Practicing Success
If $\sum\limits^{∞}_{r=1}tan^{-1}\left(\frac{1}{2r^2}\right)= t,$ then t is equal to |
$\frac{2}{3}$ 1 $\frac{\sqrt{5}}{3}$ none of these |
none of these |
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 $t=\frac{\pi}{4}$ |