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!

  • 3
    $\begingroup$ The limit can't be less than 1, there must be a typo. $\endgroup$ – anon Feb 16 '12 at 1:27
  • 2
    $\begingroup$ Eric Naslund gives a way to see that the limit is $4$ in a comment at this link. $\endgroup$ – Jonas Meyer Feb 16 '12 at 1:30
  • $\begingroup$ @anon I, too, thought so. But I have had no explaination how the question is wrong! $\endgroup$ – gaurav Feb 16 '12 at 1:34
  • 1
    $\begingroup$ @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 $\displaystyle4\left(\frac{\sqrt[2n]{(2n)!}}{2n}\right)^2\left(\frac{n}{\sqrt[n]{n!}}\right)^2$. $\endgroup$ – Jonas Meyer Feb 16 '12 at 1:42
  • 2
    $\begingroup$ The estimates in Central binomial coefficient suffice. $\endgroup$ – 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 $L$.

  • $\begingroup$ The terms tend to $1$ but the product itself tends to $0$... $\endgroup$ – anon Feb 16 '12 at 3:24
  • $\begingroup$ @anon: but we are taking the geometric mean of terms with a limit... :-) $\endgroup$ – 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 ${{2n} \choose n} ^ {1/n} \to 4$.

  • $\begingroup$ Very nice proof! $\endgroup$ – JavaMan Feb 17 '12 at 2:58
  • $\begingroup$ I see now that it's the same argument used by Eric Naslund as mentioned by Jonas Meyer. $\endgroup$ – lhf Feb 17 '12 at 10:32
  • $\begingroup$ And I thought my answer used the least background :-) (+1) $\endgroup$ – robjohn Feb 17 '12 at 11:40
  • $\begingroup$ 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 $n+1$.) $\endgroup$ – Martin Sleziak Jun 29 '15 at 17:51
  • $\begingroup$ @MartinSleziak. fixed, at last. $\endgroup$ – 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.

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

  • $\begingroup$ The OP don't ask for any integral. Please consider review your answer. $\endgroup$ – leo Feb 17 '12 at 0:56
  • $\begingroup$ Isn't this essentially the same idea as my answer cloaked in a $\log$? $\endgroup$ – robjohn Feb 17 '12 at 11:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.