A Minute Limit

Let our limit be $S$ , we have:

$\displaystyle S=\lim_{n\to \infty} n\sum_{m=1}^{n-1} \frac{\tan^{-1} \sqrt{\frac{n-m}{m}}}{n^2-mn+m^2}$

Not that if we replace $m$ by $n-m$ , the limits of the sum will be reversed, but the interval of values of $m$ will remain the same ( $[1,n-1]$ ). So we have:

$\displaystyle S= \lim_{n\to \infty} n\sum_{m=1}^{n-1} \frac{\tan^{-1} \sqrt{\frac{n-(n-m)}{(n-m)}}}{n^2-(n-m)n+(n-m)^2}$

$\displaystyle = \lim_{n\to \infty} n\sum_{m=1}^{n-1} \frac{\tan^{-1} \sqrt{\frac{m}{n-m}}}{n^2-mn+m^2}$

$\displaystyle = \lim_{n\to \infty} n\sum_{m=1}^{n-1} \frac{\cot^{-1} \sqrt{\frac{n-m}{m}}}{n^2-mn+m^2}$

Then:

$\displaystyle 2S = \lim_{n\to \infty} n\sum_{m=1}^{n-1} \frac{\tan^{-1} \sqrt{\frac{n-m}{m}} +\cot^{-1} \sqrt{\frac{n-m}{m}}}{n^2-mn+m^2}$

$\displaystyle 2S = \lim_{n\to \infty} n\sum_{m=1}^{n-1} \frac{\frac{\pi}{2}}{n^2-mn+m^2}$

$\displaystyle => S=\frac{\pi}{4} \lim_{n\to \infty} n\sum_{m=1}^{n-1} \frac{1}{n^2(1-\frac{m}{n}+(\frac{m}{n})^2)}$

$\displaystyle = \frac{\pi}{4} \lim_{n\to \infty} \frac{1}{n} \sum_{m=1}^{n-1} \frac{1}{1-\frac{m}{n}+(\frac{m}{n})^2}$

Notice that our limit now can be transformed into the Riemann Integral:

$\displaystyle = \frac{\pi}{4} \int_0^1 \frac{1}{1-x+x^2} dx$

This integral can be easily computed using substitution , to yield the value of $\frac{2\pi}{3\sqrt{3}}$ .

Therefore: $\boxed{S= \frac{\pi^2}{6\sqrt{3}}}$ .