A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{6}\]
E) \[\pi \]
Correct Answer: B
Solution :
Given series can be rewritten as \[\sum\limits_{r=1}^{\infty }{{{\tan }^{-1}}\left( \frac{1}{1+r+{{r}^{2}}} \right)}\] Now, \[{{\tan }^{-1}}\left( \frac{1}{1+r+{{r}^{2}}} \right)={{\tan }^{-1}}\left( \frac{r+1-r}{1+r(r+1)} \right)\] \[={{\tan }^{-1}}(r+1)-{{\tan }^{-1}}(r)\] \[\sum\limits_{r=0}^{n}{[{{\tan }^{-1}}(r+1)-{{\tan }^{-1}}r]}\] \[={{\tan }^{-1}}(n+1)-{{\tan }^{-1}}(0)\] \[={{\tan }^{-1}}(n+1)\] \[\Rightarrow \]\[\sum\limits_{r=0}^{\infty }{{{\tan }^{-1}}\left( \frac{1}{1+r+{{r}^{2}}} \right)}={{\tan }^{-1}}(\infty )=\frac{\pi }{2}\]
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