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October 9th, 2008, 07:45 AM   #1
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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
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October 9th, 2008, 08:11 AM   #2
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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 + ...");
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October 9th, 2008, 08:22 AM   #3
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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 ?
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October 9th, 2008, 08:59 AM   #4
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Re: Sequence sum prime inverse

Not really. The significance is that primezeta(1.39943332872633031820280721474564432790 472742948438394127476582288806249...) = 1.
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