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How do I go about evaluating this?

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

epimorphic
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asdasd
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2 Answers2

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Recall from Stirling's Approximation that, $$n! \sim \sqrt{2\pi n} n^n e^{-n}$$ $$\therefore\lim_{n\to\infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}}=\lim_{n \to \infty} \left( \frac{\sqrt{2\pi n} n^n }{n^n e^n}\right)^{1/n}=\lim_{n\to \infty}e^{-1} (2 \pi)^{1/2n}n^{1/2n}=\frac{1}{e}$$

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Apply the theorem which says that if $a_n\to L$ then $\sqrt[n]{a_1\cdots a_n} \to L$ to the limit $(1+\frac{1}{n})^{n} \to e$.

Rene Schipperus
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  • I like this one better because proving the validity of stirling's approximation is not so fast. – Asinomás Jan 04 '16 at 17:55
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    Of course the theorem being used is also not so easy to prove, notice than it works for positive real sequences. – Asinomás Jan 04 '16 at 17:58
  • Here is a possible way to prove the theorem , you have to prove that $\sum\limits_{i=1}^n\frac{\log{a_n}}{n}$ converges to $\log(L)$. but of course $\log(a_n)$ converges to $\log(L)$ so you have reduced it to proving if $a_n\to L$ then $\frac{a_1+a_2\dots a_n}{n}\to L$ which is straightforward. – Asinomás Jan 04 '16 at 18:02
  • @dREaM The theorem is easier (much easier) than the stirling approx. And not really that hard its just $\frac{a_1+\cdots +a_n}{n}\to L$ applied to $\ln a_n$ – Rene Schipperus Jan 04 '16 at 18:04
  • yeah, I gave a small sketch of proof above, I agree this is a more direct approach. – Asinomás Jan 04 '16 at 18:06
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    @dREaM We are saying the same thing, I was writing my comment and didnt see yours before I submitted. – Rene Schipperus Jan 04 '16 at 18:08