<s>[INST] How to obtain the lower bound of $\dfrac{n}{\sqrt[n]{n!}}$ by Taylor's series? We want to prove $$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}=e.$$
I have some solutions for this, but I want to find another method applying the squeeze theorem. Thus, a natrual thought is to find the upper bound and the lower bound of $a_n$.
Notice that 
$$e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$
If we substitute $x$ for $n$, then 
$$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.\tag1$$
Thus, we obtain
$$e>\frac{n}{\sqrt[n]{n!}},$$
which shows $e$ is a upper bound of $a_n$.
But how to obtain the lower bound by $(1)$? Say it again. I have other methods to deal with it. I just wonder whether there is some method depending on $(1)$ only or not.
 [/INST] \begin{align*}
e^n&=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots\\
&<(n+1)\cdot\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*}
Thus 
$$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.$$
</s>