Can this be the solution for composites?

Propositions:

It is well known by the standard Riemann Zeta function that

ζ(s)=n=11ns \zeta (s) = \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^s}

For R(s)>1 \mathfrak{R} (s) > 1 .

Also, the Prime Zeta function gives us

P(s)=p1ps P(s) = \displaystyle \sum_{p} \dfrac{1}{p^s}

For p{primes}p \in \{ \text{primes} \} and R(s)>1 \mathfrak{R} (s) > 1 .


Conclusion:

This means that the sum of sths^{\text{th}} powers of the reciprocals of all composite numbers can be given by the formula

P(s)=p1ps=n=11nsp1ps1=ζ(s)P(s)1 \begin{aligned} P'(s) &= \displaystyle \sum_{p'} \dfrac{1}{{p'}^s} \\ &= \sum_{n=1}^{\infty} \dfrac{1}{n^s} - \sum_{p} \dfrac{1}{p^s} - 1 \\ &= \zeta (s) - P(s) - 1 \end{aligned}

For p{composite numbers}p' \in \{ \text{composite numbers} \} and R(s)>1 \mathfrak{R} (s) > 1 .

Knowing that infinite sums are risky to handle. I would like your opinions on whether the conclusion drawn from the two given true propositions made above are correct or not.

Note by Tapas Mazumdar
2 years, 6 months ago

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1 vote

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if un,vnu_n,v_n are convergent sequences then so is their difference. Provided ζ(s),P(s)\zeta(s),P(s) converges which is obvious ,your conclusions are true enough

Aditya Narayan Sharma - 2 years, 6 months ago

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