Intermediate steps in showing that $ \lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}= \lim_{n\to\infty}(1+\frac{1}{n})^{n}$ I would like to understand why $ \lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}= \lim_{n\to\infty}(1+\frac{1}{n})^{n}$.
I was given a solution, but it gives no further details than 
$$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}=\lim_{n\to\infty}\sqrt[n]{\frac{n^n}{n!}}=\lim_{n\to\infty}\frac{(n+1)^{(n+1)}/(n+1)!}{n^n/n!}=\lim_{n\to\infty}(1+\frac{1}{n})^n.$$
What happened between $\lim_{n\to\infty}\sqrt[n]{\frac{n^n}{n!}}$ and $\lim_{n\to\infty}\frac{(n+1)^{(n+1)}/(n+1)!}{n^n/n!}$?
 A: If $a_n>0$ and there exists limit $\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}$, then
$$ \lim_{n\rightarrow\infty} \sqrt[n]{a_n} = \lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}$$
You can check that it's true as follows:
If $\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = g > 0$ then for every $\epsilon>0$, $\epsilon <g$ there exists $N$ such that for $n\ge N$:
$$ g-\epsilon \le \frac{a_{n+1}}{a_n} \le g+\epsilon $$
$$ (g-\epsilon)^{n-N} \le \frac{a_{N+1}}{a_N}\frac{a_{N+2}}{a_{N+1}}\dots\frac{a_n}{a_{n-1}} = \frac{a_n}{a_N} \le (g+\epsilon)^{n-N}$$
$$ \sqrt[n]{a_N (g-\epsilon)^{n-N}} \le \sqrt[n]{a_n} \le \sqrt[n]{a_N (g+\epsilon)^{n-N}} $$
$$ g-\epsilon \le \lim_{n\rightarrow\infty}\sqrt[n]{a_n} \le g+\epsilon $$
Since $\epsilon$ is arbitrary, it means that $\lim_{n\rightarrow\infty}\sqrt[n]{a_n}=g$.
If $g=0$ the proof is similar, but the lower bound is just $0$.
In your problem $a_n = \frac{n^n}{n!}$.
A: Taking logs and using Stolz-Cesàro:
$$
\log\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}=
\lim_{n\to\infty}\log\frac{n}{\sqrt[n]{n!}}=
\lim_{n\to\infty}\frac{n\log n -(\log 1 + \cdots + \log n)}n =
$$
$$
\lim_{n\to\infty}(n + 1)\log(n + 1) - n\log n - \log(n + 1) =
\lim_{n\to\infty}n\log\frac{n + 1}n = \log\lim_{n\to\infty}\left(1 + \frac 1n\right)^n.
$$
A: Let,$A=\frac{n}{{n!}^{1/n}}$,So $logA=\frac{1}{n}\sum_{r=1}^n log(\frac{n}{r})=\frac{1}{n}\sum_{r=1}^n log(\frac{1}{\frac{r}{n}})$
So as $n\to \infty$, $logA$ grows to $\int_{0}^{1} log(\frac{1}{x})dx=1$,hence $A$ grows to $exp(1)=e$
And we know that $e=lim_{n\to \infty} (1+\frac{1}{n})^{n}$
A: Well, as $n \rightarrow \infty$ we have via Stirling's approximation that:
 $$n! \sim \bigg(\frac{n}{e}\bigg)^n \cdot \sqrt{2 \pi n}$$
By plugging this in, and noting that $(2\pi n)^{\frac{1}{2n}}$ converges to $1$ when $n \rightarrow \infty$ we arrive at $\frac{n}{\frac{n}{e}} = e$ which is obviously equaly to the RHS limit. Hope this helps
