The radius of convergence of $\sum_{n\ge 0}{\log(n!)x^n}$. I want the radius of convergence of the series $\sum_{n\ge 0}{\log(n!)x^n}$. Could I use the stirling formula 
$$n!\sim_\infty \left(\frac{n}{e}\right)^n\sqrt{2 \pi n}?$$
Because then
$$\log (n!)\sim_\infty \log\left(\middle(\frac{n}{e}\middle)^n\sqrt{2 \pi n}\right)$$
Then use a ratio test to compute the limit of $$\frac{\log(n+1)!}{\log(n!)}|x|$$
 A: Note that $n!\leqslant n^n$ hence $1\leqslant\log(n!)\leqslant n\log n\leqslant n^2$ for every $n\geqslant3$. 
The radii of convergence of $\sum\limits_nx^n$ and $\sum\limits_nn^2x^n$ are both $1$ hence the radius of convergence of $\sum\limits_n\log(n!)x^n$ is $____$.
If one insists on using the ratio test, one might note that $\log(n!)\sim n\log n$ hence $\log(n!)\sim\log((n+1)!)$, that is,
$$
\frac{\log((n+1)!)}{\log(n!)}\to1,
$$
thus, indeed, the radius of convergence of $\sum\limits_n\log(n!)x^n$ is $____$.
A: Because
$$
\frac{\log((n+1)!)}{\log(n!)} = 1 + \frac{\log(n+1)}{\log(n!)},
$$
to prove that $\lim_{n\to\infty} \big( \frac{\log((n+1)!)}{\log(n!)} |x| \big) = |x|$ you only need to show that $\lim_{n\to\infty} \frac{\log(n+1)}{\log(n!)} = 0$.
This can be done, for instance, by observing that for $n\ge3$, we have $n+1 \le ((n+1)/2)^2$ and
$$
n! > n(n-1)\cdots\bigg(n-\bigg\lceil \frac n2\bigg\rceil+1\bigg) > \bigg(\frac{n+1}2\bigg)^{\lceil n/2\rceil} \ge \bigg(\frac{n+1}2\bigg)^{n/2},
$$
and so
$$
0 \le \frac{\log(n+1)}{\log(n!)} < \frac{2\log((n+1)/2)}{\frac n2\log((n+1)/2)} = \frac 4n.
$$
The squeeze theorem finishes it off.
