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This question already has an answer here:

Evaluate $$\lim_{n \rightarrow \infty~} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$

Attempt: Let $$y=\lim_{n \rightarrow \infty} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$

$$\implies \log y = \lim_{n \rightarrow \infty} \dfrac {1} {n} [\log (n+1) +\cdots+log(n+n)-log(n)] $$

How do I move forward?

Thank you very much for your help.

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marked as duplicate by YuiTo Cheng, postmortes, Lee David Chung Lin, Thomas Shelby, Parcly Taxel Jul 4 at 6:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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By a Riemann sum argument, $$ \log\frac{\left[(n+1)(n+2)\cdot\ldots\cdot(n+n)\right]^{\frac{1}{n}}}{n}=\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)$$ converges towards: $$ \int_{0}^{1}\log(1+x)\,dx = -1+\log 4,$$ hence the value of the limit is $\large\color{red}{\frac{4}{e}}$.

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    $\begingroup$ Thank you for the answer! :) $\endgroup$ – MathMan Jan 22 '15 at 21:51
  • $\begingroup$ Can you elaborate the part "converges towards $\int_{0}^{1}\log(1+x)dx=-1+\log 4$"? $\endgroup$ – Kushal Bhuyan Apr 20 '16 at 12:27
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You can't apply the theorem that $a_n \to l$ implies $b_n = (a_1a_2\cdots a_n)^{1/n}$ to this limit since the $k$-th term, i.e. $1+\tfrac{k}{n}$, depends on $n$.

If you take the natural log of the limitand(?) you get:

$\ln\left[\left(\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)\cdots\left(1+\dfrac{n}{n}\right)\right)^{1/n}\right] = \dfrac{1}{n}\displaystyle\sum_{k = 1}^{n}\ln\left(1+\dfrac{k}{n}\right)$,

which is a Riemann sum for $\displaystyle\int_{0}^{1}\ln(1+x)\,dx$.

So, as $n \to \infty$, we have $\dfrac{1}{n}\displaystyle\sum_{k = 1}^{n}\ln\left(1+\dfrac{k}{n}\right) \to \displaystyle\int_{0}^{1}\ln(1+x)\,dx = 2\ln 2 - 1$.

Exponentiating both sides gives us $(1+\tfrac{1}{n})(1+\tfrac{2}{n})\cdots(1+\tfrac{n}{n}))^{1/n} \to e^{2\ln 2 - 1} = \dfrac{4}{e}$ as $n \to \infty$.

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  • $\begingroup$ SImpler than mine ! Cheers :-) $\endgroup$ – Claude Leibovici Jan 10 '16 at 3:18
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Hint: You may use $$\lim\limits_{n \to \infty} \sqrt[n]{a_n} = \lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}$$

with $a_n = \dfrac {(n+1)(n+2)\cdots(n+n)}{n^n}$

The Riemann integral way is nice too, but if you insist on taking $\log$ you could apply Stolz-Cesaro Theorem too:

$$\lim\limits_{n \to \infty} \frac{-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)}{n} \\= \lim\limits_{n \to \infty} \left(-(n+1)\log (n+1) + \sum\limits_{k=1}^{n+1} \log \left(k+n\right)\right) - \left(-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)\right) \\ = \lim\limits_{n \to \infty} \log 2 + \log \frac{2n+1}{n+1} - n\log \left(1+\frac{1}{n}\right) = \log 4 - 1$$

Giving you the desired limit $\dfrac{4}{e}$.

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  • $\begingroup$ @Wanderer welcome:) $\endgroup$ – sciona Jan 22 '15 at 21:55
  • $\begingroup$ Dear sir, how log 2 came from nowhere? $\endgroup$ – Cloud JR Apr 30 at 17:53
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Your mistake is

$$ \dfrac {1} {n} [\log (n+1) +\cdots+\log(n+n)-\color{red}{n}\log(n)] =\frac1n\sum_{k=1}^n\log\left(1+\frac kn\right)$$ and use the Riemann sum.

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  • $\begingroup$ Thank you for pointing out :) $\endgroup$ – MathMan Jan 22 '15 at 21:51
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Jan 23 '15 at 6:27
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Let $$A_n=\frac 1n \left(\prod_{i=1}^n (n+i)\right)^{\frac 1n}$$ Using Pochhammer notation and using the corresponding expression using the gamma function, we have $$\prod_{i=1}^n (n+i)=(n+1)_n=\frac{2^{2 n} }{\sqrt{\pi }}\Gamma \left(n+\frac{1}{2}\right)$$ So, taking logarithms $$\log(A_n)=-\log(n)+\frac 1n\log \left(\frac{2^{2 n} \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}\right)$$ Now, using Stirling approximation for the gamma function, we have $$\log \left(\frac{2^{2 n} \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}\right)=n (\log (n)+2 \log (2)-1)+\frac{\log (2)}{2}-\frac{1}{24 n}+O\left(\frac{1}{n^{5/2}}\right)$$ which makes $$\log(A_n)=\left(2 \log (2)-1\right)+\frac{\log (2)}{2 n}-\frac{1}{24 n^2}+O\left(\frac{1}{n^{7/2}}\right)$$ and then the limit $\frac 4e$.

Edit

For sure, using Riemann sum as JimmyK4542 just answered makes the problem simpler. The approach I proposed shows, beside the limit itself, how it is approached.

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For some reason nobody mentioned Stirling's approximation: $$ \frac{(2n)!^{1/n}}{n!^{1/n} n} \sim \frac{(\sqrt{4\pi n}(2n/e)^{2n})^{1/n}}{(\sqrt{2\pi n}(n/e)^n)^{1/n} n} \sim \frac{(2n/e)^2}{(n/e)n} = \frac{4}{e}. $$

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You can try this easy one:

$\lim_{n\to \infty}\dfrac{1}{n}{[(n+1)(n+2)...(n+n)]}^{\frac{1}{n}}=\lim _{n\to \infty}\dfrac{1}{n}[n^n(1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n})]^{\frac{1}{n}}$

$=\lim_{n\to \infty}{\prod_{k=1}^n}(1+\frac{k}{n})^{\frac{1}{n}}$

Let $y={\prod_{k=1}^n}(1+\frac{k}{n})^{\frac{1}{n}}\implies \ln y=\frac{1}{n}\sum _{k=1}^n\ln (1+\frac{k}{n})=\int _0^ 1\ln(1+x)\operatorname{dx}$

$\implies y=\frac{4}{e}$

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  • $\begingroup$ Thanks for the prompt reply $\endgroup$ – user120386 Jan 10 '16 at 8:28
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Let $A_n$ be the given expression, then: $A_n = \displaystyle \prod_{k=1}^n\left(1+\dfrac{k}{n}\right)^{\frac{1}{n}}\Rightarrow \ln A_n =\dfrac{1}{n}\cdot \displaystyle \sum_{k=1}^n \ln\left(1+\dfrac{k}{n}\right)\Rightarrow \displaystyle \lim_{n\to \infty}A_n = \displaystyle \int_{0}^1 \ln(1+x)dx= ...$

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