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Can this be the solution for composites?

Propositions:

It is well known by the standard Riemann Zeta function that

\[ \zeta (s) = \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^s} \]

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

Also, the Prime Zeta function gives us

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

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


Conclusion:

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

\[ \begin{align} P'(s) &= \displaystyle \sum_{p'} \dfrac{1}{{p'}^s} \\ &= \sum_{n=1}^{\infty} \dfrac{1}{n^s} - \sum_{p} \dfrac{1}{p^s} - 1 \\ &= \zeta (s) - P(s) - 1 \end{align} \]

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

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
7 months, 4 weeks ago

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

Aditya Narayan Sharma - 7 months, 1 week ago

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