Prove $\lim_{n\to \infty}{\frac{(n!)^{1\over n}}{n}=\frac{1}{e}}$ by integration

Question

I have got a problem which is to prove the following for all positive integers $n\ge 2$ :
$$\lim_{n\to \infty}{\frac{(n!)^{1\over n}}{n}}=\frac{1}{e}$$ by using $$\int_{1}^{n}{\ln{x}}dx<\sum_{r=2}^{n}{\ln{r}}<\int_{1}^{n}{\ln{x}}dx+\ln{n}$$

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I know that $$\ln\frac{(n!)^{1\over n}}{n}=\frac{1}{n}\sum_{r=2}^{n}{\ln{r}}-\ln{n}$$Then $$\frac{1}{n}\int_{1}^{n}{\ln{x}}dx-\ln{n}<\frac{1}{n}\sum_{r=2}^{n}{\ln{r}}-\ln{n}<\frac{1}{n}\int_{1}^{n}{\ln{x}}dx$$ Then I know the limit as $n\to \infty$ of $\frac{1}{n}\int_{1}^{n}{\ln{x}}dx-\ln{n}=-1$.
But I dont know how to continue because the right side of the inequality tends to infinity as n tends to infinity. Any help is appreciated. Thanks!

Answer 1

HINT:

Another approach : Let $A=\dfrac{(n!)^{1/n}}n$

$$\ln A=\dfrac1n\sum_{r=1}^n\ln\dfrac rn$$

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

See also :

The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$,

Find $\lim\limits_{n \to \infty} \frac{1}{n}\sum\limits^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$