Find $\lim_{n\rightarrow\infty}\frac{n}{(n!)^{1/n}}$

Question

I am having trouble showing $$\lim_{n\rightarrow\infty}\frac{n}{(n!)^{1/n}}=e.$$

Answer 1

Let $a_{n}=n^{n}/n!$ and then $$a_{n+1}/a_{n}=\{(n +1)^{n+1}/(n+1)!\}\{n!/n^{n}\}=\left(1+\frac{1}{n}\right)^{n}$$ which tends to $e$ and therefore $a_{n}^{1/n}$ also tends to $e$.

Answer 2

Let $\displaystyle A=\frac n{\sqrt[n] n!}=\sqrt[n]{\prod _{r=1}^n\frac n r}$

$$\implies \ln A=\frac1n\sum_{r=1}^n \ln \frac nr=-\frac1n\sum_{r=1}^n \ln \frac rn=-\int_0^1\ln x\ dx $$

as $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

Answer 3

Or, using Stirling, $n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$ so $(n!)^{1/n} \sim \frac{n}{e}$. .

Answer 4

See this is nothing but the $$\lim_{n \to \infty}\left(\frac{n^n}{n!}\right)^{\frac{1}{n}}.$$ Applying $\log$ on both sides , we will have $$\frac{1}{n }\{\log n^n-\log n!\}=-\frac{1}{n }\{\log1+\log2+..\log n-n\log n\}=-\frac{1}{n }\{\log \frac{1}{n}+\log \frac{2}{n}+...+\log \frac{n}{n}\}=-\frac{1}{n}\{\sum_{k=1}^{n}\log \frac{k}{n}\}=-\int_0^1\log x dx=-\{x\log x|_0^1-x|_0^1\}=1+\lim_{\epsilon \to 0}\epsilon\log \epsilon=1$$

So the limit is $e$.