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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*k-1)/(2*k-1)*log(zeta((2*k-1)*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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