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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



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