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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.

  • 0
    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.

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    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 $[0,1]$.

15

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
    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.

  • 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