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

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

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?
>Please email me your help:
>seidovzf at
>Many thanks,

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 
log of your result.

In[5]:= Sum[Log[(p^2+1)/(p^2-1)], {p,2,Infinity}]
Out[5]= Log[2] - LogGamma[2 - I] - LogGamma[2 + I]

Fine so far. Now we do the indefinite summation, that is, sum to 
arbitrary 'n'.
In[8]:= f[n_] = Sum[Log[p^2+1]-Log[p^2-1], {p,2,n}]
Out[8]= Log[2] - 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[9]:= InputForm[ff = Limit[f[n], n->Infinity]]
((3*I)*Pi + Log[8] - 3*Log[-Gamma[2 - I]] - 3*Log[Gamma[2 + I]])/3

In[10]:= InputForm[FullSimplify[ff]]
Out[10]//InputForm= Log[Sinh[Pi]/Pi]

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], 
-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 

In[34]:= FullSimplify[g[5] - (Log[5^2+1] - Log[5^2-1])]
Out[34]= 0

Daniel Lichtblau
Wolfram Research

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