256 CHAPTER 11. POWER SERIES
Theorem 11.8.1 Suppose ∑∞i=0 ai and ∑
∞j=0 b j both converge absolutely. Then(
∞
∑i=0
ai
)(∞
∑j=0
b j
)=
∞
∑n=0
cn
where cn = ∑nk=0 akbn−k. Furthermore, ∑
∞n=0 cn converges absolutely.
Proof: It only remains to verify the last series converges absolutely. Letting pnk equal1 if k ≤ n and 0 if k > n. Then by Theorem 6.6.4 on Page 165
∞
∑n=0
|cn| =∞
∑n=0
∣∣∣∣∣ n
∑k=0
akbn−k
∣∣∣∣∣≤ ∞
∑n=0
n
∑k=0
|ak| |bn−k|=∞
∑n=0
∞
∑k=0
pnk |ak| |bn−k|
=∞
∑k=0
∞
∑n=0
pnk |ak| |bn−k|=∞
∑k=0
∞
∑n=k
|ak| |bn−k|=∞
∑k=0
|ak|∞
∑n=0
|bn|< ∞.
The above theorem is about multiplying two series. What if you wanted to consider(∑∞
n=0 an)pwhere p is a positive integer maybe larger than 2? Is there a similar theorem to
the above?
Definition 11.8.2 Define
∑k1+···+kp=m
ak1ak2 · · ·akp
as follows. Consider all ordered lists of nonnegative integers k1, · · · ,kp which have theproperty that ∑
pi=1 ki = m. For each such list of integers, form the product, ak1ak2 · · ·akp
and then add all these products.
Note that ∑nk=0 akan−k = ∑k1+k2=n ak1ak2 . Therefore, from the above theorem, if ∑ai
converges absolutely, it follows(∞
∑i=0
ai
)2
=∞
∑n=0
(∑
k1+k2=nak1ak2
).
It turns out a similar theorem holds for replacing 2 with p.
Theorem 11.8.3 Suppose ∑∞n=0 an converges absolutely. Then if p is a positive in-
teger, (∞
∑n=0
an
)p
=∞
∑m=0
cmp
wherecmp ≡ ∑
k1+···+kp=mak1 · · ·akp .
Proof: First note this is obviously true if p = 1 and is also true if p = 2 from the abovetheorem. Now suppose this is true for p and consider (∑∞
n=0 an)p+1. By the induction