Convergence of the $\sum \frac{\ln(n!)}{n^{\alpha}}$ I'm preparing to my calculus exam (it's not a homework). So I have a sum: 
$$\sum_{n=1}^{\infty} \frac{\ln(n!)}{n^{\alpha}}$$
I need to find for which values ​​of the parameter $\alpha$ the sum converges and diverges.
I tried to find some inequality to apply the comparison test, but I failed.

$$\ln n! = \ln 1 + \ln 2 + ... +\ln n < n\ln n <n^{1+\beta}$$
For all $\beta >0$. Thus, your series converges for all $\alpha>2$.
