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October 9th, 2008, 07:45 AM  #1 
Senior Member Joined: Nov 2007 Posts: 633 Thanks: 0  Sequence sum prime inverse
Let x real number, p prime number Let S = 1/2^x + 1/3^x +1/5^x + 1/7^x + 1/11^x + .....+1/p^x Find x such as S<1 
October 9th, 2008, 08:11 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Sequence sum prime inverse
For p = 2, any x > 0 works. For p = 3, x > 0.7879 works. x > 1.3994333288 works for all p. Pari/GP code: Code: primezeta(s)={ suminf(k=1, moebius(2*k1)/(2*k1)*log(zeta((2*k1)*s))+moebius(2*k)/(k*2)*log(zeta(2*k*s)) ) }; addhelp(primezeta, "primezeta(z): Sum of 1/2^s + 1/3^s + 1/5^s + 1/7^s + ..."); 
October 9th, 2008, 08:22 AM  #3 
Senior Member Joined: Nov 2007 Posts: 633 Thanks: 0  Re: Sequence sum prime inverse
Hi, Thank you for your answer. x= 1.3994333288 is slightly < to 2^1/2 which is equal 1.41421356... Is there any link between x and 2^1/2 ? 
October 9th, 2008, 08:59 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Sequence sum prime inverse
Not really. The significance is that primezeta(1.39943332872633031820280721474564432790 472742948438394127476582288806249...) = 1.


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