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

• To: mathgroup at smc.vnet.net
• Subject: [mg53968] Re: [mg53937] Product {for p=2 to infinity} (p^2+1)/(p^2-1)
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Sat, 5 Feb 2005 03:16:29 -0500 (EST)
• References: <200502040912.EAA01094@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Zak Seidov wrote:

>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?
>seidovzf at yahoo.com
>
>Many thanks,
>Zak
>
>

The result you get is consistent with a numeric approximation e.g. by
taking the product to 10^3.

Alternatively you can try to verify that the sum of logs is correct by
doing an indefinite sum. First we do the sum to infinity and recover the

In:= Sum[Log[(p^2+1)/(p^2-1)], {p,2,Infinity}]
Out= Log - LogGamma[2 - I] - LogGamma[2 + I]

Fine so far. Now we do the indefinite summation, that is, sum to
arbitrary 'n'.
In:= f[n_] = Sum[Log[p^2+1]-Log[p^2-1], {p,2,n}]
Out= Log - Log[Gamma[n]] + Log[Gamma[(1 - I) + n]] +
Log[Gamma[(1 + I) + n]] - Log[Gamma[2 + n]] - LogGamma[2 - I] -
LogGamma[2 + I]

We check that this agrees in the limit as n goes to infinity.

In:= InputForm[ff = Limit[f[n], n->Infinity]]
Out//InputForm=
((3*I)*Pi + Log - 3*Log[-Gamma[2 - I]] - 3*Log[Gamma[2 + I]])/3

In:= InputForm[FullSimplify[ff]]
Out//InputForm= Log[Sinh[Pi]/Pi]

Finally one can ask whether the indefinite sum is correct. To show this

In:= InputForm[g[n_] = FullSimplify[f[n]-f[n-1],
Assumptions->{n>2,Element[n,Integers]}]]
Out//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.

In:= FullSimplify[g - (Log[5^2+1] - Log[5^2-1])]
Out= 0

Daniel Lichtblau
Wolfram Research

```

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