Page 37 - Schaum's Outline of Theory and Problems of Advanced Calculus

28 SEQUENCES [CHAP. 2

It should be emphasized that the use of the notations 1 and 1 for limits does not in any way
imply convergence of the given sequences, since 1 and 1 are not numbers. Instead, these are
notations used to describe that the sequences diverge in specific ways.

n
2.7. Prove that lim x ¼ 0if jxj < 1.
n!1
Method 1:
We can restrict ourselves to x 6¼ 0, since if x ¼ 0, the result is clearly true. Given > 0, we must show
n
n
n
that there exists N such that jx j < for n > N. Now jx j¼jxj < when n log jxj < log .Dividing by
10
10
log
log jxj, which is negative, yields n > 10 ¼ N,proving the required result.
10
10
log jxj
Method 2:
Let jxj¼ 1=ð1 þ pÞ, where p > 0. By Bernoulli’s inequality (Problem 1.31, Chapter 1), we have
n
n
n
n
jx j¼jxj ¼ 1=ð1 þ pÞ < 1=ð1 þ npÞ < for all n > N. Thus lim x ¼ 0.
n!1
THEOREMS ON LIMITS OF SEQUENCES
2.8. Prove that if lim u n exists, it must be unique.
n!1
We must show that if lim u n ¼ l 1 and lim u n ¼ l 2 ,then l 1 ¼ l 2 .
n!1 n!1
By hypothesis, given any > 0wecan find N such that
ju n l 1 j < when n > N; ju n l 2 j < when n > N
1
1
2 2
Then
1
1
jl 1 l 2 j¼jl 1 u n þ u n l 2 j @ jl 1 u n jþju n l 2 j < þ ¼
2 2
i.e., jl 1 l 2 j is less than any positive (however small) and so must be zero. Thus, l 1 ¼ l 2 .
2.9. If lim a n ¼ A and lim b n ¼ B, prove that lim ða n þ b n Þ¼ A þ B.
n!1 n!1 n!1
We must show that for any > 0, we can find N > 0suchthat jða n þ b n Þ ðA þ BÞj < for all n > N.
From absolute value property 2, Page 3 we have
jða n þ b n Þ ðA þ BÞj ¼ jða n AÞþ ðb n BÞj @ ja n Ajþjb n Bj ð1Þ
By hypothesis, given > 0wecan find N 1 and N 2 such that
1
ja n Aj < for all n > N 1
2 ð2Þ
1
jb n Bj <
2 for all n > N 2 ð3Þ
Then from (1), (2), and (3),
1
1
jða n þ b n Þ ðA þ BÞj < þ ¼ for all n > N
2 2
where N is chosen as the larger of N 1 and N 2 . Thus, the required result follows.
2.10. Prove that a convergent sequence is bounded.
Given lim a n ¼ A,we must show that there exists a positive number P such that ja n j < P for all n. Now
n!1
ja n j¼ja n A þ Aj @ ja n AjþjAj
But by hypothesis we can find N such that ja n Aj < for all n > N, i.e.,
for all n > N
ja n j < þjAj
It follows that ja n j < P for all n if we choose P as the largest one of the numbers a 1 ; a 2 ; ... ; a N , þjAj.