Page 305 - Schaum's Outline of Theory and Problems of Advanced Calculus

296 INFINITE SERIES [CHAP. 11

1 n þ 1 1 1 1
ln .Byintegrating the inequality @ @ with respect
n þ 1 n n þ 1 x n
Consider S nþ1 S n ¼
to x from n to n þ 1, we have
1 n þ 1 1 1 1 1 n þ 1
@ ln @ or @ ln @ 0
n þ 1 n n n þ 1 n n þ 1 n
i.e., S nþ1 S n @ 0, so that S n is monotonic decreasing.
Since S n is bounded and monotonic decreasing, it has a limit. This limit, denoted by
,isequal to
0:577215 ... and is called Euler’s constant. Itis not yet known whether
is rational or not.
1 1
Y X
11.50. Prove that the infinite product ð1 þ u k Þ, where u k > 0, converges if u k converges.
k¼1 k¼1
x
x
According to the Taylor series for e (Page 275), 1 þ x @ e for x > 0, so that
n
Y
ð1 þ u k Þ¼ð1 þ u 1 Þð1 þ u 2 Þ ð1 þ u n Þ @ e e e ¼ e u 1 þu 2 þ þu n
u 2
u 1
u n
P n ¼
k¼1
Since u 1 þ u 2 þ converges, it follows that P n is a bounded monotonic increasing sequence and so has
a limit, thus proving the required result.
11.51. Prove that the series 1 1 þ 1 1 þ 1 1 þ is C 1summable to 1/2.
The sequence of partial sums is 1; 0; 1; 0; 1; 0; ... .
1 þ 0 1 S 1 þ S 2 þ S 3 1 þ 0 þ 1 2
S 1 þ S 2
Then S 1 ¼ 1; ¼ ¼ ; ¼ ¼ ; ... :
2 2 2 3 3 3
1 2 1 3 1
Continuing in this manner, we obtain the sequence 1; ; ; ; ; ; ... ; the nth term being
2 3 2 5 2
1=2 if n is even 1
2
if n is odd
T n ¼ . Thus, lim T n ¼ and the required result follows.
n=ð2n 1Þ n!1
11.52. (a)If f ðnþ1Þ ðxÞ is continuous in ½a; bŠ prove that for c in ½a; bŠ, f ðxÞ¼ f ðcÞþ f ðcÞðx cÞþ
0
1 1 1 ð x
n
00 2 f ðnÞ n ðx tÞ f ðnþ1Þ ðtÞ dt.
2! f ðcÞðx cÞ þ þ n! ðcÞðx cÞ þ n! c
(b) Obtain the Lagrange and Cauchy forms of the remainder in Taylor’s Formula. (See Page
274.)
The proof of (a)is made using mathematical induction. (See Chapter 1.) The result holds for n ¼ 0
since
ð x
0
f ðxÞ¼ f ðcÞþ f ðtÞ dt ¼ f ðcÞþ f ðxÞ f ðcÞ
C
We make the induction assumption that it holds for n ¼ k and then use integration by parts with
k
dt and u ¼ f
ðx tÞ kþ1
k!
dv ¼ ðtÞ
Then
kþ1
and du ¼ f ðtÞ dt
ðx tÞ kþ2
ðk þ 1Þ!
v ¼
Thus,

1 ð x k f kþ1 kþ1 x 1 ð x kþ1

ðx tÞ f ðkþ1Þ ðtÞ dt ¼ ðtÞðx tÞ þ ðx tÞ f ðkþ2Þ ðtÞ dt
k! C ðk þ 1Þ! C ðk þ 1Þ! C
f kþ1 kþ1 1 ð x
f
ðcÞðx cÞ
kþ1 ðkþ2Þ ðtÞ dt
ðk þ 1Þ! ðk þ 1Þ! C
¼ þ ðx tÞ
Having demonstrated that the result holds for k þ 1, we conclude that it holds for all positive integers.