TITLE: Find the limit using Riemann sum
QUESTION [6 upvotes]: Find the limit: 
$
\lim\limits_{n\to\infty}\sqrt[n]{\left(\dfrac{1 + n}{n^2} \right)\left(\dfrac{4 + 2n}{n^2} \right)...\left(\dfrac{n^2 + n^2}{n^2} \right)}
$
I tried simplifying this limit and the one I get to is:
$
\lim\limits_{n\to\infty}\dfrac{1}{n^2}\bigg(\big(2n\big)!\bigg)^{\dfrac{1}{n}}
$
I have an instruction to write the limit as a definite integral and then calculate its value. I think that there should be a way to represent the last limit as a Riemann sum and then calculate it with the integral. But I'm not sure how to get to the Riemann sum.
Looking forward for any ideas!

REPLY [2 votes]: Let $\;a_n=\dfrac{(2n)!}{n^{2n}}>0\quad$ for all $\;n\in\mathbb{N}\;.$
Since
$\exists\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}=\lim\limits_{n\to\infty}\left[\dfrac{(2n+2)!}{(n+1)^{2n+2}}\cdot\dfrac{n^{2n}}{(2n)!}\right]=$
$=\lim\limits_{n\to\infty}\left[\dfrac{(2n+1)(2n+2)}{(n+1)^2}\cdot\dfrac{n^{2n}}{(n+1)^{2n}}\right]=$
$=\lim\limits_{n\to\infty}\left[\dfrac{4n^2+6n+2}{n^2+2n+1}\cdot\dfrac1{\left(1+\frac1n\right)^{2n}}\right]=\dfrac4{e^2}$
by using an application of Stolz-Cesàro theorem, we get that
$\exists\lim\limits_{n\to\infty}\dfrac{1}{n^2}\bigg(\big(2n\big)!\bigg)^{\dfrac{1}{n}}=\lim\limits_{n\to\infty}\sqrt[n]{\dfrac{(2n)!}{n^{2n}}}=\lim\limits_{n\to\infty}\sqrt[n]{a_n}=\dfrac4{e^2}\;.$