How do I prove $\frac{n}{\sqrt[n]{n!}}$ is either monotonic and bounded, or Cauchy? I need to "study the limit behavior of the following sequence" and then compute the limit.
The sequence is $a_n=\frac{n}{\sqrt[n]{n!}} $. This can also be written as $(\frac{n^n}{n!})^{(1/n)}$
I tried to prove it was monotonic because I felt that the sequence was increasing, and I ended up with a great mess, so I tried to prove the sequence was Cauchy and got stuck:
To prove the sequence is Cauchy, for $\epsilon>0,\exists N$ such that $m,n>N$ implies $|a_m-a_n|<\epsilon$
$$|a_m-a_n|=\bigg|\bigg(\frac{m^m}{m!}\bigg)^{1/m}-\bigg(\frac{n^n}{n!}\bigg)^{1/n}\bigg|\leq\bigg|\bigg(\frac{m^m}{m!}\bigg)^{1/m}\bigg|+\bigg|\bigg(\frac{n^n}{n!}\bigg)^{1/n}\bigg|=\bigg(\frac{m^m}{m!}\bigg)^{1/m}+\bigg(\frac{n^n}{n!}\bigg)^{1/n}=\frac{m}{m!^{1/m}}+\frac{n}{n!^{1/n}}\leq\frac{m}{m^{1/m}}+\frac{n}{n^{1/n}}<m+n$$
Which seems to be really wrong, except I don't know what else to do.
Also I went on to compute the limit:
Let $s_n=\frac{n^n}{n!}$ and recall that  $\lim |\frac{s_{n+1}}{s_n}|=L=\lim|s_n|^{1/n}$
$$\bigg|\frac{s_{n+1}}{s_n}\bigg|=\bigg|\frac{(n+1)^{n+1}}{(n+1)!}*\frac{n!}{n^n}\bigg|=\frac{(n+1)^{n+1}}{(n+1)!}*\frac{n!}{n^n}=\bigg(\frac{n+1}{n}\bigg)^n=(1+\frac{1}{n})^n.$$ $$\lim|\frac{s_{n+1}}{s_n}|=\lim(1+\frac{1}{n})^n=e=L=\lim|s_n|^{1/n}=\lim a_n$$
Any help please?
 A: You might start with Stirling's approximation for the factorial.
A: An idea (Hint): define
$$a_n:=\frac{n!}{n^n}\Longrightarrow \frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}=\left(1+\frac{1}{n}\right)^{-n}\xrightarrow [n\to\infty]{} \frac{1}{e}<1$$
so by D'Alembert's test( = ratio test) , we get that the positive series
$$\sum_{n=1}^\infty \frac{n!}{n^n}\;\;\text{converges}\;\;\Longrightarrow\frac{n!}{n^n}\xrightarrow[n\to\infty]{}0$$
So that series is bounded....what can you then say about your series $\,\displaystyle{\frac{1}{\sqrt[n]{a_n}}}\,$ ...?
Another approach: Put
$$\frac{\sqrt[n]{n!}}{n}=\frac{1}{n}e^{\frac{1}{n}\log n!}=\frac{1}{n}e^{\frac{1}{n}\sum_{k=1}^n\log k}=e^{\frac{1}{n}\sum_{k=1}^n\log\frac{k}{n}}$$
But then we have a Riemann sum!:
$$\frac{1}{n}\sum_{k=1}^n\log\frac{k}{n}\xrightarrow[n\to\infty]{}\int\limits_0^1\log x\,dx=\left.\left(x\log x-x\right)\right|_0^1=-1\,\,(\text{note this is an improper integral...)}$$
So finally
$$\frac{\sqrt[n]{n!}}{n}\xrightarrow[n\to\infty]{}e^{-1}\Longrightarrow\frac{n}{\sqrt[n]{n!}}\xrightarrow[n\to\infty]{}e$$
A: From the comments on the question, it seems like you have already answered this yourself without knowing it.  You never used the fact that $|s_{n+1}/s_n|$ converges when verfiying that it is exactly equal to $(1+1/n)^n$.  Since you know the latter converges, you know the exact same sequence converges without any prior assumption.  Thus $|s_{n+1}/s_n| \to e$ and also $|a_n| \to e$ (since $a_n$ is positive, this also means $a_n \to e$).
