# Limit of $\frac{1}{\sqrt[n]{n!}}$ as $n$ approaches infinity [duplicate]

So i was trying to evalue this limit: $$\lim_{n \to \infty}\frac{1}{\sqrt[n]{n!}}, n \in \mathbb{N}$$

This, of course, by common sense is equal to zero (since factorial grows a lot faster). Is there a way to prove this limit without having to tackle with proving function growth rate. I;m not sure how that would be done, but I believe i would have to expand the factorial function to set $\mathbb{R}$ in order to compare, and that's still beyond my abilities. Thanks.

## marked as duplicate by Hans Lundmark, user147263, Jack D'Aurizio, Lord_Farin, Ali CaglayanFeb 11 '15 at 22:09

• Stirling would be a good idea, no? – servabat Feb 11 '15 at 19:12
• Can you solve it w/out the factorial? I mean if you can prove it goes to zero for that case, then it'd be true as well for n-1 under the root as approaching infinity, and so on. Apologies if I'm way off the mark, it's been a few decades since I played w/ limits. – Eric Feb 11 '15 at 19:13
• With this search you will find that this has been asked several times before: math.stackexchange.com/search?q=%5Blimits%5D+factorial+sqrt – Hans Lundmark Feb 11 '15 at 19:17
• And now, everybody, there's no need to keep adding new answers here, since there are already 13(!) answers in the question that I linked to... – Hans Lundmark Feb 11 '15 at 19:20
• @Eric: $\sqrt[n]{n} \to 1$, so that doesn't work. – Hans Lundmark Feb 11 '15 at 19:21

You can use a well-known lower bound for factorials:

$$n! ~~ \geq ~~ \sqrt{2\pi}\ n^{n+1/2}e^{-n} ~~ \geq ~~ n^{n+1/2}e^{-n} ~~ \geq ~~ n^{n}e^{-n}$$

Now simplify the expression you get when substituting the above into $\sqrt[n]{n!}$:

$$( n^{n}e^{-n})^\frac{1}{n} = \frac{n}{e}$$

use the series $e^x = 1 + \dfrac{x}{1!} + \cdots + \frac{x^n}{n!} + \cdots$ to get $$e^n > \dfrac{n^n}{n!}$$ which gives you $$\left(\frac{1}{n!}\right)^{1/n} < \frac{e}{n}$$ and letting $n \to \infty$ gives you the limit zero.

Let $k =\lfloor \frac{n}{2} \rfloor$ that is $n=2k$ or $n=2k+1$. Then $k \geq \frac{n-1}{2}$ and

$$n! \geq k(k+1) ... n \geq k^k\geq \left(\frac{n-1}{2} \right)^\frac{n-1}{2}$$

Therefore, for $n \geq 3$

$$0 \leq \frac{1}{\sqrt[n]{n!}} \leq \frac{1}{\left(\frac{n-1}{2} \right)^\frac{n-1}{2n}}\leq \frac{1}{\left(\frac{n-1}{2} \right)^\frac{1}{3}}$$

If we prove that $n!>\left(\frac{n}{e}\right)^n$, the limit is trivially zero.

Now notice that, by partial summation: $$\log\left(n!\right)=\sum_{k=1}^{n}\log k = n\log n-\sum_{k=1}^{n-1}\log\left(1+\frac{1}{k}\right)\geq n\log n-\sum_{k=1}^{n-1}\frac{1}{k}\geq n\left(\log n-1\right)$$ and the inequality is proved.