Is $\lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}} = 0$? When I was trying to solve a problem to find the radius of convergence of the power series
$$\sum \frac{2^nz^n}{n!}$$
I fully understand that the ratio test works well in this one and the radius of convergence is $\infty$.
However, knowing that the root test gives a better span of the ratio test and it is necessary to prove it, I wanted to be able to find the radius of convergence using the ratio test.
Thus obtaining the following 
$$\begin{align}
\lim_{n \to \infty} \sqrt[n]{\frac{2^nz^n}{n!}} & = |2z|\lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}} \\
\\
& = 0\\
\\
& \lt 1 \\
\\
& \Rightarrow R = \infty
\end{align}$$
must be true.
So, I was thinking that $\lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}}$ must be equal to 0.
Is there a direct proof of this ?
I just want to be a bit more algebraically savvy. 
 A: You can use Stirling's approximation: $\displaystyle n!\approx\sqrt{2\pi n}\bigg(\frac{n}{e}\bigg)^n$ as $\displaystyle n\to\infty$, so that $\displaystyle\bigg(\frac{1}{n!}\bigg)^{\frac{1}{n}}\approx \bigg(\frac{1}{\sqrt{2\pi n}\big(\frac{n}{e}\big)^n}\bigg)^{\frac{1}{n}}=\bigg(\frac{1}{(2\pi n)^{\frac{1}{2n}}\big(\frac{n}{e}\big)}\bigg)$ which clearly goes to zero as $n\to\infty$.
A: Are you aware of the inequality:
$$\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \le \liminf_{n \to \infty} a_n^{\frac{1}{n}} \le \limsup_{n \to \infty} a_n^{\frac{1}{n}} \le \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}$$
Now just notice that if $a_n = \frac{1}{n!}$, then $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1}{n+1}$$
A: You may use the fact that $n! \geq e(n/e)^{n},\,n \geq 1$.
A: We can apply the following proposition to $u_{n}=\dfrac{1}{n!}$, a proof of which can be found in this post of mine in Portuguese. I translate it below.
Proposition. Assume that for all $n$, $u_{n}>0$ and $\lim_{n\rightarrow
\infty }\dfrac{u_{n+1}}{u_{n}}=b$. Then $$\lim_{n\rightarrow \infty }\sqrt[n]{u_{n}}=b.$$ 
Proof. $\lim_{n\rightarrow
\infty }\dfrac{u_{n+1}}{u_{n}}=b$ implies that there exists a natural number $N$ such that for $n\ge N$ we have
$$
\begin{equation*}
b-\delta <\frac{u_{N+k+1}}{u_{N+k}}<b+\delta ,\qquad 0\leq k\leq n-N-1.
\end{equation*}
$$
Multiplying these $n-N$ inequalities, we get
$$
\begin{eqnarray*}
\left( b-\delta \right) ^{n-N} &=&\prod_{k=0}^{n-N-1}\left( b-\delta \right)
<\prod_{k=0}^{n-N-1}\frac{u_{N+k+1}}{u_{N+k}}<\prod_{k=0}^{n-N-1}\left(
b+\delta \right) =\left( b+\delta \right) ^{n-N}\end{eqnarray*}
$$
Since the product $\displaystyle\prod_{k=0}^{n-N-1}\dfrac{u_{N+k+1}}{u_{N+k}}=\dfrac{u_{n}}{u_{N}}$, we thus have
$$
\begin{eqnarray*}
\left( b-\delta \right) ^{n-N} &<&\frac{u_{n}}{u_{N}}<\left( b+\delta
\right) ^{n-N}\end{eqnarray*}
$$
Multiplying by $u_N$, we get
$$
\begin{eqnarray*}
\left( b-\delta \right) ^{n-N}u_{N} &<&u_{n}<\left( b+\delta \right)
^{n-N}u_{N}.
\end{eqnarray*}
$$
Hence
$$
\begin{equation*}
\left( b-\delta \right) \sqrt[n]{\left( b-\delta \right) ^{-N}u_{N}}<\sqrt[n]{u_{n}}<\left( b+\delta \right) \sqrt[n]{\left( b+\delta \right) ^{-N}u_{N}}.
\end{equation*}
$$
Since  $\lim_{n\to \infty}\sqrt[n]{\left( b+\delta \right) ^{-N}u_{N}}=1$, there exists a natural number $N^{\prime }$ such that for $n\geq N^{\prime }$, $$1-\delta<\sqrt[n]{\left( b-\delta \right) ^{-N}u_{N}}<1+\delta$$
which means that
$$
\begin{equation*}
b-\delta -b\delta +\delta ^{2}=(1-\delta
)\left( b-\delta \right) <\sqrt[n]{u_{n}}<\left( b+\delta \right) (1+\delta
)=b+b\delta +\delta +\delta ^{2}.
\end{equation*}
$$
For $\varepsilon =\delta +b\delta +\delta ^{2}$ and $n\geq \max
\{N,N^{\prime }\}$
$$
\begin{equation*}
b-\varepsilon <\sqrt[n]{u_{n}}<b+\varepsilon ,
\end{equation*}
$$
thus proving the proposition. $\qquad\square$
For $u_{n}=\dfrac{1}{n!}$, we have
$$
\begin{equation*}
\lim_{n\rightarrow \infty }\frac{u_{n+1}}{u_{n}}=\lim_{n\rightarrow \infty }
\frac{1/(n+1)!}{1/n!}=\lim_{n\rightarrow \infty }\frac{1}{n+1}=0.
\end{equation*}
$$
Consequently,
$$
\begin{equation*}
\lim_{n\rightarrow \infty }\sqrt[n]{u_{n}}=\lim_{n\rightarrow \infty }\sqrt[n]{1/n!}=0.
\end{equation*}
$$
A: $$n! >n(n-1)..(n-\lfloor \frac{n-1}{2} \rfloor)> (\frac{n}{2})^{\frac{n}{2}}$$
Thus
$$\sqrt[n]{\frac{1}{n!}}< \frac{\sqrt2}{\sqrt{n}}$$
