Kenneth Kuttler

86 CHAPTER 5. INFINITE SERIES OF NUMBERS

Theorem 5.2.3 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| ≡

(∞

∑k=1|ak|−

∑k=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

∣∣∣∣∣+∣∣∣∣∣ 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 5.2.4 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 − ln2for those who remember calculus.)

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

2n> 0

S2n−S2n−1 =1

2nThus the even partial sums are decreasing and the odd partial sums are increasing. Theeven partial sums are bounded below also. (Why?) Therefore, the limit of the even partialsums exists. However, it must be the same as the limit of the odd partial sums becauseof the last equality above. Thus limn→∞ Sn exists and so the series converges. Now I willshow later that ∑

∞k=1

12k and ∑

∞k=1

12k−1 both diverge. Include enough even terms for the

sum to exceed 7. Next add in enough odd terms so that the result will be less than 7. Nextadd enough even terms to exceed 7 and continue doing this. Since 1/k converges to 0, thisrearrangement of the series must converge to 7. Of course you could also have picked 5 or−8 just as well. In fact, given any number, there is a rearrangement of this series whichconverges to this number.

86 CHAPTER 5. INFINITE SERIES OF NUMBERSTheorem 5.2.3 Let 6: N—N be one to one and onto. Suppose )\y_, ax convergesabsolutely. Then Vi 46 (k) = Lpa1 UkProof: From absolute convergence, there exists M such thatY lau) = e al Yo a) ce.k=M+1 k=1 k=1Since @ is one to one and onto, there exists N > M such thatIt follows that it is also the case that Ye y+ |ao(x)| < €. This is because the partial sumsof the above series are each dominated by a partial sum for )'7_y,; |ax| since every index6 (k) equals some n for n > M+1. Then since € is arbitrary, this shows that the partialsums of }'dg(x) are Cauchy. Hence, this series does converge and alsoM NYa— Y ae)k=1 k=1co< y laz| <€k=M+1Hence< +© MYa Yeak=1 k=1M NYa — Yaoi)k=1 k=1Va Y ae)k=1 k=1+ < y lax] +e+ y |ae(x)| < 3€k=M+1 k=N+1N coY aon) — Y aoc)k=1 k=1Since € is arbitrary, this shows the two series are equal as claimed. §fSo what happens when series converge only conditionally?Example 5.2.4 Consider the series Yy_, (-1)* 2. Show that there is a rearrangementwhich converges to 7 although this series does converge. (In fact, it converges to —\n2for those who remember calculus.)First of all consider why it converges. Notice that if S,, denotes the n’’ partial sum, then1 1Son — Son-2 = mn mal ~~?1 1Sont1 — San. = “tel tin?1Son —S2n-1 = onThus the even partial sums are decreasing and the odd partial sums are increasing. Theeven partial sums are bounded below also. (Why?) Therefore, the limit of the even partialsums exists. However, it must be the same as the limit of the odd partial sums becauseof the last equality above. Thus lim,_,..S, exists and so the series converges. Now I willshow later that )’;"_, x and Ye 74a both diverge. Include enough even terms for thesum to exceed 7. Next add in enough odd terms so that the result will be less than 7. Nextadd enough even terms to exceed 7 and continue doing this. Since 1/k converges to 0, thisrearrangement of the series must converge to 7. Of course you could also have picked 5 or—8 just as well. In fact, given any number, there is a rearrangement of this series whichconverges to this number.