The sum of the infinite series ^-1(74)+^-1(194)+^-1(394)+^-1(674)+ is :

Question

The sum of the infinite series $\cot^{-1}\left(\dfrac74\right)+\cot^{-1}\left(\dfrac{19}{4}\right)+\cot^{-1}\left(\dfrac{39}{4}\right)+\cot^{-1}\left(\dfrac{67}{4}\right)+\cdots$ is :

Accepted Answer

The general term is $T_r=\cot^{-1}\left(\dfrac{4r^2+3}{4}\right)=\tan^{-1}\left(\dfrac{1}{r^2+\frac34}\right)$. Writing $r^2+\dfrac34=\left(r+\dfrac12\right)\left(r-\dfrac12\right)+1$,

$$T_r=\tan^{-1}\left(\frac{\left(r+\frac12\right)-\left(r-\frac12\right)}{1+\left(r+\frac12\right)\left(r-\frac12\right)}\right)=\tan^{-1}\left(r+\frac12\right)-\tan^{-1}\left(r-\frac12\right).$$

This telescopes: $\displaystyle\sum_{r=1}^{n}T_r=\tan^{-1}\left(n+\frac12\right)-\tan^{-1}\left(\frac12\right)$.

As $n\to\infty$, $\tan^{-1}\left(n+\dfrac12\right)\to\dfrac{\pi}{2}$, so the sum is $\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac12\right)$.

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