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Analysis textbook by Shanti Narayan, is asking to prove the limit $$ \lim\left[\dfrac{\left(2n\right)!}{\left(n!\right)^{2}}\right]^{1/n} \to \frac{1}{4}\quad \mbox{as}\quad n \to \infty $$ I tried but was unable to find the solution.

Even $\tt Wolfram\ Alpha$ is telling the limit to be $4$. Please help !.

Felix Marin
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Gaurav Tiwari
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    The limit can't be less than 1, there must be a typo. – anon Feb 16 '12 at 01:27
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    Eric Naslund gives a way to see that the limit is $4$ in a comment at this link. – Jonas Meyer Feb 16 '12 at 01:30
  • @anon I, too, thought so. But I have had no explaination how the question is wrong! – Gaurav Tiwari Feb 16 '12 at 01:34
  • @Jonas Meyer Thanks! :) – Gaurav Tiwari Feb 16 '12 at 01:36
  • The central binomial coefficients $\binom{2n}{n}$ are all greater than $1$. Anything greater than $1$ raised to a positive power must likewise be greater than $1$. Hence this limit cannot be less than $1$. – anon Feb 16 '12 at 01:36
  • @anon Thanks! I figured it. – Gaurav Tiwari Feb 16 '12 at 01:41
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    @gaurav: At that link you will find other methods that can be applied here. For example, when $(a_n)$ is a sequence of positive numbers such that $\lim_n \frac{a_{n+1}}{a_n}$ exists, then $\lim_n \sqrt[n]{a_n}$ exists and $\lim_n \sqrt[n]{a_n}=\lim_n \frac{a_{n+1}}{a_n}$. With $a_n=\frac{(2n)!}{(n!)^2}$, this makes it easy to check that the limit is $4$. You can also make use of $\lim \frac{\sqrt[n]{n!}}{n}=e$, by rewriting it as $\displaystyle4\left(\frac{\sqrt[2n]{(2n)!}}{2n}\right)^2\left(\frac{n}{\sqrt[n]{n!}}\right)^2$. – Jonas Meyer Feb 16 '12 at 01:42
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    The estimates in Central binomial coefficient suffice. – lhf Feb 16 '12 at 02:07
  • I meant $\lim \frac{\sqrt[n]{n!}}{n}=1/e$, but actually all you need to know is that $\lim \frac{\sqrt[n]{n!}}{n}$ exists and is nonzero, which is easier. – Jonas Meyer Feb 17 '12 at 00:24
  • related to http://math.stackexchange.com/questions/58560/elementary-central-binomial-coefficient-estimates – lhf Feb 17 '12 at 01:14

5 Answers5

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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 ${{2n} \choose n} ^ {1/n} \to 4$.

lhf
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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.

Henry
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Note that $$ \begin{align} \frac{(2(n+1))!}{(n+1)!^2} &=\frac{(2n+2)(2n+1)}{(n+1)(n+1)}\frac{(2n)!}{n!^2}\tag1\\ &=4\left(1-\frac{1}{2n+2}\right)\frac{(2n)!}{n!^2}\tag2 \end{align} $$ Bernoulli's Inequality says $$ \begin{align} \left(\frac{(2n)!}{n!^2}\right)^{1/n} &=4\left(\prod_{k=0}^{n-1}\left(1-\frac{1}{2k+2}\right)\right)^{1/n}\tag3\\ &\ge4\left(\frac12\prod_{k=1}^{n-1}\left(1-\frac{1}{k+1}\right)^{1/2}\right)^{1/n}\tag4\\ &=4\left(\frac1{2\sqrt{n}}\right)^{1/n}\tag5 \end{align} $$ Equation $(3)$ and inequality $(5)$ give $$ 4\left(\frac1{2\sqrt{n}}\right)^{1/n}\le\left(\frac{(2n)!}{n!^2}\right)^{1/n}\le4\tag6 $$ and the Squeeze Theorem yields $$ \lim_{n\to\infty}\left(\frac{(2n)!}{n!^2}\right)^{1/n}=4\tag7 $$

robjohn
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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 $$\lim_{n\to\infty} \left(\dfrac{(2n)!}{(n!)^2}\right)^{1/n}=4.$$

leo
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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 $

yaoxiao
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