Question

Evaluate: ${\sum }_{r=1}^{\infty }{\tan }^{-1}\left(\frac{2}{1+(2r+1)(2r-1)}\right)$

Answer 1

${\sum }_{r=1}^{\infty }ta{n}^{-1}\left(\frac{2}{1+(2r+1)(2r-1)}\right)$

Divide numerator and denomenator by $(2r+1)(2r-1)$

=${\sum }_{r=1}^{\infty }ta{n}^{-1}\left(\frac{\frac{2}{(2r+1)(2r-1)}}{\frac{1}{(2r+1)(2r-1)}+1}\right)$

=${\sum }_{r=1}^{\infty }ta{n}^{-1}\left(\frac{\frac{1}{2r-1}-\frac{1}{2r+1}}{\frac{1}{(2r+1)(2r-1)}+1}\right)$

$ta{n}^{-1}\frac{a-b}{1+ab}=ta{n}^{-1}a-ta{n}^{-1}b$

$\therefore$ ${\sum }_{r=1}^{\infty }ta{n}^{-1}\left(\frac{\frac{1}{2r-1}-\frac{1}{2r+1}}{\frac{1}{(2r+1)(2r-1)}+1}\right)={\sum }_{r=1}^{\infty }(ta{n}^{-1}\frac{1}{2r-1}-ta{n}^{-1}\frac{1}{2r+1})$

Open the summation we get,

$(ta{n}^{-1}1-ta{n}^{-1}\frac{1}{3})+(ta{n}^{-1}\frac{1}{3}-ta{n}^{-1}\frac{1}{5})+(ta{n}^{-1}\frac{1}{5}-ta{n}^{-1}\frac{1}{7})...........\infty$

At infinity $ta{n}^{-1}\frac{1}{\infty }=0$

And other terms all cancel out each other, only $ta{n}^{-1}1$ remains.

$\therefore$ Ans is $ta{n}^{-1}1=\frac{\pi }{4}$