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Re: What Mathematica "says" on two BBP type infinite sums

  • To: mathgroup at smc.vnet.net
  • Subject: [mg103782] Re: [mg103765] What Mathematica "says" on two BBP type infinite sums
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Mon, 5 Oct 2009 13:18:26 -0400 (EDT)
  • References: <200910051139.HAA28737@smc.vnet.net>

Alexander Povolotsky wrote:
> Hello,
> 
> What Mathematica (I don't have it) "says" on two BBP type infinite sums ?
> 
> 1) sum(((1/(exp(Pi)-log(3))/log(2))^n)/(n^3+2*n^2+2*n+7),n = 0 ... infinity)
> ISC *tells*  that
> 
> 95*sum(((1/(exp(Pi)-log(3))/log(2))^n)/(n^3+2*n^2+2*n+7),n = 0 ...
> infinity) + 8*(Pi)^2 - 146*Catalan + 20* Pi*log(2) -6*(log(2))^2 ~~ 0
> 
> Perhaps
> 
> 95*sum(1/((204461223889343/13382366326351)^n)/(n^3+2*n^2+2*n+7),n = 0
> .. infinity)
> is even more close to
>  8*(Pi)^2 - 146*Catalan + 20* Pi*log(2) -6*(log(2))^2
> ?
> Any further improvements to (...)^n factor to get closer to
> 8*(Pi)^2 - 146*Catalan + 20* Pi*log(2) -6*(log(2))^2
> ?
> 
> 2) 2*sum(1/(n^3+2*n^2+2*n+7)/(24)^n,n = 0 .. infinity)
> ISC *tells*  that this sum is close to
> Pi*sqrt(3) - 39*log(3) + 84*log(2) + 25*gamma -8*Pi*sqrt(2)
> It appears that this sum's proximity to
> Pi*sqrt(3) - 39*log(3) + 84*log(2) + 25*gamma -8*Pi*sqrt(2)
> could be improved if instead of 24
> 23.999999995011916301243901392414554490409136352246963766236377143509476137987495024510254888936149797331254797
> be used
> Any further improvements (desirably in the close form ;-) ) ?
> 
> Thanks,
> Cheers,
> Alexander R. Povolotsky

This will give a closed form for the first sum.

In[22]:= InputForm[ss = FullSimplify[
   Sum[(1/((Exp[Pi]-Log[3])*Log[2]))^n/(n^3+2*n^2+2*n+7),
     {n,0,Infinity}]]]

Out[22]//InputForm=
((9/32 + 169/(32*Sqrt[353]))*
   (2*Hypergeometric2F1[1, Root[-7 + 2*#1 - 2*#1^2 + #1^3 & , 1, 0],
      Root[-12 + 9*#1 - 5*#1^2 + #1^3 & , 1, 0], (E^Pi*Log[2] - 
Log[2]*Log[3])^
       (-1)]*(3177 - 169*Sqrt[353] + (-338 + 18*Sqrt[353])^(1/3)*
       (4*2^(1/3)*(-19 + Sqrt[353])
   Root[8388608 - 15497488*#1^3 + #1^6 & , 1, 0])) +
   Hypergeometric2F1[1, Root[-7 + 2*#1 - 2*#1^2 + #1^3 & , 2,
       0], Root[-12 + 9*#1 - 5*#1^2 + #1^3 & , 2, 0],
      (E^Pi*Log[2] - Log[2]*Log[3])^(-1)]*(6354 - 338*Sqrt[353] -
      64*(-1)^(2/3)*Root[-1 + 142528*#1^3 + #1^6 & , 2, 0] +
      Root[-68719476736 + 83833881088*#1^3 + #1^6 & , 3, 0]) +
    Hypergeometric2F1[1, Root[-7 + 2*#1 - 2*#1^2 + #1^3 & , 3, 0],
      Root[-12 + 9*#1 - 5*#1^2 + #1^3 & , 3, 0],
   (E^Pi*Log[2] - Log[2]*Log[3])^(-1)]*(6354 -
338*Sqrt[353] + 64*Root[-1 - 142528*#1^3 + #1^6 & , 6,
        0] + Root[-68719476736 + 83833881088*#1^3 + #1^6 & , 4, 0])))/42

We'll check that inverse symbolic calculator result to 30 digits.

In[23]:= tt = 8*Pi^2 - 146*Catalan + 20*Pi*Log[2] -6*Log[2]^2;
In[24]:= N[95*ss + tt, 30]
                                             -13        -42
Out[24]= -9.56534167773797380032392256045 10    + 0. 10    I

So they are certainly not equal, but they agree to 12 or so places.

Your second variant is indeed closer, agreeing to around 29 digits.

You can get the result yourself, if you go to

http://www.wolframalpha.com

and type the input below:

Sum[1/((204461223889343/13382366326351)^n)/(n^3+2*n^2+2*n+7),{n,0,Infinity}]

This appears to be a substantial computation, and it might be hit or 
miss as to whether it times out.

Daniel Lichtblau
Wolfram Research



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