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$\begingroup$

I'd like a hint to show that:

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

Thanks.

  • 1
    Note that $(n+1)(n+2)...(2n)=\frac{(2n)!}{n!}$ then apply Stirling's approximation: http://en.wikipedia.org/wiki/Stirling%27s_approximation2012-01-15
  • 0
    @ThomasAndrews: That was posted as an answer by Eric Naslund.2012-01-15
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    Doh, let my page, so didn't see that. :)2012-01-15
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    Similar limit: http://math.stackexchange.com/questions/475786/how-to-compute-lim-n-rightarrow-infty-frac1n-left-2n12n2-cdots2nn2016-08-16

5 Answers 5

1

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.

  • 0
    This is in the answer David Mitra posted a while ago.2012-01-15
  • 0
    saw it after posting..2012-01-15
20

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 $[0,1]$.

13

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

  • 0
    The link in the post seems to be dead now. Here is a [snapshot from Internet Archive](http://web.archive.org/web/20130728110356/http://math.uga.edu/~pete/243series4.pdf).2018-01-08
  • 0
    And maybe some of the posts on this site can serve as a reference, for this result. For example: [Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$](https://math.stackexchange.com/q/69386) and [Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$](https://math.stackexchange.com/q/76743).2018-01-08
7

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

7

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.

  • 2
    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 $$\lim_{n\rightarrow \infty} \frac{4}{n} \sqrt[n]{n!}.$$2012-01-15
  • 0
    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)2012-01-15
  • 0
    @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.2012-01-15
  • 0
    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.2012-01-15