Page 297 - Schaum's Outline of Theory and Problems of Advanced Calculus

288 INFINITE SERIES [CHAP. 11

n
1
1
(a)By Problem 11.27, S n ðxÞ¼ 1 x ; SðxÞ¼ lim S n ðxÞ¼ 1if < x < ;thus, the series converges in this
2 2
interval. We have n!1
n
Remainder after n terms ¼ R n ðxÞ¼ SðxÞ S n ðxÞ¼ 1 ð1 x Þ¼ x n
The series is uniformly convergent in the interval if given any > 0wecan find N dependent on ,
but not on x,suchthat jR n ðxÞj < for all n > N. Now
ln
n
n
jR n ðxÞj ¼ jx j¼ jxj < when n ln jxj < ln or n >
ln jxj
1
since division by ln jxj (which is negative since jxj < ) reverses the sense of the inequality.
2
ln ln
1
But if jxj < ; ln jxj < ln ð Þ, and n > > ¼ N. Thus, since N is independent of x,the
1
2 2 1
2
ln jxj lnð Þ
series is uniformly convergent in the interval.
ln ln
1
1
(b)In thiscase jxj @ ; ln jxj @ ln ð Þ; and n > A 1 ¼ N,sothat the series is also uniformly
2
2
1
convergent in @ x @ : ln jxj lnð Þ
1
2
2 2
1
(c) Reasoning similar to the above, with replaced by .99, shows that the series is uniformly convergent in
2
:99 @ x @ :99.
ln
(d) The arguments used above break down in this case, since can be made larger than any positive
ln jxj
number by choosing jxj sufficiently close to 1. Thus, no N exists and it follows that the series is not
uniformly convergent in 1 < x < 1.
(e) Since the series does not even converge at all points in this interval, it cannot converge uniformly in the
interval.
11.29. Discuss the continuity of the sum function SðxÞ¼ lim S n ðxÞ of Problem 11.27 for the interval
0 @ x @ 1. n!1
n
If 0 @ x < 1; SðxÞ¼ lim S n ðxÞ¼ lim ð1 x Þ¼ 1.
n!1 n!1
If x ¼ 1; S n ðxÞ¼ 0 and SðxÞ¼ 0.

1if 0 @ x < 1
0if x ¼ 1
Thus, SðxÞ¼ and SðxÞ is discontinuous at x ¼ 1 but continuous at all other points in
0 @ x < 1.
In Problem 11.34 it is shown that if a series is uniformly convergent in an interval, the sum function SðxÞ
must be continuous in the interval. It follows that if the sum function is not continuous in an interval, the
series cannot be uniformly convergent. This fact is often used to demonstrate the nonuniform convergence
of a series (or sequence).
x 2 x 2 x 2
2
11.30. Investigate the uniform convergence of x þ 2 þ 2 2 þ þ 2 n þ .
1 þ x
ð1 þ x Þ ð1 þ x Þ
2
Suppose x 6¼ 0. Then the series is a geometric series with ratio 1=ð1 þ x Þ whose sum is (see Problem
2.25, Chap. 2).
x 2 2
¼ 1 þ x
2
SðxÞ¼
1 1=ð1 þ x Þ
If x ¼ 0thesumofthe first n terms is S n ð0Þ¼ 0; hence Sð0Þ¼ lim S n ð0Þ¼ 0.
n!1
Since lim SðxÞ¼ 1 6¼ Sð0Þ, SðxÞ is discontinuous at x ¼ 0. Then by Problem 11.34, the series cannot be
x!0
uniformly convergent in any interval which includes x ¼ 0, although it is (absolutely) convergent in any
interval. However, it is uniformly convergent in any interval which excludes x ¼ 0.
This can also be shown directly (see Problem 11.93).