Arcsin Program!

Suppose $\displaystyle x = \sin(y)$ (such that $\displaystyle \sin^{-1}(x) = y$ )

Now,

$\displaystyle \tan(y) = \sqrt{1 + \frac{1}{1-\sin^2(y)}}$

$\displaystyle \tan(y) = \sqrt{\frac{2-x^2}{1-x^2}}$

or $\displaystyle y = \sin^{-1}(x) = \tan^{-1}\left(\sqrt{\frac{2-x^2}{1-x^2}}\right)$

According to the question,

$\displaystyle x = \frac{\sqrt{r} - \sqrt{r-1}}{\sqrt{r(r+1)}}$ and so the sum becomes

$\displaystyle \sum_{r = 1}^{n}{\tan^{-1}\left(\sqrt{\frac{2-{\left(\frac{\sqrt{r} - \sqrt{r-1}}{\sqrt{r(r+1)}}\right)}^2}{1-{\left(\frac{\sqrt{r} - \sqrt{r-1}}{\sqrt{r(r+1)}}\right)}^2}}\right)}$

Little bashing shows

$\displaystyle \sum_{r = 1}^{n}{\tan^{-1}\left(\frac{\sqrt{r} - \sqrt{r-1}}{1 + \sqrt{r(r-1)}}\right)}$

$\displaystyle \sum_{r = 1}^{n}{\tan^{-1}(\sqrt{r}) - \tan^{-1}(\sqrt{r-1})}$

$\displaystyle \boxed{\tan^{-1}(\sqrt{n})}$