Question

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

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

Answer 1

The correct option is B

$co{t}^{-1}\left(\frac{3}{2}\right)$

Explanation for the correct option:

On solving, we get:

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

Divide this by $3^{2r}$, we get

$\begin{array}{rcl}{S}_k& =& \sum _{r=1}^k{\tan }^{-1}\left(\frac{{\left({\displaystyle \frac{2}{3}}\right)}^r}{{\left({\displaystyle \frac{2}{3}}\right)}^{2r}.2+3}\right)\\ & =& \sum _{r=1}^k{\tan }^{-1}\left(\frac{{\left({\displaystyle \frac{2}{3}}\right)}^r}{3{\left({\displaystyle \frac{2}{3}}\right)}^{2r+1}+1}\right)\end{array}$

Now, put ${\left(\frac{2}{3}\right)}^r=t$

$\begin{array}{rcl}{S}_k& =& \sum _{r=1}^k{\tan }^{-1}\left(\frac{\left({\displaystyle \frac{t}{3}}\right)}{1+{\displaystyle \frac{2}{3}}{t}^2}\right)\\ & =& \sum _{r=1}^k{\tan }^{-1}\left(\frac{t-\left({\displaystyle \frac{2t}{3}}\right)}{1+t.{\displaystyle \frac{2}{3}}t}\right)\\ & =& \sum _{r=1}^k\left[{\tan }^{-1}\left(t\right)-{\tan }^{-1}\left(\frac{2t}{3}\right)\right]\left[\because {\tan }^{-1}\left(x\right)-{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)\right]\\ & =& \sum _{r=1}^k\left[{\tan }^{-1}{\left(\frac{2}{3}\right)}^r-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{r+1}\right]\\ & & \end{array}$

On further solving, we get:

$\begin{array}{rcl}{S}_k& =& \left[{\tan }^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{k+1}\right]\\ & \Rightarrow & S_{\infty }=\underset{k\to \infty }{lim}\left(t{\mathrm{an}}^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{k+1}\right)\\ & =& t{\mathrm{an}}^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}\left(0\right)\\ & =& t{\mathrm{an}}^{-1}\left(\frac{2}{3}\right)\left[\because {\tan }^{-1}\left(0\right)=0\right]\\ & =& co{t}^{-1}\frac{2}{3}\end{array}$

Hence, the correct answer option is (B).