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 > > > > -- > This message has been scanned for viruses and > dangerous content by MailScanner, and is > believed to be clean. > >
- References:
- Infinite series
- From: "Dr. C. S. Jog" <jogc@mecheng.iisc.ernet.in>
- Infinite series