Show $\sum_{n=2}^{\infty}\frac{1}{n(\log n)^{p}}$ is convergent. I know $\sum_{2}^{\infty}\frac{1}{n(\log n)^{p}}$ is convergent for $p >1$ and divergent for $p\le 1$.
My question is how can we show this. can we use ratio test to show this is convergent for $p >1$.
I tried as
$$\begin{aligned}\lim_{n \to \infty}\frac{n(log n)^{p}}{(n+1)(log n+1)^{p}}&=\lim_{n \to \infty} (\frac{n}{n+1})\lim_{n \to \infty} \frac{(\log n)^{p}}{(\log (n+1))^{p}}\\&=1\cdot \lim_{n \to \infty}\frac{(\log n)^{p}}{(\log (n+1))^{p}}\\&=\lim_{n \to \infty}\left(\frac{\log n}{\log n+1}\right)^{p}\end{aligned}$$
Since $\log n+1> \log n$, $\frac{\log n}{\log n+1}< 1$.
How should I proceed further and use $p > 1$ to show it is convergent.
Thanks in advance!

HINT
Apply the Condensation test
\begin{align*}
\sum_{n=2}^{\infty}\frac{2^{n}}{2^{n}\ln^{p}(2^{n})} & = \sum_{n=2}^{\infty}\frac{1}{n^{p}\ln^{p}(2)} = \frac{1}{\ln^{p}(2)}\sum_{n=2}^{\infty}\frac{1}{n^{p}}
\end{align*}
Can you take it from here?
