Convergence of $\sum_{n=1}^\infty\frac{\ln n!}{n^3}$ How can I determine the convergence of this?
$$\sum_{n=1}^\infty\frac{\ln n!}{n^3}$$
Do I need to use the comparison test or ratio test?
 A: If the general term is $\ln(n!)/n^3$, then you can prove that the series is convergent using the comparison test. In fact, since for all $\alpha\in (0,1)$ we have
$$
\lim_{n\to\infty}\frac{\ln n}{n^{\alpha}}=0,
$$
there is a constant $C_\alpha>0$ such that
$$
\frac{\ln n}{n^{\alpha}}\le C_\alpha \quad \forall n.
$$
Now, for all $n$ we have
$$
\frac{\ln(n!)}{n^3}=\sum_{k=1}^n\frac{\ln(k)}{n^3}\le \sum_{k=1}^n\frac{\ln(n)}{n^3}=\frac{\ln(n)}{n^2}=\frac{\ln(n)}{n^\alpha}\cdot\frac{1}{n^{2-\alpha}}\le \frac{C_\alpha}{n^{2-\alpha}}.
$$
Since $0<\alpha<1$, we have $1<2-\alpha<2$, and therefore the series $\sum_{n=1}^\infty\frac{1}{n^{2-\alpha}}$ is convergent. It follows from the comparison test that the series $\sum_{n=1}^\infty\ln(n!)/n^3$ is also convergent.
A: An easy way :
$$\frac{\ln(n!)}{n^3}=\\\frac{\ln(1.2.3.4.5.6...n)}{n^3}=\\ \to 0<\frac{\ln1+\ln2+\ln3+...+\ln n}{n^3}<\frac{\ln n+\ln n+\ln n+...+\ln n}{n^3}\\=\frac{n\times \ln n}{n^3}=\frac{\ln n}{n^2} \to 0$$so 
$\sum \frac{\ln(n!)}{n^3} <\sum \frac{\ln(n)}{n^2}$ and converges  
A: $$\sum_{n=1}^\infty\frac{\ln n!}{n^3}\le\sum_{n=1}^\infty\frac{\ln n^n}{n^3}=\sum_{n=1}^\infty\frac{\ln n}{n^2}$$
$$\le\sum_{n=1}^\infty\frac{\sqrt n}{n^2}=\sum_{n=1}^\infty\frac1{n^{3/2}}=\zeta(3/2)<\infty$$
By the comparison test, the original series converges.
