200 Followers Problem (repost)

Consider the Maclaurin series of $\ln (1+z)$ as follows:

$\begin{aligned} \ln (1+z) & = z - \frac {z^2}2 + \frac {z^3}3 - \frac {z^4}4 + \frac {z^5}5 - \cdots & \small \color{#3D99F6} \text{Putting }z = i \\ \ln (1+i) & = i + \frac 12 - \frac 13i - \frac 14 + \frac 15i - \cdots & \small \color{#3D99F6} \text{Taking the imaginary part} \\ \Im (\ln(1+i)) & = 1 - \frac 13 + \frac 15 - \frac 17 + \frac 19 - \cdots \end{aligned}$

Therefore,

$\begin{aligned} \sum_{m=0}^\infty \frac {(-1)^m}{2m+1} & = \Im (\ln(1+i)) \\ & = \Im \left(\ln \left(\sqrt 2 \left({\color{#3D99F6}\cos \frac \pi 4 +i\sin \frac \pi 4}\right)\right)\right) & \small \color{#3D99F6} \text{By Euler's formula} \\ & = \Im \left(\ln \left(\sqrt 2 {\color{#3D99F6}e^{\frac \pi 4 i}}\right)\right) \\ & = \Im \left(\frac 12 \ln 2 + \frac \pi 4 i\right) \\ & = \frac \pi 4 \end{aligned}$

Similarly,

$\begin{aligned} \frac 12 \ln \left(\frac {1+z}{1-z}\right) & = z + \frac {z^3}3 + \frac {z^5}5 + \frac {z^7}7 + \frac {z^9}9 + \cdots & \small \color{#3D99F6} \text{Dividing both sides by }z \\ \frac 1{2z} \ln \left(\frac {1+z}{1-z}\right) & = 1 + \frac {z^2}3 + \frac {z^4}5 + \frac {z^6}7 + \frac {z^8}9 + \cdots & \small \color{#3D99F6} \text{Putting }z=\sqrt i \\ \frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right) & = 1 + \frac 13i - \frac 15 - \frac 17i + \frac 19 + \cdots \end{aligned}$

Implying $\displaystyle \sum_{n=0}^\infty \left(\frac {(-1)^n}{4n+1} + \frac {(-1)^n}{4n+3} \right) = \Re \left(\frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right)\right) + \Im \left(\frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right)\right)$

$\begin{aligned} \frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right) & = \frac 1{2\sqrt i} \ln \left(\frac {(1+\sqrt i)(1+\sqrt i)}{(1-\sqrt i)(1+\sqrt i)}\right) \\ & = \frac 1{2\sqrt i} \ln \left(\frac {(1+2\sqrt i + i)(1+i)}{(1- i)(1+i)}\right) \\ & = \frac 1{2\sqrt i} \ln \left(\color{#3D99F6}\sqrt i + i + i \sqrt i \right) & \small \color{#3D99F6} \text{By Euler's formula} \\ & = \frac 1{2\sqrt i} \ln \left(\color{#3D99F6}e^{\frac \pi 4 i} + e^{\frac \pi 2 i} + e^{\frac {3\pi}4 i} \right) \\ & = \frac 1{2\sqrt i} \ln \left(\color{#3D99F6}\frac 1{\sqrt 2}+0-\frac 1{\sqrt 2} + i\left(\frac 1{\sqrt 2}+1+\frac 1{\sqrt 2} \right) \right) \\ & = \frac 1{2\sqrt i} \ln \left(\left(1+\sqrt 2 \right)i \right) \\ & = - \frac {i\sqrt i\ln \left(\left(1+\sqrt 2 \right)e^{\frac \pi 2i}\right)}2 \\ & = \frac {(1-i)\left(\ln \left(1+\sqrt 2 \right) + \frac \pi 2i\right)}{2\sqrt 2} \\ & = \frac 1{2\sqrt 2}\left(\frac \pi 2 + \ln(1+\sqrt 2) + i \left(\frac \pi 2 - \ln(1+\sqrt 2) \right)\right) \end{aligned}$

Therefore,

$\begin{aligned} \sum_{n=0}^\infty \left(\frac {(-1)^n}{4n+1} + \frac {(-1)^n}{4n+3} \right) & = \Re \left(\frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right)\right) + \Im \left(\frac 1{2\sqrt i} \ln \left(\frac {1+\sqrt i}{1-\sqrt i}\right)\right) \\ & = \frac 1{2\sqrt 2}\left(\frac \pi 2 + \ln(1+\sqrt 2) \right) + \frac 1{2\sqrt 2}\left(\frac \pi 2 - \ln(1+\sqrt 2)\right) \\ & = \frac \pi {2\sqrt 2} \end{aligned}$

Finally, $\displaystyle \sum_{m=0}^\infty \frac {(-1)^m}{2m+1} \sum_{n=0}^\infty \left(\frac {(-1)^n}{4n+1} + \frac {(-1)^n}{4n+3} \right) = \frac \pi 4 \cdot \frac \pi {2\sqrt 2} = \frac {\pi^2}{8\sqrt 2}$ $\implies a+b+c = 2+8+2 = \boxed{12}$ .