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We need to solve this limit $\lim _{x\to \infty \:}\left(\frac{\left(\left(2x\right)!\right)}{x^xx!}\right)^{\frac{1}{x}}$ we guess that the answer is 1, as n approach infinity 1/n => 0 so anything besides 0 to the power of 0 = 1 (except 0 of course).

We need to ensure if we had the right answer, please write some hints here.

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  • $\begingroup$ @Workaholic Yep you're right, it's our mistake. Should we now calculate it with exp(x) ? $\endgroup$ – Amine Marzouki Feb 27 '16 at 16:25
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$$ \lim _{x\to \infty \:}\left(\frac{\left(\left(2x\right)!\right)}{x^xx!}\right)^{\frac{1}{x}} $$ Using Stirling's approximation $$ (2x)! \sim (2x/e)^{2x}\sqrt{4\pi x}$$ and $$ x!\sim (x/e)^x\sqrt{2\pi x} $$ therefore the ratio $$\frac{(2x)!}{x!}\sim 2^{2 x+\frac{1}{2}} e^{-x} x^x$$ and dividing by $x^x$, we have to compute the limit $$ \lim_{x\to\infty}(2^{2 x+\frac{1}{2}} e^{-x})^{1/x}=\frac{1}{e}\lim_{x\to\infty}(2^{2 x+\frac{1}{2}} )^{1/x}=\frac{1}{e}\lim_{x\to\infty}e^{\frac{1}{x}(2x+1/2)\log 2}=\frac{e^{\log 4}}{e}=\frac{4}{e}\ . $$

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Here we rely only on Riemann sums to evaluate the limit. We can write

$$\lim _{x\to \infty }\left(\frac{\left(2x\right)!}{x^xx!}\right)^{\frac{1}{x}} =\lim_{n\to \infty}e^{\frac1n \log((2n)!-\frac1n \log(n!)-\log(n)}$$

by continuity of the Gamma function.

Then, note that we have

$$\begin{align} \frac1n \log((2n)!-\frac1n \log(n!)-\log(n)&=\frac1n \sum_{k=n+1}^{2n}\log(k/n)\\\\ &=\frac1n \sum_{k=1}^n \log(1+k/n)\\\\ &\to \int_0^1 \log(1+x)\,dx\,\,\text{as}\,\,n\to \infty\\\\ &=\log(4/e) \end{align}$$

Therefore, we arrive at the result

$$\bbox[5px,border:2px solid #C0A000]{\lim _{x\to \infty }\left(\frac{\left(2x\right)!}{x^xx!}\right)^{\frac{1}{x}}=4e^{-1}}$$

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  • $\begingroup$ +1 I like this. Not sure what "analytic continuation" is all about. A pretty easy estimate shows that the limit on $n$ gives the limit for continuous $x.$ $\endgroup$ – zhw. Feb 27 '16 at 18:48
  • $\begingroup$ @zhw. Thanks! The purpose of the "AC" reference (and continuity is really all I was using) was to ensure that the "factorial/Gamma" function didn't have some discontinuity or other non-smooth behavior at integer $n$ values. - Mark $\endgroup$ – Mark Viola Feb 27 '16 at 19:23
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Yet another idea (see Baby Rudin 3.37): $$ \lim_{x\to\infty}\left(\frac{\left(\left(2x\right)!\right)}{x^xx!}\right)^{\frac{1}{x}} = \lim_{x\to\infty}\frac{\frac{(2(x+1))!}{(x+1)^{(x+1)}(x+1)!}}{\frac{(2x)!}{x^x x!}} = \lim_{x\to\infty}\frac{(2x+2)(2x+1)x^x}{(x+1)^{x+1}(x+1)} = $$ $$ \lim_{x\to\infty}\frac{2(2x+1)}{(x+1)} \lim_{x\to\infty}\left(\frac{x}{x+1}\right)^x = \frac4e $$

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Note that $$ \frac{(2x)!}{x^x x!}=\frac{2x\cdot(2x-1)\cdots(x+2)\cdot(x+1)}{x^x}=2\cdot\left(2-\frac{1}{x}\right)\cdots\left(1+\frac{2}{x}\right)\cdot\left(1+\frac{1}{x}\right)\\=\prod_{k=1}^{x}\left(1+\frac{k}{x}\right). $$ So $$ \begin{eqnarray} \lim_{x\rightarrow\infty}\left(\frac{(2x)!}{x^x x!}\right)^{1/x}&=&\lim_{x\rightarrow\infty}\exp\left(\frac{1}{x}\sum_{k=1}^{x}\log\left(1+\frac{k}{x}\right)\right)\\&=&\exp\left(\lim_{x\rightarrow\infty}\frac{1}{x}\sum_{k=1}^{x}\log\left(1+\frac{k}{x}\right)\right)\\&=&\exp\left(\int_{0}^{1}\log(1+u)du\right)\\&=&\exp\left(\log4 - 1\right)\\&=&\frac{4}{e}. \end{eqnarray}$$

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  • $\begingroup$ Just like mine... ;-)) $\endgroup$ – Mark Viola Feb 27 '16 at 19:24

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