How to show $(1/n!)^{1/n}$ goes to $0$ as $n$ goes to infinity? How do I show $(1/n!)^{1/n}$ goes to $0$ as $n$ goes to infinity? I need this to use the spectral radius theorem to show an operator has spectrum {0}.
 A: You can show it in several ways, for example you can use that 
$$\lim_{n\to \infty} n \cdot \sqrt[n]{\frac{1}{n!}}= e$$
by elementary integration or by using that the limit of
$$\lim_{n\to \infty} \sqrt[n]{\frac{n^n}{n!}}=\lim_{n\to \infty} \frac{\frac{(n+1)^{n+1}}{(n+1)!}}{\frac{n^n}{n!}}=\frac{(n+1)^n}{n^n}=\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n=e$$
When we prove that $$\lim_{n\to \infty} n \cdot \sqrt[n]{\frac{1}{n!}}=e$$ 
we know that your limit must be zero, else $n \cdot \dots$ couldn't be bounded.
Proving $$\lim_{n\to \infty} \sqrt[n]{\frac{n^n}{n!}}=e$$ is equal to proving 
$$\lim_{n\to \infty}\sqrt[n]{\frac{n!}{n^n}} = \frac{1}{e}$$.
\begin{align*}
\lim_{n\rightarrow\infty} \sqrt[n]{\frac{n!}{n^n}}&=\frac{1}{e}\\
\iff \ln\left(\lim_{n\rightarrow \infty} \sqrt[n]{\frac{n!}{n^n}}\right)&=
\ln\left(\frac{1}{e}\right)=-1
\end{align*}
To show this one we make the following
\begin{align*}
\ln\left(\lim_{n\rightarrow \infty} \sqrt[n]{\frac{n!}{n^n}}\right)&=
\lim_{n\rightarrow \infty} \ln\left(\sqrt[n]{\frac{n!}{n^n}}\right)\\
&=\lim_{n\rightarrow \infty} \frac{1}{n} \ln\left(\frac{n!}{n^n}\right)\\
&=\lim_{n\rightarrow \infty} \frac{1}{n} \ln\left(\prod_{i=1}^n \frac{i}{n}\right)\\
&=\lim_{n\rightarrow \infty} \frac{1}{n} \left(\sum_{i=1}^n \ln\left(\frac{i}{n}\right)
\right)\\
&=\lim_{n\rightarrow \infty} \sum_{i=1}^n \frac{1}{n} \ln\left(\frac{i}{n}\right)\\
&=\int_0^1 \ln(x) \, dx\\
&=\lim_{\varepsilon \to 0} \int_\varepsilon^1 \ln(x) \, \mathrm{d}x\\
&= \lim_{\varepsilon \to 0} -1 + 1 \ln(1) - (-\varepsilon + \varepsilon \ln\varepsilon)\\
&= -1 +\lim_{\varepsilon \to 0} \varepsilon \ln\varepsilon\\
&=-1
\end{align*}
A: The function $e^x$ is an entire function and hence the Taylor series of $$e^x = 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!} + \cdots $$ has radius of convergence as $\infty$. Hence,
$$\lim_{n \to \infty} \left(\dfrac{x^n}{n!} \right)^{1/n} < 1, \,\,\,\,\, \forall x \in \mathbb{R} \implies \lim_{n \to \infty} \left(\dfrac{x}{n!^{1/n}} \right) < 1, \,\,\,\,\, \forall x \in \mathbb{R}$$
Hence,
$$\lim_{n \to \infty} \dfrac1{(n!)^{1/n}} = 0$$
A: One can use the nice result
$$ \lim_{n \to \infty} a_n^{1/n} = \lim_{n\to \infty}\frac{a_{n+1}}{a_n} . $$
$$ a_n = \frac{1}{n!}\implies \frac{a_{n+1}}{a_n}= \frac{n!}{(n+1)!}=\frac{1}{n+1} $$
$$\implies \lim_{n\to \infty}\frac{a_{n+1}}{a_n}=0.  $$
A: Note that if $n=2k$ then 
$$n! \geq k(k+1)..(2k) \geq k k k ... k =k^{k+1} \geq \left( \frac{n-1}{2} \right)^{\frac{n}{2}}$$
while if $n=2k+1$  
$$n! \geq k(k+1)..(2k)(2k+1) \geq k k k ... k =k^{k+2} \geq \left( \frac{n-1}{2} \right)^{\frac{n}{2}}$$
Thus, for all $n$ we have $n!> \left( \frac{n-1}{2} \right)^{\frac{n}{2}}$.
Hence 
$$0 < \frac{1}{\sqrt[n]{n!}}\leq \frac{1}{\sqrt[n]{\left( \frac{n-1}{2} \right)^{\frac{n}{2}}}}=\frac{\sqrt{2}}{\sqrt{n-1}}$$
Remark Both cases at the beginning can be studied at once if instead of $k$ you write $\lfloor \frac{n}{2} \rfloor$.
A: An elementary inequality* regarding the factorial is
$$\left(\frac n e\right)^n \le n! \le n^n$$
So,
$$\left(\frac 1 {n!}\right)^{1/n} \le \frac e n \to 0$$
giving the required result by the sandwich rule.

