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Determine the following limit
$$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)\ldots (n+n)} $$ Idea #1 the nth root could be rewritten as $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{\frac{(n+n)!}{n!}} $$ Noticing that $n^{-1}$ is an infinitesimal, if the n-th root were bounded, the limit would be $0$.

But hoping that $\sqrt[n]{\frac{(n+n)!}{n!}}$ is bounded is a pipe dream, this clearly tends to $\infty$.

Idea #2 $$\lim_{n\to\infty}\ln\left (\frac{1}{n}[(n+1)\ldots (n+n)]^\frac{1}{n}\right ) = \lim_{n\to\infty}\left [-\ln n+\frac{1}{n}\left (\ln (n+1)+\ln (n+2)+\ldots +\ln (n+n) \right )\right ]$$ but trying to bring it to a common denominator doesn't help either, what I would have is $$\lim_{n\to\infty}\frac{1}{n} \left [-n\ln n+\ln (n+1)+\ldots +\ln 2+\ln n\right ]$$ Which doesn't bring me any closer to simplification.
The solution of this must something entirely different. I would like a hint/point me in the direction I should go.

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3 Answers 3

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You are looking for $$ \lim_{n\to +\infty}\sqrt[n]{\frac{(2n)!}{n!\cdot n^n}}. $$ By setting $a_n=\frac{(2n)!}{n!\cdot n^n}$ we have: $$ \frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)n^n}{(n+1)(n+1)^{n+1}}=\frac{4n+2}{n+1}\cdot\frac{1}{\left(1+\frac{1}{n}\right)^n}$$ hence: $$ \lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\frac{4}{e} $$ and that implies: $$ \lim_{n\to+\infty}\sqrt[n]{\frac{(2n)!}{n!\cdot n^n}}=\color{red}{\frac{4}{e}}.$$

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    $\begingroup$ @IlanAizelmanWS: $(n+1)\cdot(n+2)\cdot\ldots\cdot(2n) = \frac{(2n)!}{n!}$. $\endgroup$ Nov 13, 2015 at 12:49
  • $\begingroup$ Yup. I got it after 3 seconds. thx :) $\endgroup$ Nov 13, 2015 at 12:49
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    $\begingroup$ that is so clean, wow, thanks a lot. $\endgroup$
    – AlvinL
    Nov 13, 2015 at 12:53
  • $\begingroup$ I got the same answer in your second line, but without $(2n+2)$, how did you get it? $\endgroup$ Nov 13, 2015 at 13:02
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    $\begingroup$ @IlanAizelmanWS: $\frac{1}{n}=\frac{1}{\sqrt[n]{n^n}}.$ $\endgroup$ Nov 13, 2015 at 13:33
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From the idea 1, we can derive the result using the Stirling's approximation $$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} $$ hence $$\frac{1}{n}\sqrt[n]{\frac{\left(2n\right)!}{n!}}\sim\frac{1}{n}\sqrt[n]{\frac{\sqrt{2n}}{\sqrt{n}}\left(\frac{2n}{e}\right)^{2n}\left(\frac{e}{n}\right)^{n}}=\frac{2^{2+1/2n}}{e}\rightarrow\frac{4}{e}. $$

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  • $\begingroup$ In case you saw it, I had a comment - which I have deleted, as it stupid... Sorry! (+1) $\endgroup$
    – peter a g
    Nov 13, 2015 at 13:27
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    $\begingroup$ @peterag Yes I saw it, but no problem. Time ago my teacher of history of mathematics told me that Euler, in some notes, wrote some trivial errors. So every mathematician, even the greater, write sooner or later something wrong ;) $\endgroup$ Nov 13, 2015 at 14:31
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Start with the expression on the right of your Idea #2. Using $\ln (n+k)=\ln n + \ln (1+k/n)$ for each $k$ shows that expression equals

$$-\ln n+\frac{1}{n}\sum_{k=1}^{n}[\ln n +\ln (1+k/n )] $$ $$= -\ln n+\frac{1}{n}[n\ln n + \sum_{k=1}^{n}\ln (1+k/n )]= \frac{1}{n}\sum_{k=1}^{n}\ln (1+k/n )$$ $$ \to \int_0^1 \ln (1+x)\, dx = 2\ln2 -1 = \ln (4/e).$$

Exponentiating back gives $4/e$ for the limit.

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  • $\begingroup$ Aha, Riemann integrability, yes? Elaborate on limits of integration please. $\endgroup$
    – AlvinL
    Nov 14, 2015 at 13:59
  • $\begingroup$ I'm just using $\sum_{k=1}^n f(k/n)\cdot (1/n) \to \int_0^1 f(x)\,dx.$ $\endgroup$
    – zhw.
    Nov 14, 2015 at 17:57
  • $\begingroup$ (+1) This is the way I would have pursued just because it uses basic tools that even an entry-level student would be able to apply. $\endgroup$
    – Mark Viola
    Apr 4, 2017 at 21:56

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