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MathGroup Archive 2005

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53976] Re: Product {for p=2 to infinity} (p^2+1)/(p^2-1)
  • From: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>
  • Date: Sat, 5 Feb 2005 03:17:36 -0500 (EST)
  • References: <ctvh4j$1jn$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Mathematica 5.1 does this:

In[1]:=FullSimplify[Product[(p^2 + 1)/(p^2 - 1), {p, 2, Infinity}]]
Out[1]=Sinh[Pi]/Pi

Steve Luttrell

"Zak Seidov" <seidovzf at yahoo.com> wrote in message 
news:ctvh4j$1jn$1 at smc.vnet.net...
> Dear Math gurus,
> I try to copy Math session:
> \!\(FullSimplify[\[Product]\+\(p = 2\)\%\[Infinity]\((\(p\^2 +
> 1\)\/\(p\^2 - \1\))\)]
> n  Sinh[\[Pi]]\/\[Pi]\),
> that is,
> Product {for p=2 to infinity} (p^2+1)/(p^2-1)=>
> sinh(pi)/pi=3.67608 -
> is it OK,
> or this should be 3/2?
> Please email me your help:
> seidovzf at yahoo.com
>
> Many thanks,
> Zak
> 



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