> I have wondered if there is an exponent

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

> k such that

>

> (1/2)^k + (1/3)^k + (1/5)^k + (1/7)^k + ... (1/p)^k + ...

>

> has a limit of one.

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]> 1.3994333287263303...

1.399433328726330318202807214745644327904727429484383941274765822888062492487247800233390475384227...

(if you need more digits)

Best,

Andrey

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