calculate this sum of limit using riemann sum, $\lim_\limits {n\to \infty} \frac{1}{n}{\left({\dfrac{(2n)!}{(n!)}}\right)}^{1/n} $
\begin{align*} &=\ln \frac{1}{n}+\ln\left ( \frac{1}{n} + \frac{1}{n}{\log 2n+\log (2n-1)+....+\log 1-(\log n+\log(n-1)+...+\log 1)} \right)\\ &=**\log \frac{1}{n}+ \frac{1}{n}\sum_{k=1}^n \ln(n+k)\\&= \frac{1}{n}\left[\sum_{k=1}^n\ln \frac{1}{n}+log(n+k)=\frac{1}{n}\sum_{k=1}^n\ln\left( 1+\frac{k}{n} \right)\right]**\\ . \end{align*}
$\frac{1}{n}*{{\log 2n+\log (2n-1)+....+\log 1-(\log n+ \log(n-1)+....+\log 1)}}$ i was confused how can i get the starred part, how to manipulate it to $\sum \ln(n+k)$? is this using property of power series? can someone explain this so clearly?
Let's compute the log and do some algebra: $$ \log \frac{1}{n}{\left({\dfrac{(2n)!}{(n!)}}\right)}^{1/n} = \log \left[ 2 \times \left( \frac{n!}{n!} \times \frac{(n+1)(n+2)\cdots(2n-1)(2n)}{2^n \times n^{n}} \right)^{1/n} \right] , $$ which is $$ \log 2 + \frac{1}{n} \log \left( \frac{n+1}{2n} \times \frac{n+2}{2n} \times \cdots \times \frac{2n}{2n} \right) , $$ and rewrinting as a sum $$ \log 2 + 2 \times \left[ \frac{1}{2n} \left( \log \frac{n+1}{2n} + \log\frac{n+2}{2n} + \cdots + \log\frac{2n}{2n} \right) \right] . $$ Inside the bracket we have a Riemann sum for $$ \int_{1/2}^1 \log x\, \mathrm{d} x = \frac{-1+\log 2}2 . $$ Therefore the log of the limit is $$ \log 2 + \log 2 - 1 $$ so the limit is $$ 4/e . $$