I'm supposed to find the limit of $\lim_{k\to \infty } \frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}}$. By using $e^{\ln }$ and logarithm-laws I managed to form it into $\frac{1}{n}\sum_{k=1}^{n}\sqrt[n]{k+n}$ or
$e^{(\frac{1}{n}\sum_{k=1}^{n}\ln (k+n))-\ln (n)} $
Not sure where to go from here, appreciate all the help
$\begin{eqnarray} \log\left(\frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}}\right)&=&\log\left(\frac{1}{n}\sqrt[n]{(n+1)(n+2)...(n+n)}\right)\\ &=&\left(\frac{1}{n}\sum_{k=1}^n\log(k+n)\right)-\log(n)\\ &=&\left(\frac{1}{n}\sum_{k=1}^n\log(k+n)\right)-\frac{n}{n}\log(n)\\ &=&\frac{1}{n}\left(\sum_{k=1}^n\log(k+n)-\log(n)\right)\\ &=&\frac{1}{n}\sum_{k=1}^n\log\left(1+\frac{k}{n}\right)\\ &\to&\int_0^1\log(x+1)\mathrm{d}x\\ &=&\log4-1\\ &=&\log(4/e) \end{eqnarray}$
Then $\frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}}=\exp(\log\left(\frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}}\right))\to e^{\log(4/e)}=4/e$