I have a formula that specializes to (c90) sum((-2)^?\2\-powers\-in(k)/k^2,k,1,inf) = %pi^2/12 inf ==== 2-powers-in(k) 2 \ (- 2) %pi (d90) > ------------------- = ---- / 2 12 ==== k k = 1 where 2_powers_in(2008) = 3, e.g.. Is this old news? It seems that (c96) sum(2^?\2\-powers\-in(k)/k^2,k,1,inf) = %pi^2/4 inf ==== 2-powers-in(k) 2 \ 2 %pi (d96) > --------------- = ----, / 2 4 ==== k k = 1 suggesting loads of other things to try. --rwg