Page 39 - Schaum's Outline of Theory and Problems of Advanced Calculus

30 SEQUENCES [CHAP. 2

2
3n þ 4n 3 þ 4=n
lim ¼ lim
n!1 2n 1 n!1 2=n 1=n
ðdÞ 2
Since the limits of the numerator and denominator are 3 and 0, respectively, the limit does not
exist.
2
3n þ 4n 3n 2 3n
Since > ¼ can be made larger than any positive number M by choosing n > N,we
2n 1 2n 2
2
3n þ 4n
can write, if desired, lim ¼1.
n!1 2n 1
4 4 4
2n 3 2 3=n 2 16

lim lim
n!1 2n þ 7 n!1 3 þ 7=n 3 81
ðeÞ ¼ ¼ ¼
2
5
2n 4n 2 2=n 4=n 5 0
lim ¼ lim ¼ 0
ð f Þ 7 3 4 7 ¼
n!1 3n þ n 10 n!1 3 þ 1=n 10=n 3
1 þ 2 10 n 10 n þ 2 2
lim ¼ lim (Compare with Problem 2.5.)
ðgÞ n n ¼
n!1 5 þ 3 10 n!1 5 10 þ 3 3
BOUNDED MONOTONIC SEQUENCES
2n 7
(a)is monotonic increasing, (b)is bounded
3n þ 2
2.15. Prove that the sequence with nth u n ¼
above, (c)is bounded below, (d)is bounded, (e) has a limit.
(a) fu n g is monotonic increasing if u nþ1 A u n , n ¼ 1; 2; 3; ... . Now
2ðn þ 1Þ 7 2n 7 2n 5 2n 7
A if and only if A
3ðn þ 1Þþ 2 3n þ 2 2n þ 5 3n þ 2
2
2
or ð2n 5Þð3n þ 2Þ A ð2n 7Þð3n þ 5Þ,6n 11n 10 A 6n 11n 35, i.e. 10A 35, which is
true. Thus, by reversal of steps in the inequalities, we see that fu n g is monotonic increasing. Actually,
since 10 > 35, the sequence is strictly increasing.
(b)Bywriting some terms of the sequence, we may guess that an upper bound is 2 (for example). To prove
this we must show that u n @ 2. If ð2n 7Þ=ð3n þ 2Þ @ 2then 2n 7 @ 6n þ 4or 4n < 11, which is
true. Reversal of steps proves that 2 is an upper bound.
(c) Since this particular sequence is monotonic increasing, the first term 1isa lower bound, i.e.,
u n A 1, n ¼ 1; 2; 3; ... . Any number less than 1isalso a lower bound.
(d)Since the sequence has an upper and lower bound, it is bounded. Thus, for example, we can write
ju n j @ 2for all n.
(e) Since every bounded monotonic (increasing or decreasing) sequence has a limit, the given sequence has
2n 7 2 7=n 2
a limit. In fact, lim ¼ lim ¼ .
n!1 3n þ 2 n!1 3 þ 2=n 3

p ffiffiffiffiffiffiffi
3u n , u 1 ¼ 1. (a)Prove that lim u n
2.16. A sequence fu n g is defined by the recursion formula u nþ1 ¼
exists. (b) Find the limit in (a). n!1
p ffiffiffiffiffiffiffi 1=2 p ffiffiffiffiffiffiffi 1=2þ1=4
3u 1 ¼ 3 3u 2 ¼ 3 ; ... .
(a) The terms of the sequence are u 1 ¼ 1, u 2 ¼ , u 3 ¼
n 1
The nth term is given by u n ¼ 3 1=2þ1=4þ þ1=2 as can be proved by mathematical induction
(Chapter 1).
Clearly, u nþ1 A u n . Then the sequence is monotone increasing.
1
By Problem 1.14, Chapter 1, u n @ 3 ¼ 3, i.e. u n is bounded above. Hence, u n is bounded (since a
lower bound is zero).
Thus, a limit exists, since the sequence is bounded and monotonic increasing.