Sums Within Sums!

$\begin{aligned} S_k & = \sum_{k=0}^\infty \frac 1{(4k+1)(4k+3)} & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \frac 12 \sum_{k=0}^\infty \left(\frac 1{4k+1} - \frac 1{4k+3} \right) \\ & = \frac 12 \color{#3D99F6} \left(\frac 11 - \frac 13 + \frac 15 - \frac 17 + \frac 19 - \frac 1{11} + \cdots \right) & \small \color{#3D99F6} \text{By Maclaurin series:} \\ & = \frac 12 \color{#3D99F6} \sum_{k=0}^\infty \frac {(-1)^k}{2k+1} & \small \color{#3D99F6} \tan^{-1} x = \sum_{k=0}^\infty \frac {(-1)^kx^k}{2k+1} \\ & = \frac {\tan^{-1} 1}2 = \frac \pi 8 \end{aligned}$

$\begin{aligned} S_n & = \sum_{n=1}^\infty \frac {(-1)^{n+1}}{n^2} \\ & = \frac 1{1^2} {\color{#3D99F6}- \frac 1{2^2}} + \frac 1{3^2}{\color{#3D99F6}- \frac 1{4^2}} + \frac 1{5^2}{\color{#3D99F6}- \frac 1{6^2}} + \cdots \\ & = \frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \color{#3D99F6} - 2 \left(\frac 1{2^2} + \frac 1{4^2} + \frac 1{6^2} + \cdots\right) \\ & = \frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \color{#3D99F6} - \frac 2{2^2} \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \cdots \right) & \small \color{#3D99F6} \text{ Riemann zeta function }\zeta(k) = \sum_{n=1}^\infty \frac 1{n^k} \\ & = \zeta(2) - \frac 12 \zeta(2) = \frac 12 \zeta(2) = \frac {\pi^2}{12} & \small \color{#3D99F6} \text{Note that }\zeta(2) = \frac {\pi^2}6 \end{aligned}$

$\begin{aligned} S_w & = \sum_{w=0}^\infty \frac {(-1)^w}{2w+1} = \tan^{-1} 1 = \frac \pi 4\end{aligned}$

Then

$\begin{aligned} \sum_{i=0}^\infty \frac {S_k}{S_ni^2} - S_w & = \sum_{i=0}^\infty \frac {\frac {\pi}8}{\frac {\pi^2}{12} \cdot i^2} - \frac \pi 4 = \frac 3{2\pi} \sum_{i=0} \frac 1{i^2} - \frac \pi 4 = \frac 3{2\pi} \cdot \frac {\pi^2}6 - \frac \pi 4 = \frac \pi 4 - \frac \pi 4 = \boxed 0 \end{aligned}$