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Analysis textbook by Shanti Narayan, is asking to prove the limit $\lim {\left({\dfrac{(2n)!}{(n!)^2}}\right)}^{1/n} \to \frac{1}{4}$ as $n \to \infty$. I tried but was unable to find the solution. Even Wolfram Alpha is telling the limit to be $4$. Please help!

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    related to http://math.stackexchange.com/questions/58560/elementary-central-binomial-coefficient-estimates2012-02-17

5 Answers 5

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There is really nno meed to use the stirling fomula to calculate this integral, we can just use stolz formula to caluate

$I_{n}=(\frac{(2n)!}{(n!)(n!)})^{\frac{1}{n}}$ Using stolz formula we can get $\quad$ $\lim log I_{n}=\lim log(\frac{2(2n+1)}{n+1})=log4 $

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    Isn't this essentially the same idea as [my answer](http://math.stackexchange.com/a/109835/) cloaked in a $\log$?2012-02-17
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A simple proof is based on the observation that $\dfrac{(2n)!}{(n!)^2}$ is the central binomial coefficient $\displaystyle{ {2n} \choose n}$.

Look at row $2n$ in the Pascal triangle. The sum of all terms is $2^{2n}= 4^n$ and so ${{2n} \choose n} \le 4^n$. Moreover, the central binomial coefficient is the largest number in that row and so $4^n \le (2n+1){{2n} \choose n}$. Hence $ \frac{4^n}{2n+1} \le {{2n} \choose n} \le 4^n $

Since $(2n+1)^{1/n} \to 1$, we conclude that ${{2n} \choose n} ^ {1/n} \to 4$.

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    @MartinSleziak. fixed, at last.2017-06-19
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If you use Stirling's approximation $n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$ then you get ${\left({\dfrac{(2n)!}{(n!)^2}}\right)}^{1/n} \approx {\left({\dfrac{\sqrt{4 \pi n} \left(\frac{2n}{e}\right)^{2n}}{{2 \pi n} \left(\frac{n}{e}\right)^{2n}}}\right)}^{1/n} = 4\left(\frac{1}{n\pi}\right)^\frac{1}{2n} \approx 4.$

It is not difficult to translate this into the language of limits.

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    @robjohn: indeed - corrected2012-02-16
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Note that $ \begin{align} \frac{(2(n+1))!}{(n+1)!^2} &=\frac{(2n+2)(2n+1)}{(n+1)(n+1)}\frac{(2n)!}{n!^2}\\ &=4\left(1-\frac{1}{2n+2}\right)\frac{(2n)!}{n!^2}\\ \end{align} $ Thus, what we are looking for is $ \begin{align} \left(\frac{(2n)!}{n!^2}\right)^{1/n} &=4\left(\prod_{k=0}^{n-1}\left(1-\frac{1}{2k+2}\right)\right)^{1/n}\\ &\to4 \end{align} $ Since the terms in the product tend to $1$, and the limit of the geometric means of non-zero terms with a limit $L$ is also $L$.

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    @$a$non: $b$ut we are taking the geometric mean of terms with a limit... :-)2012-02-16
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As a previous exercise you can prove that $\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\ln \left(\frac{n}{k}\right)=-\int_0^1\ln t\, dt=1.\tag{1}\label{eq1}$ Note that $\begin{align*} \ln\left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n} &=\frac{1}{n}\left[ \sum_{k=1}^{2n} \ln k-2\sum_{k=1}^n \ln k \right]\\ &= \frac{1}{n}\sum_{k=1}^n \ln(n+k)-\ln(k)\\ &= \frac{1}{n}\left[\sum_{k=1}^n\ln n+\sum_{k=1}^n\ln\left( 1+\frac{k}{n} \right)-\sum_{k=1}^n\ln k\right]\\ &= \frac{1}{n}\sum_{k=1}^n \ln\left(\frac{n}{k}\right)+\frac{1}{n}\sum_{k=1}^n\ln\left( 1+\frac{k}{n} \right). \end{align*}$ Note that in the last equality the second term is a Riemann's sum of $t\mapsto \ln t$ over $[1,2]$. Using \eqref{eq1} we have $\begin{align*} \lim_{n\to\infty}\ln\left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n} &= -\int_0^1\ln t\, dt+\int_1^2 \ln t\, dt\\ &=1+(-1+\ln 4), \end{align*}$ so $\lim_{n\to\infty}\ln\left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n}=\ln 4.$ By the continuity of $t\mapsto e^t$ you get $\lim_{n\to\infty} \left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n}=4.$