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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


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