Evaluate the limit: $\lim_{n\to \infty} \frac{\left[(n+1)(n+2)\ldots(n+n)\right]^{1/n}}{n}$ 
Evaluate the limit: $\lim_{n\to \infty} \frac{\left[(n+1)(n+2)\ldots(n+n)\right]^{1/n}}{n}$

I can write $$\lim_{n\to \infty} \frac{[(n+1)(n+2)\ldots(n+n)]^{1/n}}{n}$$ $$=\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{1/n}\left(1+\frac{2}{n}\right)^{1/n}\ldots \left(1+\frac{n}{n}\right)^{1/n}$$$$=e\times \sqrt{e}\times \sqrt[3]{e}\times \ldots \times \sqrt[n-1]{e}\times 1$$.
I don't know how to evaluate the final product.
 A: Taking the logarithm gives
$$ \frac{1}{n}\sum_{k=1}^n\log(n+k)-\log n=\frac{1}{n}\sum_{k=1}^n\log\left(1+\frac{k}{n}\right). $$
This is a Riemann sum which converges to $\int_0^1 \log(1+x)dx=\log 4 -1$. The desired limit is therefore $\frac{4}{e}$.
A: We have, using Stirling's formula:
$$\begin{split}\frac{\prod_{k=1}^n (n+k)}{n^n} = \frac{(2n)!}{n^n \cdot n!} &\underset{n \to \infty}{\sim} \frac{ \left(\frac{2n}{e}\right)^{2n} \sqrt{2\pi 2n}}{n^n \cdot \left(\frac{n}{e}\right)^{n} \sqrt{2\pi n}}\\
&\underset{n \to \infty}{\sim} \frac{4^n \sqrt{2}}{e^n}\end{split}$$
As such, since putting both sides to the power $\frac{1}{n}$ is allowed, we have that our given limit is $\frac{4}{e}$.
That is allowed because if $u_n = e_n v_n$ with $e_n \to 1$, then $u_n^{1/n} = e_n^{1/n} v_n^{1/n}$.
But for all big enough $n$, $\frac{1}{2} \leq e_n \leq \frac{3}{2}$, so for those same $n$: $\left(\frac{1}{2}\right)^{1/n} \leq e_n^{1/n} \leq \left(\frac{3}{2}\right)^{1/n}$, and both expressions on the sides tend to $1$, so the middle one too by the squeeze theorem.
A: Using Pochhammer symbol
$$\prod_{i=1}^n (n+i)=(n+1)_n=\frac{2^{2 n} }{\sqrt{\pi }}\Gamma \left(n+\frac{1}{2}\right)$$ So, you want to compute
$$A_n=\frac 1n\left(\frac{2^{2 n} }{\sqrt{\pi }}\Gamma \left(n+\frac{1}{2}\right)\right)^{\frac 1n}$$
Taking logarithms and using Stirling expansion
$$\log(A_n)=(2 \log (2)-1)+\frac{\log (2)}{2 n}-\frac{1}{24
   n^2}+O\left(\frac{1}{n^4}\right)$$
$$A_n=e^{\log(A_n)}=\frac 4e \left(1+\frac{\log (2)}{2 n}+\frac{3 \log ^2(2)-1}{24 n^2}+O\left(\frac{1}{n^3}\right)\right)$$ whose relative error is less than $0.1$% as soon as $n \geq 2$.
