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Lim n-> infinity (n!/n^n)1/n

Ive used the formula - Lim x->a (f(x))^g(x)=e^(Lim x->a (f(x)-1)g(x) But i keep getting one ...

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Let $A=\lim_{n\to\infty}\left(\dfrac{n!}{n^n}\right)^{1/n}$

$\displaystyle\implies\ln A=\lim_{n\to\infty}\dfrac1n\cdot\ln\prod_{r=1}^n\dfrac rn=\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\dfrac rn$

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