It seems that ∑[n>0 and n odd] n⁻² = π²/8.
∑[n>0 and n even] n⁻² = ∑[n>0] (2n)⁻² = ¼ ∑[n>0] n⁻².
But ∑[n>0] n⁻² = ∑[n>0 and n odd] n⁻² + ∑[n>0 and n even] n⁻² = π²/8 + ¼ ∑[n>0] n⁻².
Thus ¾ ∑[n>0] n⁻² = π²/8
and ∑[n>0] n⁻² = π²/6.