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I have tried to show that this limit : $$\lim\limits_{n\to \infty }\frac{n}{n!^{\frac 1 n}}=e$$

using $ \lim (1+\frac 1 n)^{\frac 1 n} , n \to \infty $ , I don't find any equivalence , however wolfram alpha says that is $e$ as shown here, then how do I evaluate it ?

amWhy
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zeraoulia rafik
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    That's not good enough. You need at least a weak form of Stirling's formula. – Ted Shifrin Apr 12 '18 at 17:44
  • As per @DeepSea's answer, this trick is useful https://math.stackexchange.com/questions/2118515/if-f-is-riemann-integrable-on-0-1-then-lim-frac1n-sum-fk-n-int/ – rtybase Apr 12 '18 at 18:08
  • @zeraouliarafik Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Apr 14 '18 at 14:13
  • This is essentially the same as https://math.stackexchange.com/questions/2685822/how-can-we-prove-that-l-lim-n-to-infty-frac-log-left-fracnnn-rig – Arnaud D. Sep 17 '18 at 10:01

3 Answers3

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As an alternative without Stirling, note that

$$\frac{n}{n!^{\frac 1 n}}=\left(\frac{n^n}{n!}\right)^{\frac 1 n}=(a_n)^\frac1n$$

and

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}=\left(1+\frac1n\right)^n\to e$$

then by ratio-root criterion $\frac{n}{n!^{\frac 1 n}}\to e$.

user
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6

We have: $\displaystyle \lim_{n \to \infty} -\dfrac{1}{n}\displaystyle \sum_{k=1}^n \ln \left(\dfrac{k}{n}\right)= -\displaystyle \int_{0}^1\ln xdx=1\implies \displaystyle \lim_{ n\to \infty} \ln\left(\dfrac{n}{n!^{1/n}}\right)=1\implies \displaystyle \lim_{n \to \infty} \dfrac{n}{n!^{1/n}}=e$.

DeepSea
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By Stirling's approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

then

$$\frac{n}{n!^{\frac 1 n}}\sim \frac{n}{(\sqrt{2 \pi n})^\frac1n\left(\frac{n}{e}\right)}\to e$$

user
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