How do i show$\lim_{n\rightarrow \infty} \frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=\frac{4}{e}$ without using integration? Without using integration show that $$\lim_{n\rightarrow \infty} \frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=\frac{4}{e}$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.
 A: By Stirling,
$$\frac1n\sqrt[n]{\frac{(2n)!}{n!}}\sim\frac1n\frac{\sqrt[2n]{4\pi n}\left(\dfrac{2n}e\right)^2}{\sqrt[n]{4\pi n}\dfrac ne}\to\frac 4e.$$
A: If we let $a_n=\frac{[(n+1)(n+2)...(2n)]}{n^n}=\frac{(2n)!}{n!n^n}$, then $\frac{[(n+1)(n+2)...(2n)]^{\frac{1}{n}}}{n}=(a_n)^{\frac{1}{n}}$. Then $\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{(2n)!}×\frac{n!}{(n+1)!}×\frac{n^n}{(n+1)^{n+1}}=\frac{(2n+1)(2n+2)}{(n+1)}×\frac{n^n}{(n+1)^{n+1}}=\frac{2(2n+1)}{(n+1)}×\frac{n^n}{(n+1)^{n+1}}=\frac{4n(1+\frac{1}{2n})}{n(1+\frac{1}{n})}×(1+\frac{1}{n})^{-n}\rightarrow4×e^{-1}$.  
Edit:explaining the last step explicitly... note that as $n\rightarrow \infty, \frac{1}{n},\frac{1}{2n}\rightarrow 0$, so we can replace $\frac{1}{2n}$ with $\frac{1}{n}$. Now hence, $\frac{4n(1+\frac{1}{2n})}{n(1+\frac{1}{n})}×(1+\frac{1}{n})^{-n}=4×(1+\frac{1}{n})^{-n} \rightarrow 4× e^{-1}$
A: By taking logs of both sides we can get:
$$L=\lim_{n\to\infty}\frac{\left[(n+1)(n+2)(n+3)...(2n)\right]^{1/n}}{n}$$
$$\ln(L)=\lim_{n\to\infty}\frac1n\ln\left[(n+1)(n+2)(n+3)...(2n)\right]-\ln(n)$$
$$\ln(L)=\lim_{n\to\infty}\frac1n\sum_{i=1}^n\ln(n+i)-\ln(n)$$
And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
$$\ln(L)=\lim_{n\to\infty}\frac1n\ln\left(\frac{(2n)!}{n!}\right)-\ln(n)$$
