Question

ge%20%5Cleft%28%20%5Cfrac%7B%20%5Csqrt%7Bn%20%2B%203%7D%20%7D%7B%28n%20%2B%202%29%20%5Csqrt%7Bn%20%2B%201%7D%20%7D%20%20-%20%20%5Cfrac%7B%20%5Csqrt%7Bn%7D%20%7D%7B%28n%20%2B%201%29%20%5Csqrt%7Bn%20%2B%202%7D%20%7D%20%20%5Cright%29%20" id="TexFormula1" title=" \rm\sum \limits_{n = 0}^{ \infty } \arcsin \large \left( \frac{ \sqrt{n + 3} }{(n + 2) \sqrt{n + 1} } - \frac{ \sqrt{n} }{(n + 1) \sqrt{n + 2} } \right) " alt=" \rm\sum \limits_{n = 0}^{ \infty } \arcsin \large \left( \frac{ \sqrt{n + 3} }{(n + 2) \sqrt{n + 1} } - \frac{ \sqrt{n} }{(n + 1) \sqrt{n + 2} } \right) " align="absmiddle" class="latex-formula">​

Answer 1

Recall that over an appropriate domain,

$\arcsin(x) \pm \arcsin(y) = \arcsin\left(x \sqrt{1-y^2} \pm y \sqrt{1-y^2}\right)$

Let $x=\frac1{n+1}$ and $y=\frac1{n+2}$ (these belong to the "appropriate domain", so the identity holds). We have

$\dfrac{\sqrt{n+3}}{(n+2)\sqrt{n+1}} = \dfrac1{n+1} \sqrt{\dfrac{(n+3)(n+1)}{(n+2)^2}} = \dfrac1{n+1} \sqrt{1 - \dfrac1{(n+2)^2}} = x \sqrt{1-y^2}$

and

$\dfrac{\sqrt n}{(n+1)\sqrt{n+2}} = \dfrac1{n+2} \sqrt{\dfrac{n(n+2)}{(n+1)^2}} = \dfrac1{n+2} \sqrt{1 - \dfrac1{(n+1)^2}} = y \sqrt{1-x^2}$

Then the sum telescopes, as

$\displaystyle \sum_{n=0}^\infty \arcsin\left(x \sqrt{1-y^2} - y \sqrt{1-x^2}\right) = \sum_{n=0}^\infty \left( \arcsin(x) - \arcsin(y) \right) \\\\ = \left(\arcsin(1) - \arcsin\left(\frac12\right)\right) + \left(\arcsin\left(\frac12\right) - \arcsin\left(\frac13\right)\right) + \cdots \\\\ = \arcsin(1) = \boxed{\frac\pi2}$