*Proof of inequality: The upper bound (which I did not use) follows trivially from the definition of the factorial. The lower bound follows from a quick and dirty evaluation of the Gamma function integral,
$$n! = \Gamma(n+1) = \int_0^\infty e^{-t} t^n dt \geq \int_n^\infty e^{-t} t^n dt \geq \int_n^\infty e^{-t} n^n dt = \left( \frac n e \right)^n$$
This is a useful bound of the factorial to know - it's weaker than the Stirling approximation, but much, much easier to prove.
A: 1.
Consider the power series of $e^x$: $\sum \frac{x^k}{k!}$. 
Plug $x=n$ and take only the $n$'th term (the others are positive): $e^n > \frac{n^n}{n!}$, which is equivalent to $\frac{1}{n!} < (\frac{e}{n})^n$. Take the $n$'th root.
(Note: I've seen this trick in some books, among them Ireland and Rosen's "A Classical Introduction to Modern Number Theory")


*

*Another way to derive the inequality is by integral:
$$\ln(n!) = \sum_{i=2}^{n} \ln i \ge \int_{1}^{n} \ln x dx = (x\ln x - x)|_{x=1}^{n} = n\ln n -n +1 \implies$$
$$n! > e(\frac{n}{e})^{n}$$
Since $\ln x$ is increasing.


*

*A third was is a complex integral. Note that $\frac{1}{n!}$ is the $n$'th coefficient of $e^x$, so:
$$\frac{1}{n!} = \frac{1}{2\pi i} \int_{C} \frac{e^z}{z^{n+1}}dz$$
When the integral is over a circle of radius $n$ and center at the origin. Now just use the parametrization $z = ne^{i\theta}$:
$$\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{e^{ne^{i\theta}}}{n^n e^{in\theta}}dz$$
Now just bound the integrand from above by $\frac{e^n}{n^n}$.
4.
All of this is an overkill, though. It follows from the fact that $n!$ grows faster then any exponential function (see N.S.'s proof).
A: Using the AGM-inequality and looking at the graph of $x\mapsto {1\over x}$ we see that
$$0<a_n:=\left({1\over n!}\right)^{1/n}\leq {1\over n}\sum_{k=1}^n{1\over k}\leq {1\over n}\left(1+\int_1^n{1\over t}\ dt\right)={1+\log n\over n}\qquad(n\geq 1)\ .$$
It follows that $\lim_{n\to\infty} a_n=0$.
A: It's a simple exercise to show $n! \ge (n/2)^{n/2}.$ Thus
$$0 \le \left (\frac{1}{n!}\right ) ^{1/n} \le \left (\frac{1}{(n/2)^{n/2}}\right ) ^{1/n} = \frac{1}{(n/2)^{1/2}} \to 0.$$
Hence the desired limit is $0.$
A: Another elementary proof
We know that for every $a > 0$, we have $a^n \ll n!$.
Hence,
$$
\left(\frac{1}{n!}\right)^\frac{1}{n} = \left(\frac{1}{a^n}\right)^{\frac{1}{n}}\times\left(\frac{a^n}{n!}\right)^{\frac{1}{n}}
$$
yields
$$
\limsup_{n\to\infty} \left(\frac{1}{n!}\right)^\frac{1}{n} \leq \frac{1}{a}\times1
$$
Finally, take $a \to \infty$.
Yet another one
$$
\left(\frac{1}{n!}\right)^\frac{1}{n} = \exp\left(-\frac{1}{n}\sum_{k=1}^n \log k
\right)$$
and
$$
\frac{1}{n}\sum_{k=1}^n\log k\geq \frac{1}{n}\sum_{\sqrt{n} \leq k \leq n} \log k \geq \frac{n-\sqrt{n}+O(1)}{n}\times\frac{1}{2}\log n \xrightarrow[n\to\infty]{} +\infty
$$
A: Note that 
$$(1/n!)^{1/n}=e^ \frac{-\log n!}{n}\to e^{-\infty} =0$$
indeed by Stolz-Cesaro
$$\lim_{n\to +\infty}\frac{\log n!}{n} =\lim_{n\to +\infty}\frac{\sum_{k=1}^n\log k}{n}=\lim_{n\to +\infty}\frac{\sum_{k=1}^{n+1}\log k-\sum_{k=1}^{n}\log k}{n+1-n}=$$
$$=\lim_{n\to +\infty} \log (n+1)=+\infty$$
