1549901369-Elements_of_Real_Analysis__Denlinger_

84 Chapter 2 • Sequences

Therefore, lim (

3

n

2

-;;) = +oo. O

n-+oo 5n +

Theorem 2.4.4 If { Xn} is a sequence of positive real numbers, then

(a) lim Xn = +oo if and only if lim (_!_) = O;

n--+oo n--+oo Xn

(b) lim Xn = 0 if and only if lim (_!_) = +oo.

n--+oo n-+oo Xn

Proof. Exercise 4. •

Example 2.4.5 If a > 1, then lim an = +oo.
n-+oo
1
Proof. Suppose a > 1. Let b = -. Then 0 < b < 1, so by Theorem 2 .3.4,
a
1
bn---; 0. Thus, by Theorem 2.4.4 (b), -b ---; +oo. That is , lim an= +oo. 0
n n--+oo

Theorem 2.4.6 (Summary of lim an) Let a ER Then
n-+oo

{

0 if lal < 1

lim an = 1 if a = 1

n-+oo

+oo if a> 1.

If a~ -1, t hen {an} diverges. •

Theorem 2.4.7 (Comparison Test) Suppose {an} and {bn} are sequences
such that '\In EN, an ~ bn.

(a) If lim an= +oo, then lim bn = +oo.

n--+oo n--+oo

(b) If lim bn = -oo, then lim an = -oo.

n--+oo n--+oo

Proof. (a) Suppose lim an = +oo. Let M > 0. Then, by Definition 2.4.1,
n-+oo
3no E N 3 n > no ::::} an > M. Since '\In E N, an ~ bn, it follows that
n >no::::} bn > M. Therefore, lim bn = +oo.
n-+oo
(b) Exercise 5. •

2n +4n
Example 2.4.8 lim = +oo.
n-+oo 3n