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.
 A: $$\ln n! = \ln 1 + \ln 2 + ... +\ln n < n\ln n <n^{1+\beta}$$
For all $\beta >0$. Thus, your series converges for all $\alpha>2$.
A: The equivalent $\ln n!\sim n\ln n$ yields readily the answer.
This equivalent is a crude form of Stirling's formula, more easily proved than the full form, as follows. 
First, $\ln n!$ is the sum of $n$ terms $\ln k$ each $\leqslant\ln n$ hence $\ln n!\leqslant n\ln n$. 
On the other hand, $\ln n!$ is greater than $(1-a)n$ terms greater than $\ln(an)$, for every $a\lt1$, hence $\ln n!\geqslant(1-a)n\ln(an)\geqslant (1-a)n\ln n+\Theta(n)$.
Since $a$ can be chosen as small as desired, the claimed equivalent follows.
A: $$\sum_{n=1}^{\infty} \dfrac{\log(n!)}{n^a} = \sum_{n=1}^{\infty} \sum_{k=1}^n \dfrac{\log(k)}{n^a} = \underbrace{\sum_{k=1}^{\infty} \sum_{n=k}^{\infty} \dfrac{\log(k)}{n^a} = \sum_{k=1}^{\infty} \log(k) \Theta\left(\dfrac1{k^{a-1}}\right)}_{\text{True for $a > 1$}} = \Theta\left(\sum_{k=1}^{\infty}  \dfrac{\log(k)}{k^{a-1}}\right)$$
$\displaystyle  \sum_{k=1}^{\infty}  \dfrac{\log(k)}{k^{a-1}}$ converges iff $a-1>1$ (and in fact equals $-\zeta'(a-1)$ for $a>1$).
Hence, $$\sum_{n=1}^{\infty} \dfrac{\log(n!)}{n^a}$$ converges for $a>2$.
