Re: Re: Product {for p=2 to infinity} (p^2+1)/(p^2-1)

*To*: mathgroup at smc.vnet.net*Subject*: [mg54000] Re: [mg53968] Re: [mg53937] Product {for p=2 to infinity} (p^2+1)/(p^2-1)*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 6 Feb 2005 00:45:27 -0500 (EST)*References*: <200502040912.EAA01094@smc.vnet.net> <200502050816.DAA22265@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 5 Feb 2005, at 08:16, Daniel Lichtblau wrote: > Finally one can ask whether the indefinite sum is correct. To show this > you would start with: > > In[31]:= InputForm[g[n_] = FullSimplify[f[n]-f[n-1], > Assumptions->{n>2,Element[n,Integers]}]] > Out[31]//InputForm= > -Log[Gamma[-I + n]] - Log[Gamma[I + n]] + > Log[Gamma[(1 - I) + n]/(-1 + n^2)] + Log[Gamma[(1 + I) + n]] > > You can now try various methods to confirm that this is your summand. > I'll leave the symbolic proof to others and just show a quick numerical > verification. > This looks like an interesting question. Let's reformulate it: prove that the function h[x_] := Log[Gamma[(1 - I) + x]/(x^2 - 1)] - Log[Gamma[-I + x]] - Log[Gamma[I + x]] + Log[Gamma[(1 + I) + x]] - Log[(x^2 + 1)/(x^2 - 1)] is 0 for all real x>1. (Actually for all complex x with Re[x]>1). FullSimplify[h[x]] does not return 0, however: FullSimplify[D[h[x],x]] 0 Now, h[x] is an analytic function in the open region Re[x]>1 (I think!) and FullSimplify[h[2]] 0 So this seems to be enough. Andrzej Kozlowski

**References**:**Product {for p=2 to infinity} (p^2+1)/(p^2-1)***From:*seidovzf@yahoo.com (Zak Seidov)

**Re: Product {for p=2 to infinity} (p^2+1)/(p^2-1)***From:*Daniel Lichtblau <danl@wolfram.com>