How to prove that $\lim \frac{1}{n} \sqrt[n]{(n+1)(n+2)... 2n} = \frac{4}{e}$

Question

I'd like a hint to show that:

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

Thanks.

Note that $(n+1)(n+2)...(2n)=\frac{(2n)!}{n!}$ then apply Stirling's approximation: http://en.wikipedia.org/wiki/Stirling%27s_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
Similar limit: http://math.stackexchange.com/questions/475786/how-to-compute-lim-n-rightarrow-infty-frac1n-left-2n12n2-cdots2nn — Martin Sleziak, Aug 16 '16 at 10:50
@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. — PinkyWay, Nov 19 '21 at 14:10
@Invisible, actually I thought I had a new solution to it but soon saw that it was already written — 2 is even prime, Nov 19 '21 at 14:11

score 23 · Answer 1 · edited Jan 15 '12 at 18:43

23

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 $[0,1]$.

edited Jan 15 '12 at 18:43

Srivatsan

26,311

answered Jan 15 '12 at 18:25

KotelKanim

2,856

score 17 · Answer 2 · edited Jul 04 '22 at 14:45

17

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

edited Jul 04 '22 at 14:45

Martin Sleziak

53,687

answered Jan 15 '12 at 18:55

David Mitra

74,748

The link in the post seems to be dead now. Here is a snapshot from Internet Archive. – Martin Sleziak Jan 08 '18 at 13:34
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)$ and Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$. – Martin Sleziak Jan 08 '18 at 13:36
I will add that the notes that used to be linked here were eventually incorporated into this text: http://alpha.math.uga.edu/~pete/2400full.pdf (Just incase, I'll include a Wayback Machine link, too.) – Martin Sleziak Jul 04 '22 at 14:46

score 10 · Accepted Answer · edited Apr 13 '17 at 12:21

10

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.

edited Apr 13 '17 at 12:21

Community

1

answered Jan 15 '12 at 18:29

Jonas Meyer

53,602

3

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

score 8 · Answer 4 · edited Jul 04 '22 at 14:47

8

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

edited Jul 04 '22 at 14:47

Martin Sleziak

53,687

answered Jan 15 '12 at 18:25

Eric Naslund

72,099

score 1 · Answer 5 · edited Apr 04 '12 at 20:12

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.

edited Apr 04 '12 at 20:12

Pedro

122,002

answered Jan 15 '12 at 20:45

Amihai Zivan

2,874

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

How to prove that $\lim \frac{1}{n} \sqrt[n]{(n+1)(n+2)... 2n} = \frac{4}{e}$

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