Question

$\underset{n\to \infty }{lim}\sum _{r=1}^n{\cot }^{-1}\left(2r^2\right)=$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. 0
• D. None of these

Answer 1

The correct option is A $\frac{\pi }{4}$

${\cot }^{-1}\left({2r}^2\right)={\tan }^{-1}\left(\frac{2}{{4r}^2}\right)={\tan }^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right)$

${\cot }^{-1}\left({2r}^2\right)={\tan }^{-1}(2r+1)-{\tan }^{-1}(2r-1)$

So, $\sum {\cot }^{-1}\left({2r}^2\right)=\sum ({\tan }^{-1}(2r+1)-{\tan }^{-1}(2r-1))$

Sum $=\sum {\cot }^{-1}\left({2r}^2\right)={\tan }^{-1}3-{\tan }^{-1}\left(1\right)+{\tan }^{-1}5-{\tan }^{-1}\left(3\right)........{\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-1)$

Sum $=\underset{n\to \infty }{Lim}\sum {\cot }^{-1}\left({2r}^2\right)={\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(1\right)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$