Calculate limit using Riemann integral/Riemann sum $$\lim_{n \to \infty}\frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}}$$
I was practicing exercise with a visible sum in those but how to get one here?
 A: Note that we can write
$$\begin{align}
\log\left(\frac1n\sqrt[n]{\frac{(2n)!}{n!}}\right)&=\frac1n\left(\log((2n)!)-\log(n!)\right)-\log(n)\\\\
&=\frac1n\left(\sum_{k=1}^{2n}\log(k)-\sum_{k=1}^n\log(k)\right)-\log(n)\\\\
&=\frac1n\sum_{k=n+1}^{2n}\log(k)-\log(n)\\\\
&=\frac1n\sum_{k=1}^{n}\log(k+n)-\log(n)\\\\
&=\frac1n\sum_{k=1}^{n}\left(\log(k+n)-\log(n)\right)\\\\
&=\underbrace{\frac1n\sum_{k=1}^n\log(1+k/n)}_{\text{Riemann Sum for}\,\,\int_0^1 \log(1+x)\,dx=2\log(2)-1}
\end{align}$$
Hence, we have
$$\lim_{n\to \infty}\left(\frac1n\sqrt[n]{\frac{(2n)!}{n!}}\right)=4e^{-1}$$
And we are done!
A: We can also use the following theorem:
If $a_{n}$ is a sequence of positive terms such that $a_{n+1}/a_{n}\to L$ then $a_{n} ^{1/n}\to L$.
Let $a_{n} = (2n)!/(n!n^{n})$ and we can see that $$\frac{a_{n+1}}{a_{n}}=\frac{(2n+2)!}{(n+1)!(n+1)^{n+1}}\cdot \frac{n!n^{n}} {(2n)!}=\frac{2(2n+1)}{n+1}\cdot\frac{n^{n}}{(n+1)^{n}}\to\frac{4}{e}$$ and hence $$a_n^{1/n}=\frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}}\to\frac{4}{e}$$
A: Use $$(2n)!=2^n \cdot n! \cdot 1 \cdot 3 \cdot 5 \ldots (2n-1)$$
Now pull $2^n$ out of the root, push $\dfrac{1}{n}$ into the root, cancel out $n!$ and then use logarithms.
You'll get :
$$A=\frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}} \implies \ln A=\ln2+\sum_{r=1}^{n}\ln\left(\frac{2r-1}{n} \right)$$
Can you proceed now (Using Integrals)? 
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\lim_{n \to \infty}\bracks{{1 \over n}\root[n]{\pars{2n}! \over n!}} & =
\lim_{n \to \infty}\braces{{1 \over n}
\bracks{\root{2\pi}\pars{2n}^{2n + 1/2}\expo{-2n} \over
\root{2\pi}n^{n + 1/2}\expo{-n}}^{1/n}}
\\[5mm] &=
\lim_{n \to \infty}\bracks{{1 \over n}\pars{2^{2n + 1/2}\,n^{n}\expo{-n}}^{1/n}} =
\expo{-1}\lim_{n \to \infty}\bracks{2^{2 + 1/\pars{2n}}} =
\bbx{\ds{4 \over e}}
\end{align}
