Find $\lim_{n \to \infty} \sqrt[n]{n!}$. I am playing around with the root/ratio test to practice with series. I just showed that $\sum \frac{1}{n!}$ converges by using the ratio test. I decided to see how things would go with the root test and I got stuck at something that I can't find on google. Right away while running the root test I encountered $\lim_{x \to +\infty}\left(\frac{1}{n!}\right)^\frac{1}{n}$. I have made the claim that $\left(n!\right)^\frac{1}{n}\geq 1 \hspace{3mm}\forall n\in \mathbb{N}.$ I have begun the proof and I think it is on the right track but some verification would be nice.
Proof:
We can see that for n=1, this obviously holds, $\left(1!\right)^1\geq 1.$ Now suppose that this is the case for some $n=k$, then we have $\left((k+1)!\right)^{\frac{1}{k+1}}= \left((k+1)k!\right)^\frac{1}{k+1}$. It is at this point that I start to have a little trouble. Can somebody give me a push in the right direction? Thanks!
 A: An alternative simple way to show that $\lim_{n\to \infty} \sqrt[n]{\frac1{n!}}=0$ may be as follows.  Recall that for any sequence $\{c_n\}$ of positive numbers, we have 
$$\limsup_{n\to \infty} \sqrt[n]{c_n}\le \limsup_{n\to \infty} \frac{c_{n+1}}{c_n}.$$
Now with $c_n\triangleq \frac{1}{n!}$, clearly $\limsup_{n\to \infty} \frac{c_{n+1}}{c_n}=0$, which implies that $\limsup_{n\to \infty} \sqrt[n]{c_n}=0.$  Therefore $\lim_{n\to \infty} \sqrt[n]{c_n}=0,$ since it's obvious that $\liminf_{n\to \infty} \sqrt[n]{c_n}\ge 0.$
A: $$
\lim_{n \to \infty}{-\ln\left(\Gamma\left(n + 1\right)\right) \over n}
=
-\lim_{n \to \infty}\Psi\left(n + 1\right)
=
-\infty
$$
$$
\color{#ff0000}{\lim_{n \to \infty}\left(1 \over n!\right)^{1/n} = 0}
$$
$\Gamma$ and $\Psi$ are the Gamma and Digamma functions, respectively.

$\mbox{Or Stirling dixit:}$
\begin{align}
\lim_{n \to \infty}\left(1 \over n!\right)^{1/n}
& =
\lim_{n \to \infty}\left(1 \over \sqrt{2\pi\,}\,n^{n + 1/2}\,
{\rm e}^{-n}\right)^{1/n}
=
{1 \over \sqrt{2\pi\,}\,}\,
\lim_{n \to \infty}{{\rm e} \over n^{1 + 1/2n}}
\\[5mm] & = 
{{\rm e} \over \sqrt{2\pi\,}\,}\,
\lim_{n \to \infty}\left(1 \over n\right)
=
\color{#ff0000}{0}
\end{align}
A: I'm not sure it holds in general. Consider $n! = n(n-1)(n-2)...2\cdot1$, at least $n/2$ terms are $>n/2$ thus;
$$n! > \left(\frac{n}{2}\right)^\frac{n}{2}$$
taking the reciprocal
$$ \frac{1}{n!} < \left(\frac{2}{n}\right)^\frac{n}{2}$$
then raising to the power $1/n$ gives
$$ \left(\frac{1}{n!}\right)^\frac{1}{n} < \left(\frac{2}{n}\right)^\frac{1}{2}$$
clearly the right hand side has $\lim_{n\rightarrow \infty} \sqrt{2/n} = 0$ thus
$$\lim_{n\rightarrow \infty} \left(\frac{1}{n!}\right)^\frac{1}{n} = 0$$
Edit: of course njguliyev's comment is a much easier way to see it can't be $\geq1$.
A: The question seems to be about $n!^{1/n}\ge 1$. This is fairly obvious, as $n!\ge1$. There is no need for induction.
A: For an elementary approach
which gives more information,
you can start with
$(1+1/n)^n
\lt e
\lt (1+1/n)^{n+1}
$.
This is proved by showing that
$(1+1/n)^n$ is an
increasing function of $n$
and
$(1+1/n)^{n+1}$ is an
decreasing function of $n$
and that they have a common limit,
good old $e$.
Using this and induction,
you can show that
$(n/e)^n \lt n! < (n/e)^{n+1}
$
so that
$n/e \lt (n!)^{1/n} < (n/e)(n/e)^{1/n}
$.
Since both $n^{1/n}$ and $a^{1/n}$
approach $1$ as $n \to \infty$,
$\dfrac{(n!)^{1/n}}{n}
\to \dfrac1{e}
$.
Here are the details.

