If Σ^∞_r=1tan^-1≤ft(1/(2r^2))= t, then t is equal to

We have,

$\sum\limits^{∞}_{r=1}tan^{-1}\left(\frac{1}{2r^2}\right)$

$⇒ t =\sum\limits^{∞}_{r=1}tan^{1}\left(\frac{2}{1+4r^2-1}\right)$

$ = \sum\limits^{∞}_{r=1}tan^{1} \begin{Bmatrix} \frac{(2r+1)-(2r-1)}{1-(2r+1)(2r-1)}\end{Bmatrix}$

$ = \sum\limits^{∞}_{r=1}\begin{Bmatrix} tan^{-1}(2r+1)-tan^{-1}(2r-1)\end{Bmatrix}$

$= \lim\limits_{n→∞}\sum\limits^{n}_{r=1}\begin{Bmatrix} tan^{-1}(2r+1)-tan^{-1}(2r-1)\end{Bmatrix}$

$= \lim\limits_{n→∞}\begin{Bmatrix} (tan^{-1}3 - tan^{-1}1) + (tan^{-1} 5 - tan^{-1}3) + ......+(tan^{-1} (2n +1) -tan^{-1} (2n-1))\end{Bmatrix}$

$⇒ t = \lim\limits_{n→∞} (tan^{-1} (2n+1)- tan^{-1}) =\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$

$∴ \tan t = 1$ but $t=\frac{\pi}{4}$