Page 298 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 11] INFINITE SERIES 289

WEIERSTRASS M TEST
11.31. Prove the Weierstrass M test, i.e., if ju n ðxÞj @ M n ; n ¼ 1; 2; 3; ... ; where M n are positive
constants such that M n converges, then u n ðxÞ is uniformly (and absolutely) convergent.
The remainder of the series u n ðxÞ after n terms is R n ðxÞ¼ u nþ1 ðxÞþ u nþ2 ðxÞþ . Now
jR n ðxÞj ¼ ju nþ1 ðxÞþ u nþ2 ðxÞþ j @ ju nþ1 ðxÞj þ ju nþ2 ðxÞj þ @ M nþ1 þ M nþ2 þ
But M nþ1 þ M nþ2 þ can be made less than by choosing n > N, since M n converges. Since N is clearly
independent of x,we have jR n ðxÞj < for n > N, and the series is uniformly convergent. The absolute
convergence follows at once from the comparison test.

11.32. Test for uniform convergence:
n
1 1 1 1
X cos nx X x X sin nx X 1
; ; ; :
ðaÞ 4 ðbÞ 3=2 ðcÞ ðdÞ 2 2
n n n n þ x
n¼1 n¼1 n¼1 n¼1

cos nx 1
(a) @ ¼ M n . Then since M n converges ð p series with p ¼ 4 > 1Þ,the series is uniformly (and
n 4 n 4
absolutely) convergent for all x by the M test.
(b)Bythe ratio test, the series converges in the interval 1 @ x @ 1, i.e., jxj @ 1.
n n

x jxj 1 1
For all x in this interval, @ ,we see that M n converges.
n n n n
3=2 ¼ 3=2 3=2 . Choosing M n ¼ 3=2
Thus, the given series converges uniformly for 1 @ x @ 1bythe M test.

1 1
sin nx
(c) @ . However, M n , where M n ¼ , does not converge. The M test cannot be used in this
n n n
case and we cannot conclude anything about the uniform convergence by this test (see, however,
Problem 11.125).

1 1 1

(d) @ , and converges. Then by the M test the given series converges uniformly for all x.
n þ x n n
2 2 2 2
n
11.33. If a power series a n x converges for x ¼ x 0 , prove that it converges (a) absolutely in the
interval jxj < jx 0 j, (b) uniformly in the interval jxj @ jx 1 j; where jx 1 j < jx 0 j.
n
n
n
(a)Since a n x 0 converges, lim a n x 0 ¼ 0 and so we can make ja n x 0 j < 1bychoosing n large enough, i.e.,
1 n!1
ja n j < n for n > N. Then
jx 0 j
n
1 1 1
X X X
n n jxj
ja n x j¼ ja n jjxj < n ð1Þ
Nþ1 Nþ1 Nþ1 jx 0 j
Since the last series in (1)converges for jxj < jx 0 j,itfollows by the comparison test that the first
series converges, i.e., the given series is absolutely convergent.
n
jx 1 j n
n
(b)Let M n ¼ . Then M n converges since jx 1 j < jx 0 j.As in part (a), ja n x j < M n for jxj @ jx 1 j,so
jx 0 j
n
that by the Weierstrass M test, a n x is uniformly convergent.
It follows that a power series is uniformly convergent in any interval within its interval of con-
vergence.
THEOREMS ON UNIFORM CONVERGENCE
11.34. Prove Theorem 6, Page 271.
We must show that SðxÞ is continuous in ½a; bŠ.