1549901369-Elements_of_Real_Analysis__Denlinger_

524 Chapter 8 • Infinite Series of Real Numbers

Then, Vk E N, the kth Maclaurin coefficient is

J(k) (0) = (a).

k! k

00

Therefore, the Maclaurin series for f is ~ (~)xk. The radius of convergence

k=O

of this series is kl~1! I((~)) I = 1, as shown in the proof of Lemma 8.7.9.

00

Now that we know that the series ~ (~)xk converges everywhere in (-1, 1),

k=O

it remains to prove that it converges to (1 + x )°' for every x in this interval. We

use the integral form of the remainder to show that Rn(x) ---> 0. By Theorem

8.7.3 (b),

Case 1 (a > 1): To eliminate the absolute value bars around the integral

in (32) we must decide whether we integrate from 0 to x or from x to 0.

If 0 < x < 1, then we integrate over the interval 0 ::::; t ::::; x , and

x-t

(^0) -1+t-< --< x and 0 < - 1 + t < - 2.
If -1 < x < 0, then we integrate over the interval x ::::; t ::::; 0, and
x - t
x ::::;
1

• t ::::; 0 (Why?) and 0 ::::; 1 + t ::::; 1 < 2.
  Plugging this information into (32) we have
  IRn(x)I ::::; (n + 1) I (n~l) 1 llax lxln 2a-ldtl
  = (n + 1) I (n~l) j lx ln 2a-l I lax dtl = (n + 1) I (n~l) j lxln+l 2a-l.
  Applying Lemma 8.7.9, we see that this expression converges to 0 as
  n---> oo. Therefore, lim Rn(x) = 0.
  n->oo