Find $\lim_{n\to \infty}\frac{n}{\sqrt[n]{n!}}$ Find the limit for $\lim_{n\to \infty}\frac{n}{\sqrt[n]{n!}}$
I have no idea how to do this - they always say use the ratio test when there are factorials but the nth root stuff makes it so ugly and i can't get anything to come out of it. Does anyone know what process I should use?
 A: Result: If $\lim_{n\to \infty}{\frac{a_{n+1}}{a_n}}=l$ then $\lim_{n\to \infty}{\sqrt[n]{a_n}}=l$
Take $(a_n)^{1/n}=\left(\frac{n^n}{n!}\right)^{1/n}$ Now consider $a_n=\frac{n^n}{n!}$ and $a_{n+1}=\frac{(n+1)^{n+1}}{(n+1)!}$.
$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\frac{(n+1)^n}{n^n}=\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$$ hence $$\lim_{n\to \infty}\frac{n}{\sqrt[n]{n!}}=e$$
A: Let $x_n = \frac{n}{\sqrt[n]{n!}}$
$\log(x_n) = \log(n) - \dfrac{\log(n!)}{n} = \dfrac{n\log(n) - \sum_{k=1}^n \log(k)}{n}$
by Stolz–Cesàro theorem, it has the same limit as $(n+1)\log(n+1) - \log(n+1) - n\log(n) = \log\left((1+\dfrac{1}{n})^n\right) \to \log(e) = 1$
A: We have
$$
\frac{n}{\sqrt[n]{n!}}=\left(\frac{n^n}{n!}\right)^{1/n}=\exp\left(\frac1n\,\log\left(\frac{n^n}{n!}\right)\right)
=\exp\left(\frac1n\,n\log n-\frac1n\,\log n!\right)
=\exp\left(\log n-\frac1n\,\sum_{k=2}^n\log k\right).
$$
Now we have
$$
\int_1^{n-1}\log t\,dt\leq\sum_{k=2}^n\log k\leq\int_1^n\log t\,dt,
$$
i.e.
$$(n-1)\log n-(n-2)\leq\sum_{k=2}^n\log k\leq n\log n - (n-1).$$
Then
$$\log n-\log n+\frac{n-1}n\leq\log n-\frac1n\,\sum_{k=2}^n\log k\leq\log n-\frac{n-1}n\,\log n+\frac{n-2}n,$$
or
$$1-\frac{1}n\leq\log n-\frac1n\,\sum_{k=2}^n\log k\leq-\frac{1}n\,\log n+1-\frac2n.$$
Thus
$$
\lim_{n\to\infty}\log n-\frac1n\,\sum_{k=2}^n\log k=1,
$$
and
\begin{align}
\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}&=\lim_{n\to\infty}\exp\left(\log n-\frac1n\,\sum_{k=2}^n\log k\right)\\ &=\exp\left(\lim_{n\to\infty}\log n-\frac1n\,\sum_{k=2}^n\log k\right)\\ &=\exp(1)=e.
\end{align}
A: Using an asymptotic expansion of $\ln(n!)$, obtained e.g. as in this post (there, by a comparison series/integral):
$$\ln(n!) \operatorname*{=}_{n\to\infty} n\ln n - n +o(n)$$
you get
$$
\sqrt[n]{n!} = e^{\frac{1}{n}\ln(n!)} = e^{\frac{1}{n}\sum_{k=1}^n \ln k} = e^{\frac{1}{n}\left( n\ln n - n + o(n)\right)}
=
e^{\ln n - 1 + o(1)} = n\cdot e^{-1+o(1)}
$$
and therefore
$$
\frac{n}{\sqrt[n]{n!}} = e^{1+o(1)} \xrightarrow[n\to\infty]{} e.
$$
