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Re: Infinite series

  • To: mathgroup at smc.vnet.net
  • Subject: [mg105721] Re: [mg105683] Infinite series
  • From: Patrick Scheibe <pscheibe at trm.uni-leipzig.de>
  • Date: Wed, 16 Dec 2009 06:20:40 -0500 (EST)
  • References: <200912151227.HAA15202@smc.vnet.net>

Hi,

with FunctionExand you get your result, but it assumes the you have
Reals

FunctionExpand[
 Sum[(-1)^m/((2*m - 3)^2*(2*m - 1)*(2*m + 1)^2), {m, 
    Infinity}] == -Pi/32]

another "not always accurate" test is

PossibleZeroQ[
 Sum[(-1)^m/((2*m - 3)^2*(2*m - 1)*(2*m + 1)^2), {m, Infinity}] + 
  Pi/32]

and Reduce returns a warning but finally decides that the equation is
true
 
Cheers
Patrick

On Tue, 2009-12-15 at 07:27 -0500, Dr. C. S. Jog 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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> 



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