A) \[\frac{\pi }{8}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{2}\]
D) \[\pi \]
Correct Answer: C
Solution :
[c] \[\because {{T}_{r}}={{\sin }^{-1}}\left( \frac{\sqrt{r}-\sqrt{(r-1)}}{\sqrt{r(r+1)}} \right)\] \[={{\tan }^{-1}}\left( \frac{\sqrt{r}-\sqrt{(r-1)}}{1+\sqrt{r}\sqrt{(r-1)}} \right)\] \[{{S}_{n}}=\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \frac{\sqrt{r}-\sqrt{(r-1)}}{1+\sqrt{r}\sqrt{(r-1)}} \right)}\] \[=\sum\limits_{r=1}^{n}{\{ta{{n}^{-1}}\sqrt{r}-ta{{n}^{-1}}\sqrt{(r-1)}\}}\] \[={{\tan }^{-1}}\sqrt{n}-{{\tan }^{-1}}\sqrt{0}={{\tan }^{-1}}\sqrt{n}-0\] \[\therefore {{S}_{\infty }}={{\tan }^{-1}}\infty =\frac{\pi }{2}\]
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