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I have to find $\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}$.

What I've got:

I express $n!$ as the product of the sequence $b_n=n$.

I know (from a previous problem) that $$\frac{1}{\sum_{i=1}^n\frac{1}{n}} \le \frac{\sqrt[n]{n!}}{n} \le \frac{\sum_{i=1}^n i}{n^2}=\frac{1}{2}$$

From this I know that the limit of the sequence is between $0$ and $\frac{1}{2}$ since $\sum_{i=1}^n\frac{1}{n}$ is divergent.

Second attempt:

I looked up the answer and I saw that the limit is $\frac{1}{e}$, so I tried expressing $\frac{\sqrt[n]{n!}}{n}$ as $$\sqrt[n]{\frac{(1-\frac{1}{n})(1-\frac{2}{n})...(1-1+\frac{1}{n})}{n^{n-1}}}$$ since I know that $(1-\frac{1}{n})^n$ converges to $\frac{1}{e}$, but this also didn't get me anywhere.

Thanks in advance!

Nikola
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1 Answers1

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Hint:

Use the following famous Stirling formula: Given $x>0$ $$ \lim_{x\to +\infty} \frac{\Gamma(x+1)}{\left(\frac{x}{e}\right)^x \sqrt{2x} }=\sqrt{\pi}. $$ Where $\Gamma$ is the bGamma function of Euler and $n! =\Gamma(n+1)$ for $n\in \mathbb{N}$

Guy Fsone
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