Re: Infinite series
- To: mathgroup at smc.vnet.net
- Subject: [mg105716] Re: [mg105683] Infinite series
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Wed, 16 Dec 2009 06:19:44 -0500 (EST)
- Reply-to: hanlonr at cox.net
Use FullSimplify Sum[(-1)^m/((2*m - 3)^2*(2*m - 1)*(2*m + 1)^2), {m, Infinity}] // FullSimplify -(Pi/32) Bob Hanlon ---- "Dr. C. S. Jog" <jogc at mecheng.iisc.ernet.in> wrote: ============= Hi: We have the following identity: \sum_{m=1}^{infinity} (-1)^m/((2m-3)^2*(2m-1)*(2m+1)^2)=-Pi/32. When we type the command, In[1]:=Sum[(-1)^m/((2*m-3)^2*(2*m-1)*(2*m+1)^2),{m,Infinity}] we get 2 1 1 -16 Pi + 2 Pi - HurwitzZeta[2, -(-)] - Zeta[2, -] 4 4 Out[1]= -------------------------------------------------- 512 The command Simplify[%] does not simplify it further. I am sure the above expression must be equal to -Pi/32, but a user would prefer this answer than the above one. Thanks and regards C. S. Jog