Find limit (Riemann sum) $\displaystyle a_n=\sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^8+k^2n^4-kn^4}}$.
I tried solving it by changing it a Riemann sum then integrating, however I couldn't manipulate the algebra to its form. 

We have:
\begin{align*}
\sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^8+k^2n^4-kn^4}} &= \frac{1}{n^2} \sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^4+k^2-k}} \\ &=  \frac{1}{n^2} \sum^{n^2}_{k=1}\sqrt{1 - \frac{n^4}{n^4+k(k-1)}} \\ &= \frac{1}{n^2} \sum^{n^2}_{k=1}\sqrt{1 - \frac{1}{1+\frac{k}{n^2}\frac{k-1}{n^2}}}\end{align*}
