1549901369-Elements_of_Real_Analysis__Denlinger_

8.6 Power Series 515

00

Theorem 8.6.19 (Abel's Theorem) Suppose f(x)

lx l < l.

L akxk for all

k=O

00 00

(a) If L ak converges, then lim f(x) = L ak.

k=O x-+1- k=O

00

(b) If L:(-l)kak converges, then

k=O

00

Proof. Suppose f(x) = L akxk for all lx l < 1.

k=O

00

(a) Suppose L ak = S, and let {Sn} denote its sequence of partial sums.

k=O

By Abel's summation by parts formula (8.5.2), for all n :::'.'. 1 and all lx l < 1,

n n

L akxk = L Sk(xk - x k+l) + Snxn+l

k=O k=O

n

= (1 - x) L Skxk + Snxn+l.

k=O

Now, {Sn} converges and xn+l --t 0, so taking the limit of both sides of this
00
equation as n --+ oo, the algebra of limits assures us that \:/ lxl < 1, I: Skxk
k=O
exists and
00
f(x) = (1-x) I: Skxk.
k=O

. c
Let c > 0. Smee Sn --t S, 3 no EN 3 n :::'.'.no::::} ISn - SI < 2· Recall that
00 1 00
\:/ lxl < 1, L xk = --, so (1-x) L x k = 1. Thus,\:/ 0 < x < 1,
k=O 1 - X k=O

lf(x) - SI= 1(1 -x) f Skxk - S(l - x) f xkl = 11-xl If Skxk - f Sxkl

k=O k=O k=O k=O

< ll -x i [k~O ISk - SI+ k=~+l ~ lx lk]

< 11-x i [M + ~ -

1

• 1
  
  
  • 1
  
  
  
  

], where M = I: ISk - SI

2 1 - X k=O

c 11-xl c

= 11-xlM + - --= 11-xlM + -. (since 0 < x < 1)

2 1-lx l 2