Question

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

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

Answer 1

Answer: (C)

Sol. $S_k={\sum }_{r=1}^k{\tan }^{-1}\left(\frac{6^r}{2^{2r+1}+3^{2r+1}}\right)$
Divide by $3^{2r}$
${\sum }_{r=1}^k{\tan }^{-1}\left(\frac{{\left(\frac{2}{3}\right)}^r}{{\left(\frac{2}{3}\right)}^{2r}\cdot 2+3}\right)$
${\sum }_{r=1}^k{\tan }^{-1}\left(\frac{{\left(\frac{2}{3}\right)}^r}{3\left({\left(\frac{2}{3}\right)}^{2r+1}+1\right)}\right)$
Let ${\left(\frac{2}{3}\right)}^r=t$
${\sum }_{r=1}^k{\tan }^{-1}\left(\frac{\frac{t}{3}}{1+\frac{2}{3}{t}^2}\right)$
${\sum }_{r=1}^k{\tan }^{-1}\left(\frac{t-\frac{2t}{3}}{1+t\cdot \frac{2t}{3}}\right)$
${\sum }_{r=1}^k\left({\tan }^{-1}(t)-{\tan }^{-1}\left(\frac{2t}{3}\right)\right)$
${\sum }_{r=1}^k\left({\tan }^{-1}{\left(\frac{2}{3}\right)}^r-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{r+1}\right)$
$S_k={\tan }^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{k+1}$
$S_{\infty }={\lim }_{k\to \infty }\left({\tan }^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}{\left(\frac{2}{3}\right)}^{k+1}\right)$
$={\tan }^{-1}\left(\frac{2}{3}\right)-{\tan }^{-1}(0)$
$\therefore S_{\infty }={\tan }^{-1}\left(\frac{2}{3}\right)={\cot }^{-1}\left(\frac{3}{2}\right)$