Page 299 - Schaum's Outline of Theory and Problems of Advanced Calculus

290 INFINITE SERIES [CHAP. 11

Now SðxÞ¼ S n ðxÞþ R n ðxÞ,sothat Sðx þ hÞ¼ S n ðx þ hÞþ R n ðx þ hÞ and thus
Sðx þ hÞ SðxÞ¼ S n ðx þ hÞ S n ðxÞþ R n ðx þ hÞ R n ðxÞ ð1Þ
where we choose h so that both x and x þ h lie in ½a; bŠ (if x ¼ b,for example, this will require h < 0).
Since S n ðxÞ is a sum of finite number of continuous functions, it must also be continuous. Then given
> 0, we can find so that
jS n ðx þ hÞ S n ðxÞj < =3 whenever jhj < ð2Þ
Since the series, by hypothesis, is uniformly convergent, we can choose N so that
jR n ðxÞj < =3 and jR n ðx þ hÞj < =3 for n > N ð3Þ
Then from (1), (2), and (3),
jSðx þ hÞ SðxÞj @ jS n ðx þ hÞ S n ðxÞj þ jR n ðx þ hÞj þ jR n ðxÞj <
for jhj < , and so the continuity is established.

11.35. Prove Theorem 7, Page 271.
If a function is continuous in ½a; bŠ, its integral exists. Then since SðxÞ; S n ðxÞ, and R n ðxÞ are continuous,
ð b ð b ð b
R n ðxÞ dx
SðxÞ¼ S n ðxÞ dx þ
a a a
To prove the theorem we must show that

b ð b ð b
ð

SðxÞ dx S n ðxÞ dx ¼ R n ðxÞ dx
a a a

can be made arbitrarily small by choosing n large enough. This, however, follows at once, since by the
uniform convergence of the series we can make jR n ðxÞj < =ðb aÞ for n > N independent of x in ½a; bŠ, and
so

b ð b ð b
ð
R n ðxÞ dx @ jR n ðxÞj dx < dx ¼

a a a b a
This is equivalent to the statements
ð b ð b ð b ð bn o
SðxÞ dx ¼ lim S n ðxÞ dx or lim S n ðxÞ dx ¼ lim S n ðxÞ dx
a n!1 a n!1 a a n!1
11.36. Prove Theorem 8, Page 271.
1
X
u n ðxÞ. Since, by hypothesis, this series converges uniformly in ½a; bŠ,wecan integrate
0
Let gðxÞ¼
n¼1
term by term (by Problem 11.35) to obtain
x 1 ð x 1
ð
X X
0
gðxÞ dx ¼ u n ðxÞ dx ¼ fu n ðxÞ u n ðaÞg
a n¼1 a n¼1
1 1
X X
¼ u n ðxÞ u n ðaÞ¼ SðxÞ SðaÞ
n¼1 n¼1
1
X
because, by hypothesis, u n ðxÞ converges to SðxÞ in ½a; bŠ.
n¼1
ð x
Differentiating both sides of gðxÞ dx ¼ SðxÞ SðaÞ then shows that gðxÞ¼ S ðxÞ, which proves the
0
theorem. a