Limit of a sequence involving root of a factorial: $\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}$

What you have is actually an indefinite integral in disguise. Let's first consider the reciprocal of what you have: \begin{eqnarray*} \lim_{n\to\infty}\frac{(n!)^{1/n}}{n} & = & e^{{\displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\ln\left(\frac{k}{n}\right)}}\\ & = & e^{{\displaystyle \int_{0}^{1}\ln xdx}}\\ & = & e^{-1}. \end{eqnarray*} Thus we get that $$ \lim_{n\to\infty}\frac{n}{(n!)^{1/n}}=e. $$

Alternatively, you could use the fact that for a sequence $(a_n)$ of positive terms, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty}\root n\of{a_n}$ and the two limits are equal.

For your problem, consider $a_n={n^n\over n!}$. Then $${a_{n+1}\over a_n}={(n+1)^{n+1}\over (n+1)!}\cdot {n!\over n^n}= {1\over n+1}\cdot\Bigl({n+1\over n}\Bigr)^n\cdot(n+1)=\Bigl(1+{1\over n}\Bigr)^n \ \ \buildrel{n\rightarrow\infty}\over\longrightarrow\ \ e. $$

Thus $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}=e$. As $\root n\of {a_n}={n\over(n!)^{1/n}}$, we have $\lim\limits_{n\rightarrow\infty}{n\over(n!)^{1/n}}=e$ as well.

Use Stirling's approximation: $ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n $ and you'll get $$ \lim_{n \rightarrow \infty} \frac{n}{(n!)^{1/n}} =\lim_{n \rightarrow \infty} \frac{n}{(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n)^{1/n}} =\lim_{n \rightarrow \infty} \frac{n}{({2 \pi n})^{1/2n} \left(\frac{n}{e}\right)} =\lim_{n \rightarrow \infty} \frac{e}{({2 \pi n})^{1/2n} }=e, $$ because $\lim_{n\to \infty} ({2 \pi n})^{1/2n}= \lim_{n\to \infty} n^{1/n}=1$.