# 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!

• Ok, now your inequality is true, but you don't need to use mathematical induction: $a>1,b>0$ imply $a^b>1$. But this inequality is not enough for convergence of the series. See the answer below for details. – njguliyev Oct 4 '13 at 9:23

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 $\geq1$.

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 $\liminf_{n\to \infty} \sqrt[n]{c_n}\ge 0.$

$$\lim_{n \to \infty}{-\ln\left(\Gamma\left(n + 1\right)\right) \over n} = -\lim_{n \to \infty}\Psi\left(n + 1\right) = -\infty$$

$$\color{#ff0000}{\lim_{n \to \infty}\left(1 \over n!\right)^{1/n} = 0}$$

$\Gamma$ and $\Psi$ are the Gamma and Digamma functions, respectively.

$\mbox{Or Stirling dixit:}$ $$\lim_{n \to \infty}\left(1 \over n!\right)^{1/n} = \lim_{n \to \infty}\left(1 \over \sqrt{2\pi\,}\,n^{n + 1/2}\, {\rm e}^{-n}\right)^{1/n} = {1 \over \sqrt{2\pi\,}\,}\, \lim_{n \to \infty}{{\rm e} \over n^{1 + 1/2n}} = {{\rm e} \over \sqrt{2\pi\,}\,}\, \lim_{n \to \infty}{1 \over n} = \color{#ff0000}{0}$$

• Yeah, the red coloring and the font sizes were really not necessary. Focus your energy on the answer, just a tip like that. Otherwise you'll spend a loooooooot of time on this website. – Patrick Da Silva Oct 4 '13 at 19:25