Evaluate $\lim\limits_{n \to \infty}\frac{n}{\sqrt[n]{n!}}$. Solution
Notice that
$$(\forall x \in \mathbb{R})~~e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$
Let $x=n$ where $n\in \mathbb{N_+}$. Then we obtain
$$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.$$
Thus, we obtain
$$e>\frac{n}{\sqrt[n]{n!}}.\tag1$$
Moreover, we can find that, for $k=0,1,\cdots,n-1.$ $$\frac{n^k}{k!}< \frac{n^n}{n!}.$$
Thus
\begin{align*}
e^n&=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)(n+2)}+\cdots\right]\\
&< (n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+\cdots\right]\\
&=(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot n\\
&=(2n+1)\cdot \frac{n^n}{n!},
\end{align*}
which shows that
$$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.\tag2$$
Combining $(1)$ and $(2)$, we have
$$e>\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}\to e(n \to \infty).$$
By the squeeze theorem,
$$\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}=e.$$
Please correct me if I'm wrong. Many thanks.
 A: Using Riemann sums:
$$ \log\frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\sum_{k=1}^{n}\log\left(\tfrac{k}{n}\right)\to \int_{0}^{1}\log(x)\,dx = -1$$
hence $\frac{n}{\sqrt[n]{n!}}$ converges to $e$.
A: Seems okay another way could be
$$\frac{n}{\sqrt[n]{n!}}=\exp({\ln n-\frac{\ln n!}{n}})=\exp\left(\frac{n\ln n-\ln n!}{n}\right)$$
($\exp(x)=e^x$)
Then by continuity we have that $$\lim_{n\to\infty}\exp\left(\frac{n\ln n-\ln n!}{n}\right)=\exp\left(\lim_{n\to\infty}\frac{n\ln n-\ln n!}{n}\right)$$
Now you can find the $\lim$ by using Stolz theorem.
$$\lim_{n\to\infty}\frac{n\ln n-\ln n!}{n}=\lim_{n\to\infty}(n+1)\ln(n+1)-\ln((n+1)!)-n\ln n+\ln n!=\lim_{n\to\infty}n\ln(n+1)-n\ln n=\lim_{n\to\infty}\ln(1+\frac1n)^n=1$$
Hence the whole $\lim$ equals to $e$
A: For the denominator, rewrite it as $ e^{\frac{\sum_{k=1}^{n} \log k}{n}}$ then approximate the sum of logs with the integral ($n \log n -n) $ to get the result.
A: You are right with your limit $e$. Using Stirling's formula you have putting $f(n)=\dfrac{n}{\sqrt[n]{n!}}$,
$$\log f(n)=\log n-\frac{n\log n-n+O(\log n)}{n}\Rightarrow \log f(n)-1=\frac{O(\log n)}{n}$$ By definition of Big $O$ notation  there is some positive constant $C$ and some $n_0$ such thar for all $n\gt n_0$ we have 
$$|\log f(n)-1|\le \frac{C\log n}{n}$$ Then $\log f(n)-1$ tends to $0$ so $f(n)$ tends to $e$ as you have show.
