I'd like a hint to show that:

$$\lim _{n\to\infty}\frac{1}{n} \sqrt[n]{(n+1)(n+2) \cdots 2n} = \frac{4}{e} .$$


  • 2
    $\begingroup$ Note that $(n+1)(n+2)...(2n)=\frac{(2n)!}{n!}$ then apply Stirling's approximation: en.wikipedia.org/wiki/Stirling%27s_approximation $\endgroup$ Jan 15, 2012 at 18:33
  • $\begingroup$ @ThomasAndrews: That was posted as an answer by Eric Naslund. $\endgroup$ Jan 15, 2012 at 18:34
  • $\begingroup$ Doh, let my page, so didn't see that. :) $\endgroup$ Jan 15, 2012 at 18:37
  • $\begingroup$ Similar limit: math.stackexchange.com/questions/475786/… $\endgroup$ Aug 16, 2016 at 10:50
  • $\begingroup$ @2isevenprime, what is the point of editing a 9 years old questio and bumping in into the fron page? There are active questions to be edited and answered. $\endgroup$
    – PinkyWay
    Nov 19, 2021 at 14:10

5 Answers 5


Taking $\log$ of the expression you get

$\frac{1}{n}\sum \log (1+\frac{k}{n}) $.

This is a Riemann sum for the function $\log(1+x)$ on the interval $[0,1]$.


This is @Jonas Meyer's idea from the link in his answer:

Let $$a_n={(n+1)(n+2)\cdots 2n\over n^n}.$$

Then $$ \lim_{n\rightarrow\infty} {a_{n+1}\over a_n}= \lim_{n\rightarrow\infty}{(2n+1)(2n+2)\over n+1}\cdot {n^n\over (n+1)^{n+1}}= \lim_{n\rightarrow\infty}{2(2n+1)\over n+1}(1+\textstyle{1\over n})^{-n}={4\over e}. $$

But, for $a_n>0$, if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}=L$, then $\lim\limits_{n\rightarrow\infty}\root n\of {a_n}=L$ (see page 3 of Pete Clark's notes here).

In this case $$\lim\limits_{n\rightarrow\infty} \root n\of {a_n} = \lim\limits_{n\rightarrow\infty}{1\over n}\root n \of {(n+1)(n+2)\cdots 2n }. $$


You could rewrite it as

$$4\left(\frac{\sqrt[2n]{(2n)!}}{2n}\right)^2\cdot\frac{n}{\sqrt[n]{n!}}$$ and use the result of this question.

  • 3
    $\begingroup$ Similar to your solution, but I prefer writing it as $\frac{(2n)!n!}{n!n!}=n!\binom{2n}{n}$. Then, as the central binomial coefficient $\binom{2n}{n}=\frac{(2n)!}{n!n!}$ is larger then the average, $\frac{2^{2n}}{2n+1}$ but smaller then the sum of all the binomial coefficients, $2^{2n}$, we see that $$\lim_{n\rightarrow \infty} \sqrt[n]{\binom{2n}{n}}=4,$$ and as limits are multiplicative, we are then left with evaluating $$\lim_{n\rightarrow \infty} \frac{4}{n} \sqrt[n]{n!}.$$ $\endgroup$ Jan 15, 2012 at 18:46
  • $\begingroup$ Also worth noting that the method in the answer you linked to can be used to solve this question. (I.e. using a power series) $\endgroup$ Jan 15, 2012 at 18:49
  • $\begingroup$ @Eric: Thank you for the comments. Your method using binomial coefficients is very interesting and could also be an addendum to your answer (not that I could vote it up a second time anyway). The intent of my hint is that it is all reduced to knowing one simpler limit. It is also a good point that the method rather than the result of the previous question could be applied. $\endgroup$ Jan 15, 2012 at 18:53
  • $\begingroup$ Funny you say that, I was actually writing it as an addendum, but then got stuck trying to think of ways to evaluate the limit of $\sqrt[n]{n!}$ without using logs and looking at the sum, and without using Stirlings formula. Then I saw the page you linked to. $\endgroup$ Jan 15, 2012 at 18:56

One possible approach is to notice the term inside the root is $\frac{(2n)!}{n!}$ and apply Stirling's approximation.


There's a theorem that is very helpful for these kind of questions.

Let $a_n$ be a sequence of positive real numbers. If $a_{n+1}/a_n$ converges, then $a_n^{1/n}$ converges to the same limit.

Continuing from here is pretty straightforward.

  • $\begingroup$ This is in the answer David Mitra posted a while ago. $\endgroup$ Jan 15, 2012 at 20:56
  • $\begingroup$ saw it after posting.. $\endgroup$ Jan 15, 2012 at 21:16

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