Charlie, I agree.

When the sum of the expression I supplied earlier is correctly evaluated from 0 to infinity, the result is indeed 2/3:

(sum_(n=0 to infinity) (1/(1+6n)^2+1/(4+6n)^2-1/(6n+2)^2-1/(6n+5)^2))/ (sum_(n=0 to infinity) (1/(6n+1)^2+1/(6n+2)^2-1/(6n+4)^2-1/(6n+5)^2)) =2/3,

since (4 (psi(1, 1/3)-psi(1, 2/3)))/(psi(1, 1/6)+psi(1, 1/3)-psi(1, 2/3)-psi(1, 5/6)) is 0.6 recurring. (psi is the polygamma function)

Edited on October 13, 2012, 10:35 am

Posted by broll on 2012-10-13 09:11:52