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I would like to understand why $ \lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}= \lim_{n\to\infty}(1+\frac{1}{n})^{n}$.

I was given a solution, but it gives no further details than $$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}=\lim_{n\to\infty}\sqrt[n]{\frac{n^n}{n!}}=\lim_{n\to\infty}\frac{(n+1)^{(n+1)}/(n+1)!}{n^n/n!}=\lim_{n\to\infty}(1+\frac{1}{n})^n.$$

What happened between $\lim_{n\to\infty}\sqrt[n]{\frac{n^n}{n!}}$ and $\lim_{n\to\infty}\frac{(n+1)^{(n+1)}/(n+1)!}{n^n/n!}$?

Mars Plastic
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nevermind_15
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4 Answers4

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If $a_n>0$ and there exists limit $\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}$, then $$ \lim_{n\rightarrow\infty} \sqrt[n]{a_n} = \lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}$$ You can check that it's true as follows:

If $\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = g > 0$ then for every $\epsilon>0$, $\epsilon <g$ there exists $N$ such that for $n\ge N$: $$ g-\epsilon \le \frac{a_{n+1}}{a_n} \le g+\epsilon $$ $$ (g-\epsilon)^{n-N} \le \frac{a_{N+1}}{a_N}\frac{a_{N+2}}{a_{N+1}}\dots\frac{a_n}{a_{n-1}} = \frac{a_n}{a_N} \le (g+\epsilon)^{n-N}$$ $$ \sqrt[n]{a_N (g-\epsilon)^{n-N}} \le \sqrt[n]{a_n} \le \sqrt[n]{a_N (g+\epsilon)^{n-N}} $$ $$ g-\epsilon \le \lim_{n\rightarrow\infty}\sqrt[n]{a_n} \le g+\epsilon $$ Since $\epsilon$ is arbitrary, it means that $\lim_{n\rightarrow\infty}\sqrt[n]{a_n}=g$.

If $g=0$ the proof is similar, but the lower bound is just $0$.

In your problem $a_n = \frac{n^n}{n!}$.

Adam Latosiński
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Taking logs and using Stolz-Cesàro: $$ \log\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}= \lim_{n\to\infty}\log\frac{n}{\sqrt[n]{n!}}= \lim_{n\to\infty}\frac{n\log n -(\log 1 + \cdots + \log n)}n = $$ $$ \lim_{n\to\infty}(n + 1)\log(n + 1) - n\log n - \log(n + 1) = \lim_{n\to\infty}n\log\frac{n + 1}n = \log\lim_{n\to\infty}\left(1 + \frac 1n\right)^n. $$

Martín-Blas Pérez Pinilla
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Let $A=\frac{n}{{n!}^{1/n}}$. So $\log A=\frac{1}{n}\sum_{r=1}^n \log(\frac{n}{r})=\frac{1}{n}\sum_{r=1}^n \log(\frac{1}{\frac{r}{n}})$.

So as $n\to \infty$, $\log A$ grows to $\int_{0}^{1} \log(\frac{1}{x})dx=1$, hence $A$ grows to $\exp(1)=e$.

And we know that $e=\lim_{n\to \infty} (1+\frac{1}{n})^{n}$.

Gary
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Soumyadip Sarkar
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Well, as $n \rightarrow \infty$ we have via Stirling's approximation that: $$n! \sim \bigg(\frac{n}{e}\bigg)^n \cdot \sqrt{2 \pi n}$$

By plugging this in, and noting that $(2\pi n)^{\frac{1}{2n}}$ converges to $1$ when $n \rightarrow \infty$ we arrive at $\frac{n}{\frac{n}{e}} = e$ which is obviously equaly to the RHS limit. Hope this helps

Victor
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  • The question is much more elementary than Stirling's formula. – David C. Ullrich Jul 22 '19 at 20:23
  • I disagree, I've used Stirling extensively throughout highschool and my early undergrad years to bypass opaque and lengthy convergence tests. The proof of this approximation is comparable in difficulty with the one presented in the other answer. But hey, if you feel like showing the OP a different piece of knowledge warrants a downvote, then do what you gotta do – Victor Jul 23 '19 at 07:54
  • "The proof of this approximation is comparable in difficulty with the one presented in the other answer. " Really? Where can I find this proof? – David C. Ullrich Jul 23 '19 at 14:17
  • Re "the other answer": Right now there are several other answers. My favorite is the one from Soumyadip Sarkar, it being it seems to me the most elementary, using nothing more than the approximation of an integral by a Riemann sum. – David C. Ullrich Jul 23 '19 at 14:43