1549901369-Elements_of_Real_Analysis__Denlinger_

2.4 Divergence to Infinity 85

Proof. By Example 2.4.5 above, lim (~)n = +oo. Now, 'tin EN,

n--+oo 3

2n +4n

Therefore, by Theorem 2.4.7, lim = +oo. D

n--+oo 3n

ALGEBRA OF INFINITE LIMITS

Theorem 2.4.9 Suppose {an}, {bn}, {en}, and {dn} are sequences such that

n--+oo lim an = +oo, n--+oo lim bn = + oo, n---too lim Cn = -00^1 and n-too lim dn = -oo. Then

(a) lim (an+ bn) = +oo;

n--+oo

(b) n--+oo lim (en+ dn) = -oo;

(d) n--+oo lim (cndn) = +oo;

(e) lim (ancn) = -oo.

n--+oo

• Proof. (a) Suppose n--+oo lim an = +oo and n--+oo lim bn = +oo. Let M > 0.

Since lim an = + oo, 3n1 E N 3 n > n1 =? an > M. Since lim bn = +oo,
n--+oo n--+oo
3 n 2 EN 3 n > n2 =? bn > M. Let no= max{n1,n2}. Then,

n > no =? an > M and bn > M

=? (an+ bn) > 2M > M.

Therefore, lim (an+ bn) = +oo.
n--+oo
(b) Exercise 6.
( c) Exercise 7.
(d) Exercise 8.
(e) Suppose lim an = + oo, and lim Cn = -oo. Let M > 0. Since
n--+oo n-+oo
lim an = +oo, 3 n1 E N 3 n > n1 =? an > M. Since lim Cn = - oo,
n--+oo n--+oo
3 n 2 EN 3 n > n2 =? Cn < - M. Let no= max{n1, n2}. Then,