Pi By What!

Let $S= \sum^{\infty}_{r=0}T_{r}$ $T_{r} = \dfrac{1}{16r^{2} + 16r + 3}$

$T_{r}= \dfrac{1}{\left(4r+1\right)\left(4r+3\right)}$

$T_{r}= \dfrac {1}{2}\left( \dfrac{1}{4r+1} - \dfrac{1}{4r+3}\right)$

$S= \dfrac{1}{2}\left(1- \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} .....\right)$ Consider $f\left(x\right) = \dfrac {1}{1+ x^{2}} = \left(1+x^{2}\right)^{-1}$

Expanding $f$ by binomial theorem for any rational index OR you may also use Taylor's series expansion $f\left(x\right) = 1- x^{2} +x^{4} - x^{6} +...$ Integrating, $\tan^{-1}\left(x\right) = x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - .....$ Put $x=1$ to obtain $\dfrac{\pi}{4} = 2S$ Therefore $a=1 , b=8$ Which gives $\sqrt{ a+b} = 3.00$

NOTE:

1) you may have used the series of $\tan^{-1}\left(x\right)$ directly,which I didn't (for the sake of completeness of solution.

2)please reshare if you like the problem ,and follow for more problems