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

Let $f (k) = \frac{1}{k\log^p(k+1)}$. When $g (k) = 2^k f (2^k)$, we have $$\sum f (k) \leqslant \sum g (k) \leqslant 2 \sum f (k),$$ so if one converges, the other converges too. We have $$g (k) = \frac {1} {(k + 1)^{p \log 2}}.$$ Then $\sum g (k)$, so $\sum f (k)$ converges when $p > 1/\log 2$.
