real analysis - For which $p$ does the series $sum_{n=2}^{infty} frac{1}{(log 1)^p+(log 2)^p+cdots +(log n)^p}$ converge?

Suppose $\displaystyle a_n=\frac{1}{(\text{log}2)^p+(\text{log3})^p+...+(\text{log}n)^p}$ for $n\geq 2$ and $p>0$. For which values of $p$ does $\sum_{n=2}^\infty a_n$ converge?

So, when $p\leq 1$, $\displaystyle a_n\geq \frac{1}{(n-1)(\text{log}n)^p}$ for $n\geq 2$. Let $\displaystyle b_n=\frac{1}{(n-1)(\text{log}n)^p}$. Then $\sum b_n$ diverges since $p\leq 1$. Therefore by comparison test $\sum a_n$ diverges when $p\leq 1$.

When $p>1$, $\displaystyle a_n\leq \frac{1}{(\text{log}n)^p}$. Let $\displaystyle c_n= \frac{1}{(\text{log}n)^p}$. If $\sum c_n$ converges then I will win. But now I am stuck. Is $\sum c_n$ converges? Then how can I prove it? If not how can I conclude the state of convergence of $\sum a_n$ for $p>1$?

Can anybody please help?