$ \lim_{n \to \infty} \frac{\sqrt[n]{n*(n+1)...(2n)}}{n}$ It tried to solve this limit
$$    \lim_{n \to \infty} \frac{\sqrt[n]{n*(n+1)...(2n)}}{n}$$
$   \frac{\sqrt[n]{n*(n+1)...(2n)}}{n} = \sqrt[n]{\frac {2n!n}{n!}} \frac{1}{n} \sim \sqrt[n]{\frac { 
 \sqrt {2 \pi  2 n}* (\frac {2n}{e})^ {2n}*n }{\sqrt {2 \pi   n}* (\frac {n}{e})^ {n} }} \frac{1}{n} = 
\sqrt[n]{\frac { 
 \sqrt {2 }* (\frac {2n}{e})^ {n}*(\frac {2n}{e})^ {n}*n }{ (\frac {n}{e})^ {n} }} \frac{1}{n} = 2^{\frac{1}{2n}}* \frac{4}{e}*n^{\frac{1}{n}} \rightarrow \frac{4}{e}$
Is it right?
 A: You can use the Riemann Integral to compute the limit.
Since
\begin{eqnarray}
&&\lim_{n \to \infty} \ln\frac{\sqrt[n]{n(n+1)...(2n)}}{n}\\
&=&\lim_{n \to \infty}\frac1n\sum_{k=1}^n\ln(1+\frac kn)\\
&=&\int_0^1\ln(1+x)dx\\
&=&2\ln2-1
\end{eqnarray}
one has
$$ \lim_{n \to \infty} \frac{\sqrt[n]{n(n+1)...(2n)}}{n}=e^{2\ln2-1}=\frac{4}{e}. $$
A: And, of course, more generally.
Let
$\begin{array}\\
f_k(n)
&=\dfrac{\sqrt[kn]{\prod_{j=1}^{kn}(n+j)}}{n}\\
g_k(n)
&=\ln(f_k(n))\\
&=\frac1{kn}\sum_{j=1}^{kn}\ln(n+j)-\ln(n)\\
&=\frac1{kn}\sum_{j=1}^{kn}(\ln(n)+\ln(1+\frac{j}{n}))-\ln(n)\\
&=\frac1{kn}(kn\ln(n)+\sum_{j=1}^{kn}\ln(1+\frac{j}{n}))-\ln(n)\\
&=\ln(n)+\frac1{kn}\sum_{j=1}^{kn}\ln(1+\frac{j}{n})-\ln(n)\\
&=\frac1{k}\frac1{n}\sum_{j=1}^{kn}\ln(1+\frac{j}{n})\\
&\to \frac1{k}\int_0^{k}\ln(1+x)dx\\
&= \frac1{k}((x+1)\ln(1+x)-x)|_0^{k}\\
&= \frac1{k}((k+1)\ln(1+k)-k)\\
&= (1+1/k)\ln(1+k)-1\\
\end{array}
$
If $k=1$ this is
$2\ln(2)-1
\approx 0.3863
$.
For large $k$ this is about
$\ln(1+k)-1
$
so
$f_k(n)
\approx \dfrac{1+k}{e}
$.
A: Using you way
$$\prod_{i=0}^n (n+i)=\frac{(2 n)!}{(n-1)!}$$
$$a_n=\frac 1n \sqrt[n]{\frac{(2 n)!}{(n-1)!}}\implies \log(a_n)=\frac 1n\log((2n)!)-\frac 1n\log((n-1)!)-\log(n)$$
Now, using Stirling approximation
$$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({p}\right)\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$apply it twice and continue with Taylor series to get
$$\log(a_n)=(2 \log (2)-1)+\frac{\log (2)}{2 n}-\frac{1}{24
   n^2}+O\left(\frac{1}{n^4}\right)$$
$$a_n=e^{\log(a_n)}=\frac{4}{e}+\frac{\log (4)}{e n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
