272 CHAPTER 5 Series and Differential Equations
nlim→∞
1
(ln(n)^2
1
n
) = nlim→∞ n
(lnn)^2
= xlim→∞
x
(lnx)^2
l’Hˆopital= lim
x→∞
d
dxx
d
dx(lnx)
2
= xlim→∞
x
(2 lnx)
l’Hˆopital= lim
x→∞
d
dxx
d
dx(2 lnx)
= xlim→∞
x
2
=∞.
This says that, asymptotically, the series
∑∞
n=2
1
(lnn)^2
is infinitely larger
than the divergent harmonic series
∑∞
n=2
1
n
implying divergence.
The next test will provide us with a rich assortment of series to test
with. (So far, we’ve only been testing against theconvergent series
∑∞
n=1
1
n^2
and thedivergentseries
∑∞
n=1
1
n
.)
Thep-Series Test. Letpbe a real number. Then
∑∞
n=1
1
np
convergesif p > 1
divergesif p≤ 1.
The p-series test is sometimes called the p-test for short; the proof
of the above is simple; just as we proved that
∑∞
n=1
1
n^2
converged by
comparing it with
∫∞
1
dx
x^2
(which converges) and that
∑∞
n=1
1
n
diverged by
comparing with
∫∞
1
dx
x^2
(which diverges), we see that
∑∞
n=0
1
np
will have