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  4. The value of displaystyle∑ r =1∞ cot -1(( r 2/2)+(15/8)) is equal to

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

Inverse Trigonometric Functions

Solution:

$\operatorname{Sum}=\displaystyle\sum_{ r =1}^{\infty} \cot ^{-1}\left(\frac{ r ^2}{2}+\frac{15}{8}\right)=\displaystyle\sum_{ r =1}^{\infty} \tan ^{-1}\left(\frac{1}{\frac{ r ^2}{2}+\frac{15}{8}}\right)=\displaystyle\sum_{ r =1}^{\infty} \tan ^{-1}\left(\frac{2}{ r ^2+\frac{15}{4}}\right)$
$=\displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left(\frac{2}{4+r^2-\frac{1}{4}}\right)=\sum_{r=1}^{\infty} \tan ^{-1}\left(\frac{2}{4+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right)=\displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left(\frac{\frac{1}{2}}{\left(\frac{r+\frac{1}{2}}{2}\right)\left(\frac{r-\frac{1}{2}}{2}\right)}\right)$
$=\displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left\{\frac{\frac{\left(r+\frac{1}{2}\right)}{2}-\frac{\left(r-\frac{1}{2}\right)}{2}}{\left.1+\frac{\left(r+\frac{1}{2}\right)}{2} \cdot \frac{\left(r-\frac{1}{2}\right)}{2}\right)}\right\}=\underset{n \rightarrow \infty}{\text{Lim}} \displaystyle\sum_{r=1}^n\left(\tan ^{-1}\left(\frac{r+\frac{1}{2}}{2}\right)-\tan ^{-1}\left(\frac{r-\frac{1}{2}}{2}\right)\right)$
$=\underset{n \rightarrow \infty}{\text{Lim}}\left(\tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{1}{4}\right)+\left(\tan ^{-1} \frac{5}{4}-\tan ^{-1} \frac{3}{4}\right)+\left(\tan ^{-1} \frac{7}{4}-\tan ^{-1} \frac{5}{4}\right)......... \left(\tan ^{-1}\left(\frac{n}{2}+\frac{1}{4}\right)-\tan ^{-1}\left(\frac{n}{2}-\frac{1}{4}\right)\right)$
$=\underset{n \rightarrow \infty}{\text{Lim}} \left(\tan ^{-1}\left(\frac{n}{2}+\frac{1}{4}\right)-\tan ^{-1} \frac{1}{4}\right)=\frac{\pi}{2}-\tan ^{-1} \frac{1}{4}=\cot ^{-1} \frac{1}{4}=\tan ^{-1} 4$