Page 43 - Schaum's Outline of Theory and Problems of Advanced Calculus

34 SEQUENCES [CHAP. 2

n
or S n ¼ að1 r Þ
1 r
n a
If jrj < 1; lim S n ¼ lim að1 r Þ by Problem 7:
ðaÞ ¼
n!1 1 r 1 r
n!1
(b)If jrj > 1, lim S n does not exist (see Problem 44).
n!1
2.26. Prove that if a series converges, its nth term must necessarily approach zero.
Since S n ¼ u 1 þ u 2 þ þ u n , S n 1 ¼ u 1 þ u 2 þ þ u n 1 we have u n ¼ S n S n 1 .
If the series converges to S,then
lim u n ¼ lim ðS n S n 1 Þ¼ lim S n lim S n 1 ¼ S S ¼ 0
n!1 n!1 n!1 n!1

1
X
n 1
2.27. Prove that the series 1 1 þ 1 1 þ 1 1 þ ¼ ð 1Þ diverges.
n¼1
Method 1:
n
lim ð 1Þ 6¼ 0, in fact it doesn’t exist. Then by Problem 2.26 the series cannot converge, i.e., it diverges.
n!1
Method 2:
The sequence of partial sums is 1; 1 1; 1 1 þ 1; 1 1 þ 1 1; ... i.e., 1; 0; 1; 0; 1; 0; 1; ... .Since this
sequence has no limit, the series diverges.

MISCELLANEOUS PROBLEMS

u 1 þ u 2 þ þ u n
2.28. If lim u n ¼ l, prove that lim ¼ l.
n
n!1 n!1
v 1 þ v 2 þ þ v n
Let u n ¼ v n þ l. We must show that lim ¼ 0if lim v n ¼ 0. Now
n
n!1 n!1
v 1 þ v 2 þ þ v n v 1 þ v 2 þ þ v P v Pþ1 þ v pþ2 þ þ v n
n ¼ n þ n
so that

jv 1 þ v 2 þ þ v P j jv Pþ1 jþjv Pþ2 jþ þ jv n j
@
v 1 þ v 2 þ þ v n
þ ð1Þ
n n n

Since lim v n ¼ 0, we can choose P so that jv n j < =2for n > P. Then
n!1
=2 þ =2 þ þ =2 ðn PÞ =2
< <
jv Pþ1 jþjv Pþ2 jþ þ jv n j
n n ¼ n 2 ð2Þ
After choosing P we can choose N so that for n > N > P,

<
jv 1 þ v 2 þ þ v P j
n 2 ð3Þ
Then using (2) and (3), (1)becomes

v 1 þ v 2 þ þ v n < ¼ for n > N
n 2 þ 2
thus proving the required result.
2 1=n
2.29. Prove that lim ð1 þ n þ n Þ ¼ 1.
n!1
2 1=n
Let ð1 þ n þ n Þ ¼ 1 þ u n where u n A 0. Now by the binomial theorem,