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 >