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