Page 40 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 2] SEQUENCES 31

p ffiffiffiffiffiffiffi p ffiffiffiffiffiffi
(b)Let x ¼ required limit. Since lim u nþ1 ¼ lim 3u n ,we have x ¼ 3x and x ¼ 3. (The other
n!1 n!1
possibility, x ¼ 0, is excluded since u n A 1:Þ
n 1 n lim ð1 1=2 n Þ
1
Another method: lim 3 1=2þ1=4þ þ1=2 ¼ lim 3 1 1=2 ¼ 3 n!1 ¼ 3 ¼ 3
n!1 n!1
2.17. Verify the validity of the entries in the following table.
Monotonic Monotonic Limit
Sequence Bounded Increasing Decreasing Exists
2; 1:9; 1:8; 1:7; .. . ; 2 ðn 1Þ=10 ... No No Yes No
n 1
1; 1; 1; 1; .. . ; ð 1Þ ; ... Yes No No No
1 1 1 1 n 1 =ðn þ 1Þ; ... Yes No No Yes (0)
2 ; ; ; ; ... ; ð 1Þ
5
3 4
n
2
2
:6;:66;:666; ... ; ð1 1=10 Þ; ... Yes Yes No Yes ( )
3 3
n
1; þ2; 3; þ4; 5; ... ; ð 1Þ n; .. . No No No No
n

1
is monotonic, increasing, and bounded,
n
2.18. Prove that the sequence with the nth term u n ¼ 1 þ
and thus a limit exists. The limit is denoted by the symbol e.
n
1
¼ e, where e ffi 2:71828 ... was introduced in the eighteenth century by
Note: lim 1 þ
n
n!1
Leonhart Euler as the base for a system of logarithms in order to simplify certain differentiation
and integration formulas.
By the binomial theorem, if n is a positive integer (see Problem 1.95, Chapter 1),
n nðn 1Þ 2 nðn 1Þðn 2Þ 3 nðn 1Þ ðn n þ 1Þ n
2! 3! n!
ð1 þ xÞ ¼ 1 þ nx þ x þ x þ þ x
Letting x ¼ 1=n,
n

1 1 nðn 1Þ 1 nðn 1Þ ðn n þ 1Þ 1
u n ¼ 1 þ ¼ 1 þ n þ 2 þ þ n
n n 2! n n! n

1 1 1 1 2
2! n 3! n n
¼ 1 þ 1 þ 1 þ 1 1
1 1 2 n 1
n! n n n
þ þ 1 1 1
Since each term beyond the first two terms in the last expression is an increasing function of n,it follows that
the sequence u n is a monotonic increasing sequence.
It is also clear that
n
1 1 1 1 1 1 1
< 3
1 þ < 1 þ 1 þ þ þ þ < 1 þ 1 þ þ 2 þ þ n 1
n 2! 3! n! 2 2 2
by Problem 1.14, Chapter 1.
Thus, u n is bounded and monotonic increasing, and so has a limit which we denote by e. The value of
e ¼ 2:71828 ... .
x
1
¼ e, where x !1 in any manner whatsoever (i.e., not necessarily along
x
2.19. Prove that lim 1 þ
x!1
the positive integers, as in Problem 2.18).
n x nþ1
1 1 1
If n ¼ largest integer @ x,then n @ x @ n þ 1 and 1 þ @ 1 þ @ 1 þ .
n þ 1 x n
,
n nþ1
1 1 1

Since ¼ e
n þ 1 ¼ lim 1 þ n þ 1 1 þ n þ 1
lim 1 þ
n!1 n!1