How can one prove $\lim \frac{1}{(n!)^{\frac 1 n}} = 0$?

Question

I have tried bounding the terms by $\dfrac 1 {2^{\frac 1 n}}$, but this clearly cannot be made as small as possible.

Answer 1

Elementary solution: To calculate $n!$, instead of multiplying the numbers from $1$ to $n$, multiply $(1 * n) * (2 * (n - 1)) * (3 * (n - 2)) ...$ Each product is greater than or equal to $n$; there may be one number left over in the middle which is greater than $\sqrt n$.

That makes $n! \geq (\sqrt n)^n$, and the limit follows immediately.

Or you could multiply only the larger half of the numbers, and $n! > (n/2)^{n/2}.$

Answer 2

By Stirling's formula,

$$ n!\sim_{n\rightarrow+\infty}\sqrt{2\pi n}\left(\frac ne\right)^n, $$

so

$$ (n!)^{-1/n}=\left(\frac{\sqrt{2\pi n}\left(\frac ne\right)^n}{n!}\right)^{1/n}(2\pi n)^{-1/2n}\frac en =\left(\frac{\sqrt{2\pi n}\left(\frac ne\right)^n}{n!}\right)^{1/n}\exp\left(-\frac{1}{2n}\ln(2\pi n)\right)\frac en \xrightarrow[n\rightarrow+\infty]{}0, $$

since $-\frac{1}{2n}\ln(2\pi n)$ tends to $0$ as $n$ tends to infinity, and $\left(\frac{\sqrt{2\pi n}\left(\frac ne\right)^n}{n!}\right)^{1/n}$ tends to 1 as $n$ tends to infinity.

Answer 3

Use Stirling's approximation and find for $n\rightarrow \infty$ $$\frac{1}{(n!)^{\frac 1 n}} = \frac{e}{n} + O(\ln(n)n^{-2})$$

Answer 4

This can also be done directly without using Stirlings formula.

Denoting $a_n = \frac{1}{(n!)^{1/n}}$ we have $\log a_n = -\frac{\log n!}{n} = -\log n -\frac{1}{n}\sum_{i=1}^n \log (i/n)$

which gives

$\log(a_n n) =-\frac{1}{n}\sum_{i=1}^n \log (i/n)$

The term on the rhs is a Riemann sum for the integral $\int_1^0\log(x)dx = 1$. Thus for ${n\to \infty}$ we have $\log(a_n n) \to 1$ and therefore $a_n \to 0$. We can also read off the asymptotic behavior $a_n \sim \frac{e}{n}$.

Answer 5

Starting like Winther, but using Cesaro-Stolz: $$ \lim_{n\to\infty}\log a_n = \lim_{n\to\infty}-\frac{\log n!}{n} = -\lim_{n\to\infty}\frac{\log 1 + \log 2 + \cdots + \log n}{n} = $$

$$ =-\lim_{n\to\infty}\frac{\log 1 + \log 2 + \cdots + \log n}{n} =-\lim_{n\to\infty}\frac{\log(n+1)}{(n+1)-n} =-\lim_{n\to\infty}\log(n+1)=-\infty $$ and $$ \lim_{n\to\infty}a_n =0.$$

Answer 6

Let $(a_n)$ a bounded sequence of positive real numbers. We know that

$$\liminf\frac{a_{n+1}}{a_n} \le \liminf \sqrt[n]{a_n} \le \limsup\sqrt[n]{a_n} \le\limsup\frac{a_{n+1}}{a_n}$$

Let $a_n=\frac{1}{n!}$. Since $\liminf\frac{a_{n+1}}{a_n}=\limsup\frac{a_{n+1}}{a_n}=0$, we have $\lim_{n \to \infty} \sqrt[n]{a_n}=0 $

Answer 7

$$ \frac{n^n}{n!} \le \sum_{k=0}^\infty \frac{n^k}{k!} = e^n $$ $$ \therefore \frac1{(n!)^{1/n}} \le \frac en \to 0 $$

Answer 8

By the AM-GM inequality and the Harmonic sum approximation,

$${1\over (n!)^{1\over n}}=\sqrt[n]{{1\over1}\cdot{1\over2}\cdot\cdots\cdot{1\over n}}\le{1\over n}\left({1\over1}+{1\over2}+\cdots+{1\over n}\right)\approx{\ln n\over n}\to0$$

Answer 9

Here's a very crude but (I think) natural method, which doesn't require any really slick ideas or big results.

Unpacking definitions, we see that we're trying to show $$ \forall m : \exists N : \forall n\ge N :\quad n!\ge m^n $$ Since $m^n$ consists of $n$ factors, all $m$, and $n!$ consists of $n$ factors, some smaller than $m$ and some larger, it seems like we want the largeness of the larger-than-$m$ factors in $n!$ to make up for there being not quite $n$ of them. And indeed, $$ n! = \prod_{k=1}^n k \ge \prod_{k=m+1}^n k \ge (m+1)^{n-m} \ge m^n $$ where the (desired) final inequality is equivalent to $$ \left(1+\frac1m\right)^n \ge (m+1)^m $$ which is true for sufficiently large $n$, since the LHS tends to $\infty$ as $n\to\infty$ (with $m$ fixed).