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