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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- References:
- Infinite series
- From: "Dr. C. S. Jog" <jogc@mecheng.iisc.ernet.in>
- Infinite series