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Possible Duplicate:
Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$

I have a litle discution with a friend about the folowing limit: $$\lim_{n\to\infty} \frac{\sqrt[n]{n!}}{n}$$ I would solve it like this: $$\lim_{n\to\infty} \sqrt[n]{\frac{n!}{n^{n}}} =0$$ or $$\lim_{n\to\infty} \frac{\sqrt[n]{n}\sqrt[n]{n-1}\cdots\sqrt[n]{1}}{n}=\frac{1*1*1\cdots}{\infty}=0$$ and in this 2º way would there be a problem with $1^{\infty}$? I would say that no, because there is no functions involved, since as much as I know this undetermination is because you would whant to avoid the situation such as $f(x)^{g(x)}$ where $f(x)\to1$ and $g(x)\to\infty$ Could anyone clarify this for me?

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    $\begingroup$ Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$, = this question. $\endgroup$ – Martin Sleziak Oct 28 '12 at 9:54
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    $\begingroup$ You argument about $1\cdot 1\cdot 1 \cdots =1$ is (in principle) very similar to: Proof of 1=0 by mathematical induction?. $\endgroup$ – Martin Sleziak Oct 28 '12 at 9:56
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    $\begingroup$ The problem in your inference is that, you say $\sqrt[n]{t}\rightarrow1$ for any $t\in[1,n]$, However, you have in total $n$ terms, which will have a effect when $n\rightarrow\infty$ because your denominator is only $n$. $\endgroup$ – AmFCG Oct 28 '12 at 10:06
  • $\begingroup$ ok, got it. Thankyou $\endgroup$ – Mykolas Oct 28 '12 at 10:22