## Re: [PrimeNumbers] Re: zeta limit

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• ... 1/2^k + 1/3^k + ... + 1/p^k + ... just equals to Sum[MoebiusMu[n]*Log[Zeta[n*k]]/n, {n, 1, +Infinity}] which converges rapidly; so it can be used to
Message 1 of 5 , Apr 6, 2007
> I have wondered if there is an exponent
> k such that
>
> (1/2)^k + (1/3)^k + (1/5)^k + (1/7)^k + ... (1/p)^k + ...
>
> has a limit of one.

1/2^k + 1/3^k + ... + 1/p^k + ...

just equals to

Sum[MoebiusMu[n]*Log[Zeta[n*k]]/n, {n, 1, +Infinity}]

which converges rapidly; so it can be used to estimate the waned value of k as
1.3994333287263303...

Best,

Andrey

[Non-text portions of this message have been removed]
• ... 1.399433328726330318202807214745644327904727429484383941274765822888062492487247800233390475384227... (if you need more digits) Best, Andrey [Non-text
Message 2 of 5 , Apr 6, 2007
> 1.3994333287263303...

1.399433328726330318202807214745644327904727429484383941274765822888062492487247800233390475384227...

(if you need more digits)

Best,

Andrey

[Non-text portions of this message have been removed]
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