156 CHAPTER 6. INFINITE SERIES

Definition 6.2.1 A series ∑∞k=m ak is converges absolutely if ∑

∞k=m |ak| converges.

If the series does converge but does not converge absolutely, then it is said to convergeconditionally.

Theorem 6.2.2 If ∑∞k=m ak converges absolutely, then it converges.

Proof: Let ε > 0 be given. Then by assumption and Theorem 6.1.8, there exists nε

such that whenever q ≥ p ≥ nε , ∑qk=p |ak|< ε.Therefore, from the triangle inequality, ε >

∑qk=p |ak| ≥

∣∣∣∑qk=p ak

∣∣∣ . By Theorem 6.1.8, ∑∞k=m ak converges.

In fact, the above theorem is really another version of the completeness axiom. Thusits validity implies completeness. You might try to show this.

One of the interesting things about absolutely convergent series is that you can “addthem up” in any order and you will always get the same thing. This is the meaning of thefollowing theorem. Of course there is no problem when you are dealing with finite sumsthanks to the commutative law of addition. However, when you have infinite sums strangeand wonderful things can happen because these involve a limit.

Theorem 6.2.3 Let θ : N→ N be one to one and onto. Suppose ∑∞k=1 ak converges

absolutely. Then ∑∞k=1 aθ(k) = ∑

∞k=1 ak.

Proof: From absolute convergence, there exists M such that ∑∞k=M+1 |ak|< ε. Since θ is

one to one and onto, there exists N ≥M such that {1,2, · · · ,M}⊆{θ (1) ,θ (2) , · · · ,θ (N)} .It follows that it is also the case that ∑

∞k=N+1

∣∣aθ(k)∣∣ < ε. This is because the partial sums

of the above series are each dominated by a partial sum for ∑∞k=M+1 |ak| since every index

θ (k) equals some n for n ≥ M + 1. Then since ε is arbitrary, this shows that the partialsums of ∑aθ(k) are Cauchy. Hence, this series does converge and also∣∣∣∣∣ M

∑k=1

ak −N

∑k=1

aθ(k)

∣∣∣∣∣≤ ∞

∑k=M+1

|ak|< ε

Hence ∣∣∣∣∣ ∞

∑k=1

ak −∞

∑k=1

aθ(k)

∣∣∣∣∣≤∣∣∣∣∣ ∞

∑k=1

ak −M

∑k=1

ak

∣∣∣∣∣+∣∣∣∣∣ M

∑k=1

ak −N

∑k=1

aθ(k)

∣∣∣∣∣+

∣∣∣∣∣ N

∑k=1

aθ(k)−∞

∑k=1

aθ(k)

∣∣∣∣∣< ∞

∑k=M+1

|ak|+ ε +∞

∑k=N+1

∣∣aθ(k)∣∣< 3ε

Since ε is arbitrary, this shows the two series are equal as claimed.So what happens when series converge only conditionally?

Example 6.2.4 Consider the series ∑∞k=1 (−1)k 1

k . Show that there is a rearrangementwhich converges to 7 although this series does converge. (In fact, it converges to − ln2.)

First of all consider why it converges. Notice that if Sn denotes the nth partial sum, then

S2n −S2n−2 =1

2n− 1

2n−1< 0

S2n+1 −S2n−1 = − 12n+1

+1

2n> 0

S2n −S2n−1 =1

2n