Re: Infinite series
- To: mathgroup at smc.vnet.net
- Subject: [mg105698] Re: [mg105683] Infinite series
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Wed, 16 Dec 2009 06:16:20 -0500 (EST)
- References: <200912151227.HAA15202@smc.vnet.net>
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
FullSimplify and Limit agree with you.
In[7]:= ss = Sum[(-1)^m/((2*m-3)^2*(2*m-1)*(2*m+1)^2),{m,Infinity}]
Out[7]//InputForm= (-16*Pi + 2*Pi^2 - HurwitzZeta[2, -1/4] - Zeta[2,
5/4])/512
In[8]:= FullSimplify[ss]
Out[8]//InputForm= -Pi/32
In[9]:= ssn = Sum[(-1)^m/((2*m-3)^2*(2*m-1)*(2*m+1)^2),{m,n}]
Out[9]//InputForm=
-(-4*(-1)^n - 4*(-1)^n*n + 16*(-1)^n*n^2 - 16*(-1)^n*n^3 + Pi -
8*n^2*Pi + 16*n^4*Pi + 2*(-1)^n*LerchPhi[-1, 1, 3/2 + n] -
16*(-1)^n*n^2*LerchPhi[-1, 1, 3/2 + n] +
32*(-1)^n*n^4*LerchPhi[-1, 1, 3/2 + n])/(32*(-1 + 4*n^2)^2)
In[10]:= Limit[ss2, n->Infinity]
Out[10]//InputForm= ss2
Daniel Lichtblau
Wolfram Research
- References:
- Infinite series
- From: "Dr. C. S. Jog" <jogc@mecheng.iisc.ernet.in>
- Infinite series