Question

`\rm\sum \limits_(n = 0)^( \infty ) \arcsin \large \left( ( √(n + 3) )/((n + 2) √(n + 1) ) - ( √(n) )/((n + 1) √(n + 2) ) \right)`​

Answer 1

Recall that over an appropriate domain,

`\arcsin(x) \pm \arcsin(y) = \arcsin\left(x √(1-y^2) \pm y √(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

`(√(n+3))/((n+2)√(n+1)) = \frac1{n+1} \sqrt{((n+3)(n+1))/((n+2)^2)} = \frac1{n+1} \sqrt{1 - \frac1{(n+2)^2}} = x √(1-y^2)`

and

`(\sqrt n)/((n+1)√(n+2)) = \frac1{n+2} \sqrt{(n(n+2))/((n+1)^2)} = \frac1{n+2} \sqrt{1 - \frac1{(n+1)^2}} = y √(1-x^2)`

Then the sum telescopes, as

`\displaystyle \sum_(n=0)^\infty \arcsin\left(x √(1-y^2) - y √(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}`