Let $${S_k} = \sum\limits_{r = 1}^k {{{\tan }^{ - 1}}\left( {{{{6^r}} \over {{2^{2r + 1}} + {3^{2r + 1}}}}} \right)} $$. Then $\mathop {\lim }\limits_{k \to \infty } {S_k}$ is equal to :

S_k = $$\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{6^r}(3 - 2)} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)$$<br><br>= $$\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{2^r}\,.\,{3^{r + 1}} - {3^r}{2^{r + 1}}} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)$$ <br><br>= $$\sum\limits_{r = 1}^k {{{\tan }^{ - 1}}} \left( {{{{{\left( {{3 \over 2}} \right)}^{r + 1}} - {{\left( {{3 \over 2}} \right)}^r}} \over {1 + {{\left( {{3 \over 2}} \right)}^{r + 1}}{{\left( {{3 \over 2}} \right)}^r}}}} \right) $$ <br><br>= $$\sum\limits_{r = 1}^k {\left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{r + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^r}} \right]} $$ <br><br>= $${\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^2} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^1}$$ <br><br>+ $${\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^2}$$ <br><br>+ $${\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^4} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3}$$ <br>. <br>. <br>. <br>+ $${{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^k}}$$ <br><br>= $${{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^1}}$$ <br><br>$\therefore$ $\mathop {\lim }\limits_{k \to \infty } {S_k}$ <br><br>= $$\mathop {\lim }\limits_{k \to \infty } \left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^1}} \right]$$ <br><br>= $${{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}$$ <br><br>= ${{\pi \over 2} - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}$ <br><br>= ${\cot ^{ - 1}}\left( {{3 \over 2}} \right)$