Let S_k=_r=1^k^-1(6^r2^2r+1+3^2r+1).

Question

Let $S_k=\displaystyle\sum_{r=1}^{k}	an^{-1}\left(\dfrac{6^r}{2^{2r+1}+3^{2r+1}}ight)$. Then $\displaystyle\lim_{k	o\infty}S_k$ is equal to

Accepted Answer

Sk = $$\sum\limits_{r = 1}^k  {{{\tan }^{ - 1}}} \left( {{{{6^r}(3 - 2)} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)$$

= $$\sum\limits_{r = 1}^k  {{{\tan }^{ - 1}}} \left( {{{{2^r}\,.\,{3^{r + 1}} - {3^r}{2^{r + 1}}} \over {\left( {1 + {{\left( {{3 \over 2}} \right)}^{2r + 1}}} \right){2^{2r + 1}}}}} \right)$$

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

= $$\sum\limits_{r = 1}^k {\left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{r + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^r}} \right]} $$

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

+ $${\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^2}$$

+ $${\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^4} - {\tan ^{ - 1}}{\left( {{3 \over 2}} \right)^3}$$

.

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

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

$$ \therefore $$ $$\mathop {\lim }\limits_{k \to \infty } {S_k}$$

= $$\mathop {\lim }\limits_{k \to \infty } \left[ {{{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^{k + 1}} - {{\tan }^{ - 1}}{{\left( {{3 \over 2}} \right)}^1}} \right]$$

= $${{{\tan }^{ - 1}}\left( \infty  \right) - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}$$

= $${{\pi  \over 2} - {{\tan }^{ - 1}}\left( {{3 \over 2}} \right)}$$

= $${\cot ^{ - 1}}\left( {{3 \over 2}} \right)$$

Solutions · 1