1.∑r=0∞tan−1(r(r+1)+((r+1)!)2)\displaystyle\sum_{r=0}^{\infty} \tan^{-1}\left(\dfrac{r}{(r+1) + ((r+1)!)^2}\right)r=0∑∞tan−1((r+1)+((r+1)!)2r) is equal toa.3π4\dfrac{3\pi}{4}43πb.π2\dfrac{\pi}{2}2πc.cot−13\cot^{-1} 3cot−13d.tan−12\tan^{-1} 2tan−12Login to continueOnly logged in users canattempt or see the solution.
