Re: Infinite series
- To: mathgroup at smc.vnet.net
- Subject: [mg105713] Re: Infinite series
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Wed, 16 Dec 2009 06:19:10 -0500 (EST)
On 12/15/09 at 7:27 AM, jogc at mecheng.iisc.ernet.in (Dr. C. S. Jog)
wrote:
>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}]
>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.
There are a variety of reasons Simplify often does not achieve
what you are looking for. But the obvious thing to try when you
don't get what you want with Simplify is FullSimplify which
works in this case. That is:
In[1]:= Sum[(-1)^m/((2*m - 3)^2*(2*m - 1)*(2*m + 1)^2), {m, Infinity}]
Out[1]= 1/512 (-HurwitzZeta[2, -(1/4)] - Zeta[2, 5/4] - 16 Pi +
2 Pi^2)
In[2]:= % // FullSimplify
Out[2]= -Pi/32