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