Page 38 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 2] SEQUENCES 29

1
2.11. If lim b n ¼ B 6¼ 0, prove there exists a number N such that jb n j > jBj for all n > N.
2
n!1
Since B ¼ B b n þ b n ,we have: (1) jBj @ jB b n jþjb n j.
1
Now we can choose N so that jB b n j¼ jb n Bj < jBj for all n > N, since lim b n ¼ B by hypothesis.
2
1
1
Hence, from (1), jBj < jBjþjb n j or jb n j > jBj for all n > N. n!1
2 2
2.12. If lim a n ¼ A and lim b n ¼ B, prove that lim a n b n ¼ AB.
n!1 n!1 n!1
We have, using Problem 2.10,
ja n b n ABj¼ja n ðb n BÞþ Bða n AÞj @ ja n jjb n BjþjBjja n Aj ð1Þ
@ Pjb n BjþðjBjþ 1Þja n Aj
But since lim a n ¼ A and lim b n ¼ B,given any > 0wecan find N 1 and N 2 such that
n!1 n!1

jb n Bj < for all n > N 1 ja n Aj < for all n > N 2
2P 2ðjBjþ 1Þ
1
1
Hence, from (1), ja n b n ABj < þ ¼ for all n > N, where N is the larger of N 1 and N 2 . Thus, the
2 2
result is proved.
1 1 a n A
2.13. If lim a n ¼ A and lim b n ¼ B 6¼ 0, prove (a) lim ¼ ,(b) lim ¼ .
B B
n!1 b n n!1 b n
n!1 n!1
(a)We must show that for any given > 0, we can find N such that

1
1 jB b n j
< for all n > N

¼ ð1Þ
B
b n jBjjb n j
2
1
By hypothesis, given any > 0, we can find N 1 ,suchthat jb n Bj < B for all n > N 1 .
2
1
Also, since lim b n ¼ B 6¼ 0, we can find N 2 such that jb n j > jBj for all n > N 2 (see Problem 11).
2
n!1
Then if N is the larger of N 1 and N 2 ,wecan write (1)as
1 B
1 2
1 jb n Bj 2
< ¼ for all n > N
¼
B 1
b n jBjjb n j
2
jBj jBj
and the proof is complete.
(b)From part (a) and Problem 2.12, we have

a n 1 1 1 A
lim ¼ lim a n ¼ lim a n lim ¼ A ¼
n!1 b n n!1 b n n!1 n!1 b n B B
This can also be proved directly (see Problem 41).
2.14. Evaluate each of the following, using theorems on limits.
2
3n 5n 3 5=n 3 þ 0 3
lim ¼ lim
ðaÞ 2 2 ¼ ¼
n!1 5n þ 2n 6 n!1 5 þ 2=n 6=n 5 þ 0 þ 0 5
( 3 ) ( 3 2 ) ( 2 )
n n þ n þ 2n 1 þ 1=n þ 2=n
lim nðn þ 2Þ ¼ lim ¼ lim
n þ 1 n þ 1
ðbÞ 2 2 2
n!1 n!1 ðn þ 1Þðn þ 1Þ n!1 ð1 þ 1=nÞð1 þ 1=n Þ
1 þ 0 þ 0
¼ 1
¼
ð1 þ 0Þ ð1 þ 0Þ
p ffiffiffiffiffiffiffiffiffiffiffi
p ffiffiffi
ffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffi n 1
p p ffiffiffi p p ffiffiffi
ffiffiffi ¼ 0
n þ 1 þ
ffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffi
n n
ðcÞ lim ð n þ 1 n Þ¼ lim ð n þ 1 n Þ p p ffiffiffi ¼ lim p p
n!1 n!1 n!1
n þ 1 þ n þ 1 þ