8.2 Nonnegative Series 461Thus, for all integers n > no,1:: 0 f :S CEo ak) - an. (5)
With the help of Figure 8.1 (b) we see that
n n
l~ f 2: s_u, P) = L f(k) = L ak.
k=no+l k=no+l
Thus,so,
n
L ak :S 1:: 0 f + ano·
k=no(6)
00
Part 1 ( =>): Suppose I: an converges. Then its sequence of partial sums is
n=no
bounded. So, by (5), the monotone increasing sequence { l~ f} ~=no is bounded.
Thus, n~CXJ lim J,n no f exists, from which we can easily prove that the improper
integral Jno r= f converges.
Part 2 ( ~): Suppose the improper integral l: f converges. Then the sequence
{l:: f }
00
0 converges, so it is bounded. Thus, by (6), the monotone increasing
n=nosequence of partial sums { f= an}
00
is bounded, hence converges. That is,
k=no n=no
00
L an converges. •
n=no
00
Example 8.2.4 Use the integral test to prove that the series ~ 2 n
0
di-
n=l L., n + l
verges.
x 10 - x^2
Solution. Let J(x) = x 2 +
10
. Then f'(x) = 2 , so f'(x) < 0
(x^2 + 10).
when x 2: 4. Thus, f is continuous and monotone decreasing on [4, oo) and
lim x = 0. Moreover, the improper integral 1
00x-+oo x2 + 10 4 x^2 + x 1 0 dx diverges,
since
lim lb 2 x dx = lim [~ ln(x^2 + 10)] b = +oo.
b-+oo 4 X + 10 b-+oo (^2 4)