Compute the limit $$ \lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n $$

How can this be done? The best I could do was rewrite the limit as $$ \lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} $$

Following that log suggestion in the comments below: \begin{align} &\ln \left(\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \ln \left( \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \left(\ln \left(\frac{n+1}n \right)^{\frac 1n} + \ln\left(\frac{n+2}n \right)^{\frac 1n} + \cdots + \ln\left(\frac{n+n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \sum_{i=1}^n \ln \left(\frac{n+i}n \right)^{\frac 1n} \\ &= \lim_{n \to \infty} \frac 1n \sum_{i=1}^n \ln \left(1 + \frac in \right) \\ &= \int_1^2 \ln x \, dx \\ &= (x \ln x - x)\vert_1^2 \\ &= (2 \ln 2 - 2)-(1 \ln 1-1) \\ &= (\ln 4-2)-(0-1) \\ &= \ln 4-1 \\ &= \ln 4 - \ln e \\ &= \ln \left( \frac 4e \right) \end{align}

but I read from somewhere that the answer should be $\frac 4e$.

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    $\begingroup$ Your integral is the log of the answer $\endgroup$ – robjohn May 14 '18 at 18:15
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    $\begingroup$ @CarlMummert From the accepted answer of this meta question, we can see that this is a well-written question (except the use of $\frac1n$ as a power), and "additional indication of source would not make a difference".You may re-read AmateurMathGuy's reply to your comment to know why they are not required. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 14 '18 at 18:19
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    $\begingroup$ You voted to close because "Being homework is a sufficient reason to vote to close a question", but HW questions are allowed. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 14 '18 at 18:19
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    $\begingroup$ @CarlMummert I do not see how my question "has no real context, and does not fit the quality standards that many users expect." I put in the work and effort as this Math SE site suggests. $\endgroup$ – GarlicBread May 15 '18 at 3:58
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    $\begingroup$ Agreed. This is supported by the Close Queue Review result. All reviewers left this question open. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 15 '18 at 9:00

Hint: let $a_n=(n+1)\cdots(2n)/n^n$, and use the fact that $a_{n+1}/a_n\to L$ implies $\sqrt[n]{a_n}\to L$ (i.e., the ratio test implies the root test).


If you know Sterling's Approximation: $$ n! \sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n}, $$ then you could approach it as follows:

$$ \lim_{n \to \infty} \frac{\sqrt[n]{(n + 1)(n + 2)\cdots (2n)}}{n} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{(2n)!}{n!}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{\frac{\left(\frac{2n}{e}\right)^{2n} \sqrt{4\pi n}}{\left(\frac{n}{e}\right)^n \sqrt{2\pi n}}} = \lim_{n \to \infty} \frac{1}{n} \sqrt[n]{2^{2n} \left(\frac{n}{e}\right)^n \sqrt{2}} = \lim_{n \to \infty} \frac{1}{n} \cdot 2^2 \cdot \frac{n}{e} \cdot 2^{\frac{1}{2n}} = \frac{4}{e}. $$


Now take a logarithm and realize you have just gotten a Riemann sum. (Interval: $[1,2]$, widths: $1/n$, heights $\ln(1+k/n)$.)

  • $\begingroup$ I'm not sure how to rewrite the limit of the Riemann sum as its corresponding definite integral, although I think I managed to arrive at the Riemann sum in my work. $\endgroup$ – GarlicBread May 14 '18 at 17:41

If you accept using the following limit that has been asked and proved (for example here) many times here on MSE

  • $\lim_{n\rightarrow \infty}\frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$,

then your limit is easily derived:

$$\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n = \frac{ (2n!)^{\frac{1}{n}} }{ (n!)^{\frac{1}{n}}\cdot n} = 4\frac{n}{ (n!)^{\frac{1}{n}}} \left( \frac{ (2n!)^{\frac{1}{2n}} }{ 2n} \right)^2 \stackrel{n\rightarrow\infty}{\longrightarrow}4\cdot e \cdot \frac{1}{e^2}=\frac{4}{e}$$


Note that $\displaystyle\frac{\sqrt[n]{n+1}}{n}\neq\left(\frac{n+1}{n}\right)^{1/n}$

Also, this is just for clarification since some nice suggestions have already been given but also because the question mark remains over your last equality

With this is mind and going from your first step,










This $+\operatorname{ln}(n)$ is independent of $p$ and summed $n$ times, then divided by $n$, so $\operatorname{ln}(n)-\operatorname{ln}(n)=0$


This is what Eric meant and the argument varies from $1$ to $2$

It becomes a little clearer if we let $m = 1/n$

$$\lim_{n\to\infty}\left(\frac{1}{n}\sum_{p=1}^n\operatorname{ln}\left(1+\frac{p}{n}\right)\right)=\lim_{m\to\ 0}\left(\left(\sum_{p=1}^{1/m}\operatorname{ln}(1+pm)\right)m\right)=\int_1^2\operatorname{ln}(x)dx=(x\operatorname{ln}(x)-x)\Big\rvert_1^2$$

$$\therefore \operatorname{ln}(S)=2(\operatorname{ln}(2)-1)+1=\operatorname{ln}(4)-1=\operatorname{ln}\left(\frac{4}{e}\right)$$




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