# Determine the convergence/divergence of $\sum_{n=1}^{\infty}\frac{\ln{n!}}{n^3}$

Does the series$$\sum_{n=1}^{\infty}\frac{\ln{n!}}{n^3}$$ converge or diverge? I initially thought about using the ratio test but then I got the ratio is $$1$$, so the test is inconclusive here.

I was thinking maybe using the comparison test? But I am not too sure which series to compare with. Any help would be appreciated.

• Do you know Stirling's Approximation for $\ln (n!)$? – Clayton Oct 2 '18 at 1:13
• Stirling is unnecessarily fancy! – Ted Shifrin Oct 2 '18 at 1:41

notice $$\ln n! = \ln 1 + \ln 2 + \ln 3 + .. + \ln n$$
• So, in particular, $\ln n! \le n\ln n$. – Ted Shifrin Oct 2 '18 at 1:41
Given $$\sum_{n=1}^\infty\dfrac{\ln(n!)}{n^3}$$
$$\lim_{n\rightarrow\infty} \left(n\left(\dfrac{\dfrac{\ln(n!)}{n^3}}{\dfrac{\ln((n+1)!)}{(n+1)^3}}-1\right)\right)=2$$
Since $$L>1$$ the series converges.