Sum this infinite alternating Series

Let $S = 1-\dfrac 15 + \dfrac 17 - \dfrac 1{11} + \dfrac 1{13} - \cdots$ . Consider the following Maclaurin series:

$\begin{aligned} \ln \left(\frac {1+x}{1-x}\right) & = \frac 12 \left(x + \frac {x^3}3 + \frac {x^5}5 + \frac {x^7}7 + \frac {x^9}9 + \frac {x^{11}}{11} + \frac {x^{13}}{13} + \cdots \right) & \small \color{#3D99F6} \text{Putting }x = \omega = e^{\frac {2\pi}3 i} \text{, 3rd root of unity.} \\ \ln \left(\frac {1+e^{\frac {2\pi}3 i}}{1-e^{\frac {2\pi}3 i}}\right) & = \frac 12 \left(\omega + \frac {\omega^3}3 + \frac {\omega^5}5 + \frac {\omega^7}7 + \frac {\omega^9}9 + \frac {\omega^{11}}{11} + \frac {\omega^{13}}{13} + \cdots \right) \\ & = \frac 12 \left( \omega + \frac 13 + \frac {\omega^2}5 + \frac {\omega}7 + \frac 19 + \frac {\omega^2}{11} + \frac {\omega}{13} + \cdots \right) & \small \color{#3D99F6} \text{Taking the imaginary part.} \\ \Im \left(\ln \left(\frac {1+e^{\frac {2\pi}3 i}}{1-e^{\frac {2\pi}3 i}}\right) \right) & = \frac 12 \left(\sin \frac {2\pi}3 + \frac {\sin 0}3 + \frac {\sin \frac {4\pi}3}5 + \frac {\sin \frac {2\pi}3}7 + \frac {\sin 0}9 + \frac {\sin \frac {4\pi}3}{11} + \frac {\sin \frac {2\pi}3}{13} + \cdots \right) \\ & = \frac 12 \left(\frac {\sqrt 3}2 + 0 - \frac {\sqrt 3}2\cdot \frac 15 + \frac {\sqrt 3}2 \cdot \frac 17 + 0 - \frac {\sqrt 3}2 \cdot \frac 1{11} + \frac {\sqrt 3}2 \cdot \frac 1{13} - \cdots \right) \\ & = \frac {\sqrt 3}4 \left( 1 - \frac 15 + \frac 17 - \frac 1{11} + \frac 1{13} - \cdots \right) \\ & = \frac {\sqrt 3}4 S \end{aligned}$

Now, we have:

$\begin{aligned} S & = \frac 4{\sqrt 3} \Im \left(\ln \left(\frac {1+e^{\frac {2\pi}3 i}}{1-e^{\frac {2\pi}3 i}}\right) \right) = \frac 4{\sqrt 3} \Im \left(\ln \left(\frac {3+\sqrt 3 i}{1 - \sqrt 3 i}\right) \right) = \frac 4{\sqrt 3} \Im \left(\ln \left(\frac {(3+\sqrt 3 i)(1+\sqrt 3 i)}{(1 - \sqrt 3 i)(1+\sqrt 3 i)}\right) \right) \\ & = \frac 4{\sqrt 3} \Im \left(\ln \left(\frac {4\sqrt 3 i}4\right) \right) = \frac 4{\sqrt 3} \Im \left(\frac {\ln 3}2 + \ln i \right) = \frac 4{\sqrt 3} \Im \left(\frac {\ln 3}2 + \ln e^{\frac \pi 2i} \right) = \frac 4{\sqrt 3} \Im \left(\frac {\ln 3}2 + \frac \pi 2i \right) \\ & = \frac \pi {2\sqrt 3} \end{aligned}$

Therefore, $a+b = 2+ 3 =\boxed{5}$ .

It is simply the integral from 0 to 1 of the function (1-x^4)/(1-x^6).........