1549901369-Elements_of_Real_Analysis__Denlinger_

110 Chapter 2 • Sequences

Therefore, Xnk ---+ L.

Part 2 ( {::::): Exercise 3. •

Example 2.6.9 Find lim (1 +
2

1

) n

n---+cx::i n

Solution: By writing out the first few terms of the sequence, we have

(^1 + 2 ~)^1 , (1 + ~) 4 '^2 (1 + ~) 6 ,^3 (1 + ~) 8 4 , (1 + ~) 10 5 ' ....

We can rewrite these terms as

(

1 )

10

1+-

10

Thus, { ( 1 +

2

~ n is a suOOequence of { J ( 1+ ff} Hence,

lim (1 + 2-)n = lim J(1 + .!.)n =

n->oo 2n n->oo n

lim (1 + .!.)n = yle. D

n->oo n

Corollary 2.6.10 (Application to Proving Divergence)
(a) If a sequence has two subsequences that converge to different limits,
then it diverges.
(b) If a sequence has a divergent subsequence, then it diverges.

Examples 2.6.11 The following sequences diverge:

{

(a) The sequence 1+(-l)n}.

2

= {O, 1, 0, 1, 0, 1, 0, 1, 0, 1, · · ·} diverges. It

has a subsequence {O, 0, 0, · · ·} converging to 0, and a subsequence {1, 1, 1, · · ·}
converging to l.

{

(b) The sequence l,l,2,^1 1 1 1 1}

2

, 3,

3

,4,

4

,5,

5

, ... ,n,;,, ... diverges. It has

a subsequence {1, 2, 3, 4, 5, · · ·} that diverges. D

Theorem 2.6.12 (a) A sequence diverges to +oo {::} every subsequence di-
verges to +oo.

(b) A sequence diverges to -oo {::} every subsequence diverges to -oo.