Convergence of $a_n = 1/\log(n!)$ 
Test the following series for convergence
$$
a_n = \frac{1}{\log(n!)}
$$

I found this example in my calculus book, and I am unable to solve it. I don't know how to start, what should I aim for, or what should I prove. I'd appreciate any help or suggestions.
 A: We have using the squeeze theorem
$$0\le \frac1{\ln(n!)}\le \frac1{\ln n}\xrightarrow{n\to\infty}0$$
so the sequence $(a_n)$ is convergent. If you want  the convergence of the series $\sum a_n$, we can use the Stirling appoximation
$$n!\sim\left(\frac ne\right)^n\sqrt{2\pi n}$$
so we can see that 
$$\ln(n!)\sim Cn\ln n$$
where $C$ is a constant and then by the integral test we get the divergence of this series.
A: To show convergence of $a_n=\frac{1}{\log n!}$, we could make use of Stirling's Formula or perhaps more simply, use the inequalities
$$\left(\frac{n}{2}\right)^{n/2}<n!<n^n \tag 1$$
From $(1)$ we can see that
$$\bbox[5px,border:2px solid #C0A000]{\frac{1}{n\log n}<\frac{1}{\log n!}<\frac{1}{\frac n2\log \frac n2}} \tag 2$$
So, from the squeeze theorem we have from $(2)$ that $1/\log n! \to 0$, which was to be shown.

Note that $(2)$ facilitates determination of the convergence or divergence of the series $\sum_{n=2}^{\infty}\frac{1}{\log n!}$.  To see this, we note that
$$\lim_{L\to \infty}\int_2^{L}\frac{1}{x\log x}\,dx=\lim_{L\to \infty}\log \log L-\log \log 2=\infty$$
and thus from the integral test, the series $\sum_{n=2}^{\infty}\frac{1}{\log n!}$ diverges.
A: For the series $\sum a_n,$ note that for $n>1,$
$$n! < n^n\implies a_n = 1/\ln n! > 1/\ln (n^n) = 1/(n\ln n).$$
Because $\sum_{n=2}^{\infty} 1/(n\ln n)$ diverges, so does $\sum a_n$
