Question: 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.

Answer: 
By Stirling,

$$\frac1n\sqrt[n]{\frac{(2n)!}{n!}}\sim\frac1n\frac{\sqrt[2n]{4\pi n}\left(\dfrac{2n}e\right)^2}{\sqrt[n]{4\pi n}\dfrac ne}\to\frac 4e.$$
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:
$$\ln(L)=\lim\_{n\to\infty}\frac1n\ln\left(\frac{(2n)!}{n!}\right)-\ln(n)$$