Question

Let $S_k=\sum _{r=1}^k{\tan }^{-1}\left(\frac{6^r}{2^{2r+1}+3^{2r+1}}\right)$, then $\underset{k\to \infty }{lim}{S}_k$ is equal to :

• A. ${\tan }^{-1}\left(\frac{3}{2}\right)$
• B. ${\cot }^{-1}\left(\frac{3}{2}\right)$
• C. $\frac{\pi }{2}$
• D. ${\tan }^{-1}\left(3\right)$

Answer 1

The correct option is C $\frac{\pi }{2}$

$\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{6^r}{(1+{\left(\frac{3}{2}\right)}^{2r+1})2^{2r+1}}\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{2^r.3^{r+1}-3^r{2}^{r+1}}{(1+{\left(\frac{3}{2}\right)}^{2r+1})2^{2r+1}}\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{{\left(\frac{3}{2}\right)}^{r+1}-{\left(\frac{3}{2}\right)}^r}{1+{\left(\frac{3}{2}\right)}^{r+1}{\left(\frac{3}{2}\right)}^r}\right)=\sum _{r=1}^{\infty }[{\tan }^{-1}{\left(\frac{3}{2}\right)}^{r+1}-{\tan }^{-1}{\left(\frac{3}{2}\right)}^r]=\frac{\pi }{2}-{\tan }^{-1}\frac{3}{2}={\cot }^{-1}\frac{3}{2}$