Finding limit using inequalities: $\liminf \frac{a_{n+1}}{a_n} \le \liminf (a_n)^ {1/n}\le\limsup (a_n)^ {1/n}\le \limsup \frac{a_{n+1}}{a_n}$ The purpose of this exercise is to prove that $\lim \frac{n}{(n!)^{1/n}}=e$ when $n$ goes to infinity. 
In order to find the limit, the following inequality is used when $n$ goes to infinity with ${a_n}$ a sequence of positive terms: 
$$ \liminf \frac{a_{n+1}}{a_n} \le \liminf (a_n)^ {1/n}\le\limsup (a_n)^ {1/n}\le \limsup \frac{a_{n+1}}{a_n}$$
What is the proof of this inequality? 
 A: 
We will prove the right-hand side inequality $\limsup_{n\to \infty}(a_n)^{1/n}\le \limsup_{n\to \infty}\frac{a_{n+1}}{a_n}$ since we can prove the left-hand side inequality following analogously.  We will assume that $a_n\ge 0$ for all $n$ and that the $\limsup_{n\to \infty}\frac{a_{n+1}}{a_n}=L<\infty$.


If $L=\limsup_{n\to \infty}\frac{a_{n+1}}{a_n}$, then for all $\epsilon>0$, there exists a number $N>0$ such that whenever $n>N$
$$\frac{a_{n+1}}{a_n}\le \sup_{m\ge n}\frac{a_{m+1}}{a_m}\le L+\epsilon \tag 1$$
From $(1)$ we have for $n>N$
$$\frac{a_n}{a_N}=\underbrace{\left(\frac{a_{N+1}}{a_N}\right)\left(\frac{a_{N+2}}{a_{N+1}}\right)\left(\frac{a_{N+3}}{a_{N+2}}\right)\cdots \left(\frac{a_{n}}{a_{n-1}}\right)}_{n-N\,\,\text{terms}}\le (L+\epsilon)^{n-N}\tag 2$$
From $(2)$, we can write
$$(a_n)^{1/n}\le (L+\epsilon)^{1-N/n}(a_N)^{1/n} \tag 3$$
Taking the $\limsup_{n\to \infty}$ on both sides of $(3)$ reveals that for all $\epsilon>0$
$$\limsup_{n\to \infty}(a_n)^{1/n}\le L+\epsilon \tag4$$
Since $\epsilon$ is arbitrary, then $(4)$ implies
$$\limsup_{n\to \infty}(a_n)^{1/n}\le L=\limsup_{n\to \infty}\frac{a_{n+1}}{a_n}$$
which proves the right-hand side inequality. 

Let $a_n=\frac{n^n}{n!}$.  Then we see that
$$\begin{align}
\lim_{n\to \infty}\frac{a_{n+1}}{a_n}&=\lim_{n\to \infty}\frac{(n+1)^{n+1}}{n^n}\frac{n!}{(n+1)!}\\\\
&=\lim_{n\to \infty}\left(1+\frac1n\right)^n\\\\
&=e\tag 5
\end{align}$$
Note that $(5)$ implies that $\liminf_{n\to \infty}\frac{a_{n+1}}{a_n}=\limsup_{n\to \infty}\frac{a_{n+1}}{a_n}=e$.
Since $\liminf_{n\to \infty}\frac{a_{n+1}}{a_n}\le \liminf_{n\to\infty}(a_n)^{1/n}\le \limsup_{n\to\infty}(a_n)^{1/n}\le \limsup_{n\to \infty}\frac{a_{n+1}}{a_n}$, then we have
$$e=\liminf_{n\to\infty}(a_n)^{1/n}\le \limsup_{n\to\infty}(a_n)^{1/n}=e$$
which implies that 
$$\lim_{n\to\infty}(a_n)^{1/n}=\lim_{n\to \infty}\frac{n}{(n!)^{1/n}}=e$$
And we are done!
