It tried to solve this limit
$$ \lim_{n \to \infty} \frac{\sqrt[n]{n\cdot(n+1)\cdots(2n)}}{n}$$ $ \frac{\sqrt[n]{n\cdot(n+1)\cdots(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?
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 $$ \lim_{n \to \infty} \frac{\sqrt[n]{n(n+1)...(2n)}}{n}=e^{2\ln2-1}=\frac{4}{e}. $$
As suggested in the comments, we can also use that
$$\frac{a_{n+1}}{a_n}\to L \implies \sqrt[n]{a_n}\to L$$
then by
$$\sqrt[n]{a_n}=\sqrt[n]{\frac{n\cdot(n+1)\cdots(2n)}{n^n}}$$
we have
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)\cdot(n+2)\cdots(2n+2)}{(n+1)^{n+1}}\frac{n^n}{n\cdot(n+1)\cdots(2n)}=\frac{(2n+1)(2n+2)}{n(n+1)}\frac1{\left(1+\frac1n\right)^n} \to \frac 4 e$$
therefore $\sqrt[n]{a_n}\to \frac 4 e $.
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.
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} $.
$$ \lim\limits_{n\to\infty}\frac{\sqrt[n]{n(n+1)\dots(2n-1)}}{n}$$
$$ \frac{1}{n}\exp\left(\sum_{i=0}^{n-1}\frac{\log(n+1)}{n}\right)=\frac{1}{n}\exp\left(\frac{\log n}{n}+\frac{\log (n+1)}{n}+\dots +\frac{\log(2n-1)}{n}\right) $$
If we can get form like $$\exp\left(\frac{1}{n}\sum_{i=1}^{n-1}\log(1+i/n) \right)=\exp(\int_1^2\ln(1+i)di)\implies e^{?}$$
$$ \frac{1}{n}\sum_{i=0}^{n-1}\left(\log(n)+\log(\frac{n+i}{n})\right)=\frac{1}{n}(n-1+1)\log n+\frac{1}{n}\sum_{i=0}^{n-1}\log(1+\frac{i}{n}) $$ Find limit when $n\to\infty$, that's Riemann wanted. The next step we can calculate integral $$\exp\left(\int_0^1\ln(1+t)dt\right)=\frac{4}{e}$$ That's becaues $a_n=\frac{i}{n},i\to n-1$ has upper bound and lower bound, substitute with $t\in[0,1]$
