question_answer

The sum \[\sum\limits_{n=1}^{\infty }{{{\tan }^{-1}}\left( \frac{3}{{{n}^{2}}+n-1} \right)}\] is equal to

A) \[\frac{3\pi }{4}-{{\cot }^{-1}}2\]       

B)        \[\frac{3\pi }{4}+{{\cot }^{-1}}2\]

C) \[\frac{\pi }{2}+{{\cot }^{-1}}3\]         

D)        \[\frac{\pi }{2}+ta{{n}^{-1}}2\]

Correct Answer: A

Solution :

[a] \[=\sum\limits_{n=1}^{\infty }{{{\tan }^{-1}}}\frac{(n+2)-(n-1)}{1+(n-1)\,\,(n+2)}\] \[={{\tan }^{-1}}0+{{\tan }^{-1}}1+{{\tan }^{-1}}2\] \[=\frac{\pi }{4}+{{\tan }^{-1}}2\] \[=\frac{\pi }{4}+\frac{\pi }{2}-{{\cot }^{-1}}2\] \[=\frac{3\pi }{4}-{{\cot }^{-1}}2\]