1549901369-Elements_of_Real_Analysis__Denlinger_

516 Chapter 8 • Infinite Series of Real Numbers

Choose 6 < min { 1,

2

~ }. Then,

l-6<x<l=?l-x<6
€ € €
=? (1-x)M + 2 < 2MM + 2 = c
=? lf(x) - SI < €.
00
Therefore, lim f ( x) = S = I: ak.
X--tl - k=O
00
(b) Suppose I: (-l)kak converges. Then, letting y = -x,
k=O
00 00
lim f(x ) = lim f(-y) = lim I: ak(-y)k = lim I: (-l)kakyk
X--t -1 + y--t 1 - y--> 1 -k=O y--t 1 - k=O
00
= I: (-l)kak by (a). •
k=O

00

Corollary 8.6.20 Suppose f(x) = I: akxk for all !xi < p.

k=O

00 00

(a) If I: akpk converges, then lim f(x) = I: akpk.

k=O X--tp- k=O

00

(b) If I: ( -1 )k akpk converges, then

k=O

Proof. Exercise 14. •

The following example will show how Theorem 8.6. 19 is used in practice.

Examples 8.6.21 (a) In Example 8.6.18 (a) we showed that ln(l + x) =

00 ( l)k k+l

L -k +xl for all x in (-1, 1). This series diverges when x = -1 but

k=O

converges when x = 1 (verify). According to Abel's theorem (8 .6.19),

00 (-l)k

2:-k-= lim ln(l + x) = ln2

k=Q +^1 X--tl-

since ln(l + x ) is continuous at x = l. Thus, the Maclaurin series for ln(l + x)
is a valid representation on the interval ( -1, l]. As a bonus we have another
derivation of the sum of the alternating harmonic series (see Exercise 8.3.7),

1 1 1 1

1-- + - - - + - - · · · = ln2.

2 3 4 5