20090112, 01:48  #1 
May 2004
New York City
1000010001011_{2} Posts 
Find the Value
It's well known that:
Product_{[prime p > 1]}{1/(11/p^s)} = Sum_{[integer n > 0]}{1/n^s} = zeta(s) for real s > 1, and also that for s=2 this equals pi^2 / 6. Using this or any other way, find the value of the product for s=2 if the primes > 1 are replaced by the composites > 1. (This isn't hard.) You'll find, for s=2, the product over primes > the product over composites. Is this true for all values of s > 1, or is there some s where the two products are equal? 
20090113, 22:18  #2 
Dec 2008
you know...around...
2^{2}×11×17 Posts 
If my humble 5minuteprogram is right, there should be a crossover point between s=1 and s=2, somewhere near sqrt(2).
When s=1.395, the product over the composites gets bigger than the one with the primes. Haven't got the time right now to check s=1.4 far enough, but I guess the product over the composites will get bigger eventually as well. 
20090114, 17:13  #3 
Dec 2008
you know...around...
2^{2}·11·17 Posts 
I spent another half hour or so during my lunch break today, and if my extrapolations are even roughly correct, then the point where both products are equal should be about s=1.39773 ± 0.0001.
The easiest way to calculate the series of composites for me is Product (n>=2) {1/(11/n^s)}  Zeta(s), so I don't think I can get any further ATM (maybe another decimal digit or two at most). But it's an interesting problem nonetheless. 
20090114, 19:36  #4 
"William"
May 2003
New Haven
2,371 Posts 

20090114, 21:02  #5  
Dec 2008
you know...around...
748_{10} Posts 
Quote:
The answer is 12/Pi² or 1.2158542037080532573265535585... Also, just noticed the typo in my previous post. It's not ...  Zeta(s) but ... / Zeta(s), of course! Last fiddled with by mart_r on 20090114 at 21:05 

20090205, 15:16  #6 
May 2004
New York City
108B_{16} Posts 
Nice work. I'd just like to confirm: did you determine that the crossover
point is the only one? There are no others as s > 1 ? 
20090205, 17:04  #7 
Dec 2008
you know...around...
2^{2}·11·17 Posts 

20090702, 19:46  #8 
May 2004
New York City
10213_{8} Posts 
Suppose the limit is Exactly 1.4.
What else (...) does that imply? 
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