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

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

For $p' \in \{ \text{composite numbers} \}$ 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.

#Calculus

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Comments

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 - 4 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$