Question

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