Find limit (Riemann sum)

Question

$\displaystyle a_n=\sum^{n^2}_{k=1}\sqrt{\frac{k^2-k}{n^8+k^2n^4-kn^4}}$.

I tried solving it by changing it a Riemann sum then integrating, however I couldn't manipulate the algebra to its form.

Answer 1

Fixing $t_k=\frac{k}{n^2}$ for all $0 \leq k \leq n^2$ you can write:

$$a_n=\frac{\left(1-\frac{1}{1+t_k t_{k-1}}\right)^{\frac{1}{2}}}{n^2}$$

Thanks to inequality $t_{k-1} \leq t_k \leq t_{k+1}$ you can squeeze the integral and you reduce to calculate the following:

$$\lim_n a_n=\int_0^1 \left(1-\frac{1}{1+x^2}\right)^{\frac{1}{2}} dx=\int_0^1 \frac{x}{(1+x^2)^{\frac{1}{2}}}dx=1$$

In fact for example $t_k \geq t_{k-1}$ implies $t_k^2 \geq t_{k-1}t_k$ and thus $1+t_k^2 \geq t_{k-1}t_k+1$ and so $\frac{1}{1+t_k^2} \leq \frac{1}{t_{k-1}t_k+1}$.

Then: $$\left(1-\frac{1}{1+t_k^2}\right)^{\frac{1}{2}} \geq \left(1-\frac{1}{t_{k-1}t_k+1}\right)^{\frac{1}{2}}$$

Answer 2

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*}