Harmony Squared

$\begin{aligned} S_A & = \left(\frac 12\right)^2 + \left(\frac 13 + \frac 16 \right)^2 + \left(\frac 14 + \frac 18 + \frac 1{12} \right)^2 + \cdots \\ & = \sum_{k=1}^\infty H_k^2(k+1)^{-2} \\ & = \color{#3D99F6} s_h (2,2) & \small \color{#3D99F6} \text{where }s_h (m,n) \text{ is Euler sum.} \\ & = \color{#3D99F6} \frac {11}{360} \pi^4 & \small \color{#3D99F6} \text{See note 1.} \end{aligned}$

$\begin{aligned} S_B & = \left(\frac 11\right)^2 + \left(\frac 12 + \frac 14 \right)^2 + \left(\frac 13 + \frac 16 + \frac 19 \right)^2 + \cdots \\ & = \sum_{k=1}^\infty H_k^2k^{-2} & \small \color{#3D99F6} \text{See note 2.} \\ & = s_h (2,2) + {\color{#3D99F6}2s_h(1,3)} + \zeta (4) \\ & = \frac {11}{360}\pi^4 + {\color{#3D99F6}2\times \frac {\pi^4}{360}} + \frac {\pi^4}{90} & \small \color{#3D99F6} \text{See note.} \\ & = \frac {17}{360} \pi^4 \end{aligned}$

Therefore, $\dfrac {S_A}{S_B} = \dfrac {11}{17}$ , then $\dfrac {A+B}2 = \dfrac {11+17}2 = \boxed{14}$ .

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See reference: Euler sum

Note 1: $\ \displaystyle \small s_h (2,2) = \frac 32 \zeta(4) + \frac 12 [\zeta(2)]^2 = \frac {11}{360}\pi^4 \quad ...(28)$

Note 2: $\ \displaystyle \small \sum_{k=1}^\infty H_k^2k^{-n} = s_h (2,n) + 2s_h(1,n+1) + \zeta (n+2) \quad ...(7)$

Note 3: $\ \displaystyle \small s_h (1,n) = \frac 12 n \zeta(n+1) - \frac 12 \sum_{k=1}^{n-2}\zeta(n-k)\zeta(k+1) \quad ... (21) \\ \ \displaystyle \small \implies s_h(1,3) = \frac 32 \zeta(4) - \frac 12 \zeta(2)\zeta(2) = \frac {\pi^4}{360}$