From: Michael Borcherds <michael@ge...>  20120730 15:05:00

Thanks, using specfn is an improvement as sum(1/x^2,x,1,infinity); then works! However these still don't :) sum(1/(x+1)^2,x,1,infinity); sum(1/x^2,x,2,infinity); thanks, Mike On 30 July 2012 15:40, Arthur Norman <acn1@...> wrote: > On Mon, 30 Jul 2012, Michael Borcherds wrote: > >> Hi, >> >> Is there any way to get Reduce to calculate this? >> Sum(1/x^2, x, 1, infinity); >> >> thanks! >> >> Mike >> > > You may not like this answer, but consider > > load_package specfn; > zeta(2); > 2 > pi >  > 6 > > > > Arthur > 