Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'd like a hint to show that:

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


share|cite|improve this question
Note that $(n+1)(n+2)...(2n)=\frac{(2n)!}{n!}$ then apply Stirling's approximation: – Thomas Andrews Jan 15 '12 at 18:33
@ThomasAndrews: That was posted as an answer by Eric Naslund. – Jonas Meyer Jan 15 '12 at 18:34
Doh, let my page, so didn't see that. :) – Thomas Andrews Jan 15 '12 at 18:37
Thanks for the answers – Jr. Jan 15 '12 at 19:56
up vote 5 down vote accepted

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.

share|cite|improve this answer
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 $$\lim_{n\rightarrow \infty} \frac{4}{n} \sqrt[n]{n!}.$$ – Eric Naslund Jan 15 '12 at 18:46
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) – Eric Naslund Jan 15 '12 at 18:49
@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. – Jonas Meyer Jan 15 '12 at 18:53
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. – Eric Naslund Jan 15 '12 at 18:56

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 $[0,1]$.

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer
This is in the answer David Mitra posted a while ago. – Jonas Meyer Jan 15 '12 at 20:56
saw it after posting.. – Amihai Zivan Jan 15 '12 at 21:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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