Is there a standard way to compute $\lim\limits_{n\to\infty}(\frac{n!}{n^n})^{1/n}$? I'm computing the radii of convergence for some complex power series. For one I need to compute
$$\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}.$$
I know the answer is $\frac{1}{e}$, so the radius is $e$. But how could you compute this by hand? I tried taking the logarithms and raising $e$ by this logarithm, but it didn't lead me to the correct limit. (This is just practice, not homework.)
 A: $$\frac{a_{n+1}}{a_n} \to L \implies (a_n)^{1/n} \to L$$
A: There are two formulas to compute radius of convergence of the series $\sum\limits_{n=1}^\infty{c_n}z^n$
$$
\frac{1}{R}=\lim\limits_{n\to\infty}|c_n|^{1/n}=\lim\limits_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|.
$$
Use the second one.
A: One can indeed consider logarithms (among other methods). To see this, call the $n$th term $x_n$, then
$$
\log(x_n)=n^{-1}\log(n!)-\log(n)=n^{-1}\sum_{k=1}^n\log(k/n).
$$
This is a Riemann sum, hence, when $n\to\infty$,
$$
\log(x_n)\to\int_0^1\log(x)\mathrm dx=\left[x\log(x)-x\right]_0^1=-1,
$$
that is, $\lim\limits_{n\to\infty}x_n=1/\mathrm e$.
A: Stirling's formula for $n!$ works wonders here, but I guess that may not qualify as "computing by hand"?
Just for the sake of completeness, Stirling's formula states that
$$n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$
or:
$$\displaystyle\lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n} = 1$$
Substituting the RHS from the first equation and taking the limit as $n \to \infty$ pretty much yields the solution instantly.
