How to find the limit of this hard sequence? How to find this limits
$$\lim_{n \to \infty}\sum_{i=1}^{n}\frac{n^2-(i-1)^2}{n\left\{[2n(a+b)]^2+[a(2i-1)]^2\right\}}$$
Where $a$ and $b$  are real positive constants.
My friend asked this question,but I have no idea how to do it.
 A: Unless I'm missing something, this is a simple Riemann sum. Just write $$\lim_{n\to\infty}~\sum_{i=1}^n\frac{n^2-(i-1)^2}{n\left\{[2n(a+b)]^2+[a(2i-1)]^2\right\}}~=~\lim_{n\to\infty}~\sum_{i=1}^n\frac{n^2-i^2}{n\left\{[2n(a+b)]^2+[a(2i)]^2\right\}},$$ since, as $n\to\infty,~$ the influence of the $-1$ on the value of the series tends to $0,~$ so its presence or absence is irrelevant. $($Even if it would have been $\pm1,000,~$ or even $\pm1,000,000,~$ it still would not have made any difference$).~$ Now drag the $\dfrac1n$ outside the series, since its value doesn't depend on the iterator i, and simplify the remaining fraction by dividing both its numerator and denominator by $n^2.~$ We then have $S=\displaystyle\int_0^1\frac{1-x^2}{4(a+b)^2~+~4a^2x^2}~dx,~$ whose evaluation is trivial.
A: I'm sorry I post this as an answer when it is only a lower bound, but I do not have enough reputation to comment.
First note that, after the swap $i \to n-i+1$ the sum is
$$
\frac{1}{n} \sum_{i=1}^n \frac{i(2n-i)}{4n^2(a+b)^2+a^2(2n+1-2i)^2}
$$
By AM-GM, the $n$-th term is greater than or equal to
\begin{align*}
\left (\prod_{i=1}^n \frac{i(2n-i)}{4n^2(a+b)^2+a^2(2n+1-2i)^2} \right )^{1/n} &= \left [\frac{(2n)!}{2} \right ]^{1/n} \left (\prod_{i=1}^n \frac{1}{4n^2(a+b)^2+a^2(2n+1-2i)^2} \right )^{1/n} \\ &\geq \left [\frac{(2n)!}{2} \right ]^{1/n} \left (\prod_{i=1}^n \frac{1}{4n^2[(a+b)^2+a^2]} \right )^{1/n} \\ &= \left [\frac{(2n)!}{2} \right ]^{1/n} \frac{1}{4n^2[(a+b)^2+a^2]}
\end{align*}
which tends to
$$
\frac{1}{e^2[(a+b)^2+a^2]}
$$
unless I was wrong using Stirling.
So, taking in account Daniel Fischer's useful comment, one should have that your limit (if it exists) is between
$$
\frac{1}{e^2[(a+b)^2+a^2]} \text{ and } \frac{1}{4(a+b)^2}.
$$
