<s>[INST] 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.
 [/INST] 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.
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