Bill Gosper wrote:
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?
Well, if we put r = 2-powers-in(k) and sum over r first, we get sum {r=0..oo} (-2)^r 4^-r sum {k=1,3,..oo} 1/k^2 which is just the product of two sums, one of which is 1-1/2+1/4-1/8... = 1/2+1/8+... = 2/3 and the other of which is famously pi^2/8. So it's certainly straightforwardly equivalent to old news :-). -- g