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

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

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

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

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

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

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

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

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

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

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