{n} ≡ ∑
_{k=1} ^{n} a _{k} is bounded
independent of n and lim_{n→∞} b _{n} = 0, with b _{n} ≥ b _{n+1} for all n. Then

converges.

Proof: This follows quickly from Theorem 10.1.8 . Indeed, letting

≤ C, and using
the partial summation formula above along with the assumption that the

b _{n} are
decreasing,

||∑q || || q∑−1 ||
|| anbn||= ||bqAq − bpAp −1 + An (bn − bn+1)||
|n=p | | n=p |

q∑−1
≤ C (|bq|+ |bp|) + C (bn − bn+1)
n=p

= C (|bq|+|bp|)+ C (bp − bq)

and by assumption, this last expression is small whenever p and q are sufficiently large.
■

Definition 10.4.2 If b _{n} > 0 for all n, a series of the form ∑
_{k}

^{k} b _{k}
or ∑
_{k} ^{k−1} b _{k} is known as an alternating series.

The following corollary is known as the alternating series test.

Corollary 10.4.3 (alternating series test) If lim_{n→∞} b _{n} = 0, with b _{n} ≥ b _{n+1} ,
then ∑
_{n=1} ^{∞}

^{n} b _{n} converges.

Proof: Let a _{n} =

^{n} . Then the partial sums of

∑
_{n} a _{n} are bounded and so
Theorem

10.4.1 applies.

■

In the situation of Corollary 10.4.3 there is a convenient error estimate available.

Theorem 10.4.4 Let b _{n} > 0 for all n such that b _{n} ≥ b _{n+1} for all n and
lim_{n→∞} b _{n} = 0 and consider either ∑
_{n=1} ^{∞}

^{n} b _{n} or ∑
_{n=1} ^{∞} ^{n−1} b _{n} . Then

| |
||∑∞ n N∑ n ||
|| (− 1) bn − (− 1) bn|| ≤ |bN+1 |,
| n=1 n=1 |
||∑∞ n−1 N∑ n−1 ||
|| (− 1) bn − (− 1) bn|| ≤ |bN+1 |
n=1 n=1
See Problem 8 on Page 696 for an outline of the proof of this theorem along with
another way to prove the alternating series test.

Example 10.4.5 How many terms must I take in the sum, ∑
_{n=1} ^{∞}

^{n}
to be closer than to ∑
_{n=1} ^{∞} ^{n} ?

From Theorem 10.4.4 , I need to find n such that

≤ and then

n − 1 is the
desired value. Thus

n = 3 and so

| |
||∑∞ n --1--- ∑2 n --1--|| 1-
|| (− 1) n2 + 1 − (− 1) n2 + 1|| ≤ 10
n=1 n=1

Definition 10.4.6 A series ∑
a _{n} is said to converge absolutely if ∑

converges. It is said to converge conditionally if ∑
fails to converge but ∑
a _{n}
converges.

Thus the alternating series or more general Dirichlet test can determine convergence
of series which converge conditionally.