Solve this limit usinig Stolz's theorem. Any help?!
$$\lim\limits_{n\rightarrow \infty} \frac{n}{\sqrt[n]{n!}}$$
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1See also: Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$ and the linked posts, Finding the limit of $\frac {n}{\sqrt[n]{n!}}$ and the linked posts – Martin Sleziak Aug 16 '15 at 12:13
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Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Aug 17 '15 at 08:20
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This provides another way.
Let $a_{n} > 0$ for all $n \geq 1$. Then it can be shown that $$\liminf a_{n+1}/a_{n} \leq \liminf (a_{n})^{1/n} \leq \limsup (a_{n})^{1/n} \leq \limsup a_{n+1}/a_{n}.$$ Thus if there is some $l \in \mathbb{R}$ such that $a_{n+1}/a_{n} \to l$ then $(a_{n})^{1/n} \to l$.
Let $a_{n} := n^{n}/n!$ for all $n \geq 1$. Then $(a_{n})^{1/n} = (n^{n}/n!)^{1/n}$ for all $n \geq 1$. But, since $$ \frac{a_{n+1}}{a_{n}} = \frac{(n+1)^{n+1}}{(n+1)!}\cdot \frac{n!}{n^{n}} = (1+\frac{1}{n})^{n} \to e $$ as $n$ grows, we have $(a_{n})^{1/n} \to e$ as $n$ grows.
Yes
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The proof of inequality stated in this answer can be found here and also in other posts which are linked there. – Martin Sleziak Aug 16 '15 at 17:06
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