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I've done these two proves:

$\left(\frac{\sqrt[n]{n!}}{n}\right)_{n}\rightarrow \frac{1}{e}$

$\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}-1 \right)_{n} \rightarrow 0$

And now I've to prove this statement, using the previous statements if are necessary:

$\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}} \right)_{n}^{n}\rightarrow e$

All I could done to solve the limit is transformate the expression in this way: $e^{lim_{n \to \infty} n\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}\right)}$ and theoretically the limit of the exponent has to be 1, but I dont't know how to continue to prove it.

Thanks in advance.

Martin Sleziak
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Marcos Amorós Ibáñez
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3 Answers3

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Alternatively, by (a weak) Stirling approximation, as $n \to \infty$, $$ \ln n!=n\ln n-n+O(\ln n) $$ one may write, as $n \to \infty$, $$ \begin{align} \left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}} \right)^{n}&=\left(\frac{\sqrt[n]{n+1}\times\sqrt[n+1]{(n+1)!}}{\sqrt[n]{(n+1)!}} \right)^{n} \\\\&=(n+1)\times ((n+1)!)^{-\frac1{n+1}} \\\\&=e^{\ln(n+1)}\times e^{\large-\frac{\ln[(n+1)!]}{n+1}} \\\\&=e^{\ln(n+1)}\times e^{\large-\ln(n+1)+1+O\big(\frac{\ln n}n\big)} \\\\&=e\times e^{O\big(\frac{\ln n}n\big)} \\\\& \to e \end{align} $$ as expected.

Olivier Oloa
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    Thank you for your answer, but at the moment I can't use Stirling's aproximation as a tool to solve limits of successions. – Marcos Amorós Ibáñez Dec 05 '16 at 22:31
  • Observe that, as $n \to \infty$, $\ln\left(n!\right) = \sum_{k=1}^n{\ln\left(k\right)} \sim \int_1^n{\ln\left(x\right), dx} = \left[ x\ln\left(x\right) - x \right]_1^n = n\ln\left(n\right) - n + 1$ is sufficient. – Olivier Oloa Dec 05 '16 at 22:34
  • Thanks again! But I have not yet given differential or integral calculus in my real analysis class. So I can only use basic tools of successions to solve the problem. – Marcos Amorós Ibáñez Dec 05 '16 at 22:48
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It is sufficient to show that $$ \lim_{n\to\infty}\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}-\frac{n+1}{n}\right)=0.$$ Since $\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=\frac1e$, one has \begin{eqnarray} \lim_{n\to\infty}\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}-\frac{n+1}{n}\right)&=&\lim_{n\to\infty}\left(\frac{(n+1)\frac{\sqrt[n+1]{(n+1)!}}{n+1}}{n\frac{\sqrt[n]{n!}}{n}}-\frac{n+1}{n}\right)\\ &=&\lim_{n\to\infty}\left(\frac{n+1}{n}\right)\left(\frac{\frac{\sqrt[n+1]{(n+1)!}}{n+1}}{\frac{\sqrt[n]{n!}}{n}}-1\right)\\ &=&0 \end{eqnarray} and hence $$　\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}＝1+\frac{1}{n}＋o(\frac1{n}).　$$ So $$　\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}\right)^n＝\left(1+\frac1{n}+o(\frac1{n})\right)^n.　$$ Therefore $$　\lim_{n\to\infty}\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}\right)^n＝e.　$$

xpaul
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  • I think there is a subtle flaw here. You have only proved that $$\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}＝1+\frac{1}{n}＋o(1)$$ and not $$\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}＝1+\frac{1}{n}＋o(\frac1{n})$$ – Paramanand Singh Dec 07 '16 at 11:54
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$\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}} \right)^{n}=\frac{\sqrt[n+1]{(n+1)!^{n}}}{n!}=\sqrt[n+1]{\frac{(n+1)!^{n}}{n!^{n+1}}}=\sqrt[n+1]{\frac{(n+1)^{n}}{n!}}$

We know that if $lim_{n \to \infty} \frac{a_{n}}{a_{n-1}} $ exists, then $lim_{n \to \infty} \sqrt[n]{a_{n}}=lim_{n \to \infty} \frac{a_{n}}{a_{n-1}} $

So, applying this to $a_{n}:=\frac{(n+1)^{n}}{n!}$ we obtained:

$\frac{(n+1)^{n}}{n!} \frac{(n-1)!}{n^{n-1}}=\frac{n+1}{n}\left(\frac{n+1}{n}\right)^{n-1}=\left( 1+\frac{1}{n}\right)^{n} \rightarrow e$

Marcos Amorós Ibáñez
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