<s>[INST] Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$ 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 $\frac 4e$.
 [/INST] 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:
$$\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n = \frac{ (2n!)^{\frac{1}{n}} }{ (n!)^{\frac{1}{n}}\cdot n} = 4\frac{n}{ (n!)^{\frac{1}{n}}} \left( \frac{ (2n!)^{\frac{1}{2n}} }{ 2n} \right)^2 \stackrel{n\rightarrow\infty}{\longrightarrow}4\cdot e \cdot \frac{1}{e^2}=\frac{4}{e}$$
</s>