Recall that over an appropriate domain,
[tex]\arcsin(x) \pm \arcsin(y) = \arcsin\left(x \sqrt{1-y^2} \pm y \sqrt{1-y^2}\right)[/tex]
Let [tex]x=\frac1{n+1}[/tex] and [tex]y=\frac1{n+2}[/tex] (these belong to the "appropriate domain", so the identity holds). We have
[tex]\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}[/tex]
and
[tex]\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}[/tex]
Then the sum telescopes, as
[tex]\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}[/tex]
