# Calculating the limit of $[(2n)!/(n!)^2]^{1/n}$ as $n$ tends to $\infty$

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!

• The limit can't be less than 1, there must be a typo. – anon Feb 16 '12 at 1:27
• Eric Naslund gives a way to see that the limit is $4$ in a comment at this link. – Jonas Meyer Feb 16 '12 at 1:30
• @anon I, too, thought so. But I have had no explaination how the question is wrong! – gaurav Feb 16 '12 at 1:34
• @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 1:42
• The estimates in Central binomial coefficient suffice. – lhf Feb 16 '12 at 2:07

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 $L$.

• The terms tend to $1$ but the product itself tends to $0$... – anon Feb 16 '12 at 3:24
• @anon: but we are taking the geometric mean of terms with a limit... :-) – robjohn Feb 16 '12 at 9:08

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

• Very nice proof! – JavaMan Feb 17 '12 at 2:58
• I see now that it's the same argument used by Eric Naslund as mentioned by Jonas Meyer. – lhf Feb 17 '12 at 10:32
• And I thought my answer used the least background :-) (+1) – robjohn Feb 17 '12 at 11:40
• I see how to get $4^n\le(2n+1)\binom{2n}n$ immediately from binomial theorem (I have $2n$ summands, and this is a common bound for all of them.) I do not see immediately $4^n\le(n+1)\binom{2n}n$. What am I missing? (It probably does not matter here that much, but you linked to this post in a question asking about the inequality with $n+1$.) – Martin Sleziak Jun 29 '15 at 17:51
• @MartinSleziak. fixed, at last. – lhf Jun 19 '17 at 22:10

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.

• I think it should be $\displaystyle4\left(\frac{1}{n\pi}\right)^{\frac{1}{2n}}$ – robjohn Feb 16 '12 at 2:20
• @robjohn: indeed - corrected – Henry Feb 16 '12 at 8:12

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

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 $\quad$ $\lim log I_{n}=\lim log(\frac{2(2n+1)}{n+1})=log4$

• Isn't this essentially the same idea as my answer cloaked in a $\log$? – robjohn Feb 17 '12 at 11:48