Page 288 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 11] INFINITE SERIES 279

COMPARISON TEST AND QUOTIENT TEST
11.5. If 0 @ u n @ v n ; n ¼ 1; 2; 3; ... and if v n converges, prove that u n also converges (i.e., establish
the comparison test for convergence).
Let S n ¼ u 1 þ u 2 þ þ u n ; T n ¼ v 1 þ v 2 þ þ v n .
Since v n converges, lim T n exists and equals T,say. Also, since v n A 0; T n @ T.
n!1
Then S n ¼ u 1 þ u 2 þ þ u n @ v 1 þ v 2 þ þ v n @ T or 0 @ S n @ T:
Thus S n is a bounded monotonic increasing sequence and must have a limit (see Chapter 2), i.e., u n
converges.

1
X 1
1 1
2 3 n
11.6. Using the comparison test prove that 1 þ þ þ ¼ diverges.
n¼1
We have 1 A 1
2
1
1 þ A 1 1 1
2 3 4 þ ¼ 2
4
1 þ þ þ A 1 1 1 1 1
1
1
1
8
8
4 5 6 7 8 þ þ þ ¼ 2
8
1 1 1 1 A 1 1 1 1 1
9
8 þ þ 10 þ þ 15 16 þ 16 þ 16 þ þ 16 (8 terms) ¼ 2
etc. Thus, to any desired number of terms,
1 1 1 1 1 1 þ A 1 1 1

2 3 4 5 6 7 2 2 2
1 þ þ þ þ þ þ þ þ þ
Since the right-hand side can be made larger than any positive number by choosing enough terms, the given
series diverges.
1
X 1
By methods analogous to that used here, we can show that , where p is a constant, diverges if
n p
n¼1
p @ 1 and converges if p > 1. This can also be shown in other ways [see Problem 11.13(a)].
1
X ln n
11.7. Test for convergence or divergence 3 .
2n 1
n¼1
1 1 ln n n 1
Since ln n < n and @ ; we have @ ¼ :
3
3
2n 1 n 3 2n 1 n 3 n 2
1
X 1
Then the given series converges, since converges.
n 2
n¼1
u n
11.8. Let u n and v n be positive. If lim ¼ constant A 6¼ 0, prove that u n converges or diverges
n!1 v n
according as v n converges or diverges.

u n
By hypothesis, given > 0wecan choose an integer N such that A < for all n > N. Then for

n ¼ N þ 1; N þ 2; ... v n
u n
< A < or ðA Þv n < u n < ðA þ Þv n ð1Þ
v n
Summing from N þ 1to 1 (more precisely from N þ 1to M and then letting M !1),
1 1 1
X X X
v n @
ðA Þ u n @ ðA þ Þ v n ð2Þ
Nþ1 Nþ1 Nþ1
There is no loss in generality in assuming A > 0. Then from the right-hand inequality of (2), u n
converges when v n does. From the left-hand inequality of (2), u n diverges when v n does. For the cases
A ¼ 0or A ¼1, see Problem 11.66.