To use the integral test, we need to compute
Let
, then
. With this change of variables, when
then
and when
then
does as well. Applying the substitution,
Any restrictions on
will come if we cannot substitute the infinite upper limit and get zero. We have the following for the indefinite integral of
:
![{\displaystyle \int {\frac {1}{u^{p}}}{\textrm {d}}u={\begin{cases}{\frac {1}{1-p}}{\frac {1}{u^{p-1}}},&p>1\\\ln u,&p=1\\{\frac {1}{1-p}}u^{1-p},&p<1.\end{cases}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9a3e5c6555601a3239e0bcc813d0e001b554a8b0)
The only case in which plugging in and infinite limit will result in zero is for
. Since the Right hand side of the equality converges whenever
, by the integral test, the series converges for
.