For what $p$ does the series $\sum \frac1{n^p \ln(n)}$ converge "Find the values of $p$ s.t. the following series converges: $\sum_{n=2}^{\infty} \frac{1}{n^p \ln(n)}$"
I am trying to do this problem through using the Integral Test to find the values of $p$. I know that for $p = 0$, the series diverges so I will only be considering values of $p \neq 0$.
The function $f(x) = (x^p \ln(x))^{-1}$ satisfies the criterion of the Integral Test, but I am having a difficult time integrating the function.
We have $$\int_{n=2}^{\infty} \frac{1}{x^p \ln(x)} dx$$ A u-substitution with $u = \ln(x)$ will not help us and neither would setting it to $x^p$. Someone suggested letting $x = e^u$, but...i'm not so sure where we would go with that. 

Comparison test is best here. But here's how to use that substitution idea. Notice how it comes down to comparison in the end.
Using $x=e^u$ we have $dx=e^u du$. Thus $$\int_2^\infty  \frac{dx}{x^p \ln x} = \int_{\ln2}^\infty \frac{e^u}{u e^{p u}} du = \int_{\ln2}^\infty e^{(1-p) u} \frac{1}{u} du.$$
It is easy to see that for $p<1$ the integrand is unbounded, so we have divergence. For $p=1$ the antiderivative is known and we again diverge. For $p>1$ we can bound the integrand above by $e^{(1-p)u}$, and the integral of this is easily found to be convergent.
