The sum of the infinite series $${\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 2 - 1}}{{\sqrt 6 }}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt {12} }}} \right) + ..... + ..... + {\sin ^{ - 1}}\left( {\frac{{\sqrt n - \sqrt {\left( {n - 1} \right)} }}{{\sqrt {\left\{ {n\left( {n + 1} \right)} \right\}} }}} \right) + .....$$                     is

Solution :
$$\eqalign{ & \because {T_r} = {\sin ^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {\left( {r - 1} \right)} }}{{\sqrt {r\left( {r + 1} \right)} }}} \right) \cr & = {\tan ^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {\left( {r - 1} \right)} }}{{1 + \sqrt r \sqrt {\left( {r - 1} \right)} }}} \right) \cr & {S_n} = \sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {\frac{{\sqrt r - \sqrt {\left( {r - 1} \right)} }}{{1 + \sqrt r \sqrt {\left( {r - 1} \right)} }}} \right)} \cr & = \sum\limits_{r = 1}^n {\left\{ {{{\tan }^{ - 1}}\sqrt r - {{\tan }^{ - 1}}\sqrt {\left( {r - 1} \right)} } \right\}} \cr & = {\tan ^{ - 1}}\sqrt n - {\tan ^{ - 1}}\sqrt 0 = {\tan ^{ - 1}}\sqrt n - 0 \cr & \therefore {S_\infty } = {\tan ^{ - 1}}\infty = \frac{\pi }{2} \cr} $$