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.