Page 302 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 11] INFINITE SERIES 293

Now
n
R n ðxÞ¼ ðR n R nþ1 Þx þðR nþ1 R nþ2 Þx nþ1 þðR nþ2 R nþ3 Þx nþ2 þ
n
n
¼ R n x þ R nþ1 ðx nþ1 x Þþ R nþ2 ðx nþ2 x nþ1 Þþ
n 2
¼ x fR n ð1 xÞðR nþ1 þ R nþ2 x þ R nþ3 x þ Þg
Hence, for 0 @ x < 1,
2
jR n ðxÞj @ jR n jþð1 xÞðjR nþ1 jþjR nþ2 jx þjR nþ3 jx þ Þ ð1Þ
Since a k converges by hypothesis, it follows that given > 0wecan choose N such that jR k j < =2for
all k A n. Then for n > N we have from (1),

2
jR n ðxÞj @ þð1 xÞ þ x þ x þ ¼ þ ¼ ð2Þ
2 2 2 2 2 2
3
2
since ð1 xÞð1 þ x þ x þ x þ Þ ¼ 1 (if 0 @ x < 1).
Also, for x ¼ 1; jR n ðxÞj ¼ jR n j < for n > N.
Thus, jR n ðxÞj < for all n > N, where N is independent of the value of x in 0 @ x @ 1, and the
required result follows.
Extensions to other power series are easily made.
11.43. Prove Abel’s limit theorem (see Page 272).
1
X k
As in Problem 11.42, assume the power series to be a k x ,convergent for 0 @ x @ 1.
k¼1
1 1
X k X
Then we must show that lim a k x ¼ a k .
k¼0 k¼0
x!1
k
This follows at once from Problem 11.42, which shows that a k x is uniformly convergent for
k
0 @ x @ 1, and from Problem 11.34, which shows that a k x is continuous at x ¼ 1.
Extensions to other power series are easily made.
x 3 x 5 x 7
1
11.44. (a) Prove that tan x ¼ x þ þ where the series is uniformly convergent in
3 5 7
1 @ x @ 1.
1 1 1
(b) Prove that ¼ 1 þ þ .
4 3 5 7
2
(a)ByProblem 2.25 of Chapter 2, with r ¼ x and a ¼ 1, we have
1 2 4 6
1 þ x 2 ¼ 1 x þ x x þ 1 < x < 1 ð1Þ
Integrating from 0 to x, where 1 < x < 1, yields
x 3 5 7
ð
dx 1 x x x
¼ tan x ¼ x þ þ ð2Þ
0 1 þ x 2 3 5 7
using Problems 11.33 and 11.35.
Since the series on the right of (2)converges for x ¼ 1, it follows by Problem 11.42 that the series
is uniformly convergent in 1 @ x @ 1 and represents tan 1 x in this interval.
(b)By Problem 11.43 and part (a), we have
!
x 3 x 5 x 7 1 1 1
lim tan 1 x ¼ lim x þ þ or ¼ 1 þ þ
3 5 7 4 3 5 7
x!1 x!1
1 1 e
ð x 2
11.45. Evaluate 2 dx to 3 decimal place accuracy.
0 x