Here's the proof that
$(1+1/n)^n
\lt e
\lt (1+1/n)^{n+1}
$.
Let
$a_n = (1+1/n)^n$
and $b_n = (1+1/n)^{n+1}
$.
We will prove that
$a_n$ is an increasing sequence
and
$b_n$ is an decreasing sequence.
Since $a_n < b_n$,
this implies,
for any positive integers $n$ and $m$
with $m < n$ that
$a_m < a_n < b_n < b_m$.
We use the very ingenious proof in [1],
which uses the
arithmetic-geometric mean inequality
(AGMI),
which we will use in the form
$((v_1+v_2+...v_n)/n)^n 
\gt v_1v_2...v_n
$ (all $v_i$ positive)
with equality if and only if
all the $v_i$ are equal.
For $a_n$,
consider $n$ values of $1+1/n$
and $1$ value of $1$.
By the AGMI,
$((n+2)/(n+1))^{n+1} > (1+1/n)^n
$,
or $(1+1/(n+1))^{n+1} > (1+1/n)^n
$,
or $a_{n+1}
\gt a_n
$.
For $b_n$,
consider $n$ values of $1-1/n$
and $1$ value of $1$.
By the AGMI,
$(n/(n+1))^{n+1}
\gt (1-1/n)^n$
or
$(1+1/n)^{n+1} < (1+1/(n-1))^n
$,
or
$b_n < b_{n-1}
$
so
$b_{n+1} < b_{n}
$.
Since
$b_n-a_n
=(1+1/n)^{n+1}-(1+1/n)^n
=(1+1/n)^{n}(1+1/n-1)
=(1+1/n)^{n}/n
\to 0
$,
$a_n$ and $b_n$
have a common limit,
which I will call $e$.

Here's the proof that
$(1+1/n)^n
\lt e
\lt (1+1/n)^{n+1}
$
implies that
$(n/e)^n
\lt n!
\lt (n/e)^{n+1}
$.
Suppose
$n! > (\dfrac{n}{e})^n
$.
We want to show that
$(n+1)! > (\dfrac{n+1}{e})^{n+1}
$.
Then
$\begin{array}\\
(n+1)!
&=(n+1)n!\\
&>(n+1) (\dfrac{n}{e})^n\\
&=\dfrac{(n+1)n^n}{e^n}\\
\end{array}
$
so we want
$\dfrac{(n+1)n^n}{e^n}
\gt \dfrac{(n+1)^{n+1}}{e^{n+1}}
$
or
$e \gt \dfrac{(n+1)^n}{n^n}
=(1+1/n)^n
$
which is shown above.
Similarly,
suppose
$n! < (\dfrac{n}{e})^{n+1}
$.
We want to show that
$(n+1)! < (\dfrac{n+1}{e})^{n+2}
$.
Then
$\begin{array}\\
(n+1)!
&=(n+1)n!\\
&<(n+1) (\dfrac{n}{e})^{n+1}\\
&=\dfrac{(n+1)n^{n+1}}{e^{n+1}}\\
\end{array}
$
so we want
$\dfrac{(n+1)n^{n+1}}{e^{n+1}}
\lt \dfrac{(n+1)^{n+2}}{e^{n+2}}
$
or
$e \lt \dfrac{(n+1)^{n+1}}{n^{n+1}}
=(1+1/n)^{n+1}
$
which is shown above.
[1] N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58
(1951), 563.
A: We have that
$$\sqrt[n]{n!}=e^{\frac{\log (n!)}n} \to \infty$$
indeed by Stolz-Cesaro theorem
$$\frac{\log ((n+1)!)-\log(n!)}{n+1-n}=\log (n+1) \to \infty \implies \frac{\log (n!)}n \to \infty$$
