How to show $(1/n!)^{1/n}$ goes to $0$ as $n$ goes to infinity?

Question

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}.

Answer 1

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*}

Answer 2

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, $$\lim_{n \to \infty} \dfrac1{(n!)^{1/n}} = 0$$

Answer 3

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 $\lim_{n\to\infty} a_n=0$.

Answer 4

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 $\lfloor \frac{n}{2} \rfloor$.

Answer 5

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.

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*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.

Answer 6

One can use the nice result

$$ \lim_{n \to \infty} a_n^{1/n} = \lim_{n\to \infty}\frac{a_{n+1}}{a_n} . $$

$$ a_n = \frac{1}{n!}\implies \frac{a_{n+1}}{a_n}= \frac{n!}{(n+1)!}=\frac{1}{n+1} $$

$$\implies \lim_{n\to \infty}\frac{a_{n+1}}{a_n}=0. $$

Answer 7

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 $$ \frac{1}{n}\sum_{k=1}^n\log k\geq \frac{1}{n}\sum_{\sqrt{n} \leq k \leq n} \log k \geq \frac{n-\sqrt{n}+O(1)}{n}\times\frac{1}{2}\log n \xrightarrow[n\to\infty]{} +\infty $$

Answer 8

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")

1. 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.

1. 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).