Question

The value of $\sum _{r=1}^{\infty }{\cot }^{-1}\left(\frac{r^2}{2}+\frac{15}{8}\right)$ is equal to

• A. ${\tan }^{-1}1$
• B. ${\tan }^{-1}2$
• C. ${\tan }^{-13}$
• D. ${\tan }^{-1}4$

Answer 1

Answer: (D)

$\mathrm{Sum}=\sum _{r=1}^{\infty }{\cot }^{-1}\left(\frac{r^2}{2}+\frac{15}{8}\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{\frac{r^2}{2}+\frac{15}{8}}\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{2}{r^2+\frac{15}{4}}\right)$
$=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{2}{4+r^2-\frac{1}{4}}\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{2}{4+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{\frac{1}{2}}{\left(\frac{r+\frac{1}{2}}{2}\right)\left(\frac{r-\frac{1}{2}}{2}\right)}\right)$
$=\sum _{r=1}^{\infty }{\tan }^{-1}\left\{\frac{\frac{\left(r+\frac{1}{2}\right)}{2}-\frac{\left(r-\frac{1}{2}\right)}{2}}{1+\frac{\left(r+\frac{1}{2}\right)}{2}\cdot \frac{\left(r-\frac{1}{2}\right)}{2})}\right\}=\underset{n\to \infty }{\mathrm{Lim}}\sum _{r=1}^n\left({\tan }^{-1}\left(\frac{r+\frac{1}{2}}{2}\right)-{\tan }^{-1}\left(\frac{r-\frac{1}{2}}{2}\right)\right)$
$=\underset{n\to \infty }{\mathrm{Lim}}\left({\tan }^{-1}\frac{3}{4}-{\tan }^{-1}\frac{1}{4}\right)+\left({\tan }^{-1}\frac{5}{4}-{\tan }^{-1}\frac{3}{4}\right)+\left({\tan }^{-1}\frac{7}{4}-{\tan }^{-1}\frac{5}{4}\right).........\left({\tan }^{-1}\left(\frac{n}{2}+\frac{1}{4}\right)-{\tan }^{-1}\left(\frac{n}{2}-\frac{1}{4}\right)\right)$
$=\underset{n\to \infty }{\mathrm{Lim}}\left({\tan }^{-1}\left(\frac{n}{2}+\frac{1}{4}\right)-{\tan }^{-1}\frac{1}{4}\right)=\frac{\pi }{2}-{\tan }^{-1}\frac{1}{4}={\cot }^{-1}\frac{1}{4}={\tan }^{-1}4$