8.2 Nonnegative Series 463COMPARISON TESTSTheorem 8.2.6 (Comparison Test) Suppose Lan and L bn are nonneg-
ative series, and ::lno E N 3 n 2: no =} an :::; bn. (When this holds we say that
L bn dominates L an.) Then(a) if L bn converges, then so does Lan.
(b) If Lan diverges, then so does L bn.
Proof. Exercise 2. •Examples 8.2. 7 Use the comparison test to prove that the following series
converge (or diverge):
00
n+8
(a) L n3 - 5n + 7
n=l(b) f n+6
n=l Jn3 + 2Solution.
(a) It appears that as n gets large, the terms of the series (a) are something
like ~. Thus, we will try to show that this series converges by comparing it
n
1
with L 2 , which is a convergent p-series (p = 2). Observe that '<In 2: 8,
n
n+8 n+n 2 2 4
-----< =--< =-
n3 - 5n + 7 n^3 - 5n n^2 - 5 n^2 - n^2 /2 n^2 ·
[Note: n 2: 8 =} n^2 > 10 =} n^2 /2 > 5 =} n^2 - 5 > n^2 - n^2 /2.)
(^00 4 00) n+8
Since L 2 converges, the comparison test assures us that L 3
n=l n n=l n -^5 n +^7
converges.
1
(b) As n gets large, the terms of the series (b) are something like fa.
Thus, we will try to prove that the series (b) diverges by comparing it with
L Jn, which is a divergent p-series (p = ~). Observe that Vn 2: 2,
00 1
Now, L
n=l Jn+ 1
n+6 n 1
---> - --===
Jn3 + 2 Jn3 + n^2 - Jn+ 1 ·
f: ~, which diverges. Therefore, by the comparison
n=2 yn
(^00) n + 6
test, L v'n3+2 diverges. D
n=l n^3 + 2