166 CHAPTER 6. INFINITE SERIES

It isn’t hard to see where this comes from. Formally write the following in the caser = 0:

(a0 +a1 +a2 +a3 · · ·)(b0 +b1 +b2 +b3 · · ·)

and start multiplying in the usual way. This yields

a0b0 +(a0b1 +b0a1)+(a0b2 +a1b1 +a2b0)+ · · ·

and you see the expressions in parentheses above are just the cn for n = 0,1,2, · · · . There-fore, it is reasonable to conjecture that ∑

∞i=r ai ∑

∞j=r b j = ∑

∞n=r cn and of course there would

be no problem with this in the case of finite sums but in the case of infinite sums, it isnecessary to prove a theorem. The following is a special case of Merten’s theorem.

Theorem 6.6.7 Suppose ∑∞i=r ai and ∑

∞j=r b j both converge absolutely2. Then(

∞

∑i=r

ai

)(∞

∑j=r

b j

)=

∞

∑n=r

cn

where cn = ∑nk=r akbn−k+r.

Proof: Let pnk = 1 if r ≤ k ≤ n and pnk = 0 if k > n. Then cn = ∑∞k=r pnkakbn−k+r.

Also,∞

∑k=r

∞

∑n=r

pnk |ak| |bn−k+r|=∞

∑k=r

|ak|∞

∑n=r

pnk |bn−k+r|

=∞

∑k=r

|ak|∞

∑n=k

|bn−k+r|=∞

∑k=r

|ak|∞

∑n=k

∣∣bn−(k−r)∣∣= ∞

∑k=r

|ak|∞

∑m=r

|bm|< ∞.

Therefore, by Theorem 6.6.5

∞

∑n=r

cn =∞

∑n=r

n

∑k=r

akbn−k+r =∞

∑n=r

∞

∑k=r

pnkakbn−k+r

=∞

∑k=r

ak

∞

∑n=r

pnkbn−k+r =∞

∑k=r

ak

∞

∑n=k

bn−k+r =∞

∑k=r

ak

∞

∑m=r

bm

6.7 Exercises1. Determine whether the following series converge absolutely, conditionally, or not at

all and give reasons for your answers.

(a) ∑∞n=1 (−1)n 2n+n

n2n

(b) ∑∞n=1 (−1)n 2n+n

n22n

(c) ∑∞n=1 (−1)n n

2n+1

(d) ∑∞n=1 (−1)n 10n

n!

(e) ∑∞n=1 (−1)n n100

1.01n

(f) ∑∞n=1 (−1)n 3n

n3

(g) ∑∞n=1 (−1)n n3

3n

(h) ∑∞n=1 (−1)n n3

n!

(i) ∑∞n=1 (−1)n n!

n100

2Actually, it is only necessary to assume one of the series converges and the other converges absolutely. Thisis known as Merten’s theorem and may be read in the 1974 book by Apostol listed in the bibliography.