For what values of $p$ does $\sum_{k = 1}^{\infty} \frac{1}{k\log^p(k+1)}$ converge? Find all $p\geq 0$ such that the following series converges $\sum_{k = 1}^{\infty} \frac{1}{k\log^p(k+1)}$.
Proof: the general term for the series is $\frac{k^p}{k^p\log^p(k+1)^n} = \frac{1}{k\log^p(k+1)^n}$.
By comparison, $\frac{1}{\log^pn}\leq \frac{1}{\log^p(n+1)}
$.
And it's convergent when $p>1$ thus $\sum_{k = 1}^{\infty} \frac{1}{k\log^p(k+1)}$ is convergent when $p>1$ and divergent when $0 < p < 1$.
Can anyone please verify this?
Any suggestion would help. I was trying to use the integral test, but I don't know how to do it.
Thank you

By Cauchy condensation test all gets reduced to the study of 
$$\sum_{k = 1}^{\infty} \frac{1}{k^p}$$
whence we get the conclusions according to the values of $p$.
Q.E.D.
