Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given $S_k=\sum _{r=1}^k{\tan }^{-1}\left(\frac{6^r}{2^{2r+1}+3^{2r+1}}\right)$

$\Rightarrow S_k=\sum _{r=1}^k{\tan }^{-1}\left(\frac{2^r·3^r}{2^{2r}·2+3^{2r}·3}\right)$

Divide by $3^{2r}$ we get, 

$S_k=\sum _{r=1}^k{\tan }^{-1}\left(\frac{{\left(\frac{2}{3}\right)}^r}{{\left(\frac{2}{3}\right)}^{2r}.2+3}\right)$

$S_k=\sum _{r=1}^k{\tan }^{-1}\left(\frac{{\left(\frac{2}{3}\right)}^r}{3\left({\left(\frac{2}{3}\right)}^{2r+1}+1\right)}\right)$

Let ${\left(\frac{2}{3}\right)}^r=t$

$S_k=\sum _{r=1}^k{\tan }^{-1}\left(\frac{\frac{t}{3}}{1+\frac{2}{3}{t}^2}\right)$

$S_k=\sum _{r=1}^k{\tan }^{-1}\left(\frac{t-\frac{2t}{3}}{1+t.\frac{2t}{3}}\right)$

$S_k=\sum _{r=1}^k\left({\tan }^{-1}\left(t\right)-{\tan }^{-1}\left(\frac{2t}{3}\right)\right)$

$S_k=\sum _{r=1}^k\left({\tan }^{-1}{\left(\frac{2}{3}\right)}^r-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{r+1}\right)$

$S_k=\left({\tan }^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}{\left(\frac{2}{3}\right)}^2\right)+\left({\tan }^{-1}{\left(\frac{2}{3}\right)}^2-{\tan }^{-1}{\left(\frac{2}{3}\right)}^3\right)+...+\left({\tan }^{-1}{\left(\frac{2}{3}\right)}^k-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{k+1}\right)$

$S_k={\tan }^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{k+1}$

Then, $S_{\infty }=\underset{k\to \infty }{\lim }\left({\tan }^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{k+1}\right)$

$={\tan }^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}\left(0\right)$

$\therefore S_{\infty }={\tan }^{-1}\left(\frac{2}{3}\right)={\cot }^{-1}\left(\frac{3}{2}\right), \left(\because {\tan }^{-1}x={\cot }^{-1}\left(\frac{1}{x}\right)\right)$