What is the Mobius sum $\sum_{n=1}^\infty \frac{(-1)^{n+1}|\mu(n)|}{n^s}$?

Question

It can be observed that,

$$A(s) := \sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}$$

$$B(s) := \sum_{n=1}^\infty \frac{|\mu(n)|}{n^s} = \frac{\zeta(s)}{\zeta(2s)}$$

$$C(s) := \sum_{n=1}^\infty \frac{(-1)^{n+1}\mu(n)}{n^s} = \frac{2^s+1}{2^s-1}\frac{1}{\zeta(s)}$$

Q: Is it true that,

$$D(s) := \sum_{n=1}^\infty \frac{(-1)^{n+1}|\mu(n)|}{n^s} = \frac{2^s-1}{2^s+1}\frac{\zeta(s)}{\zeta(2s)}$$ hence, $$A(s)\,B(s) = C(s)\,D(s) = \frac{1}{\zeta(2s)}?$$

Answer 1

Yes, it is true. Let $\mathbb{P}$ be the set of all positive primes. For $T\subseteq\mathbb{P}$, write $\prod(T)$ for the product of elements in $T$. Let $s\in\mathbb{C}$ with $\text{Re}(s)>1$. We have $$D(s)=\sum_{T\subseteq \mathbb{P}}\,\frac{(-1)^{\prod(T)+1}}{\big(\prod(T)\big)^s}=\sum_{T\subseteq \mathbb{P}\setminus\{2\}}\left(\frac{1}{\big(\prod(T)\big)^s}-\frac{1}{\big(2\,\prod(T)\big)^s}\right)\,.$$ Hence, $$D(s)=\left(1-\frac{1}{2^s}\right)\,\sum_{T\subseteq \mathbb{P}\setminus\{2\}}\,\frac{1}{\big(\prod(T)\big)^s}=\frac{2^s-1}{2^s+1}\,\sum_{T\subseteq \mathbb{P}\setminus\{2\}}\left(\frac{1}{\big(\prod(T)\big)^s}+\frac{1}{\big(2\,\prod(T)\big)^s}\right)\,.$$ Therefore, $$D(s)=\frac{2^s-1}{2^s+1}\,\sum_{T\subseteq \mathbb{P}}\,\frac{1}{\big(\prod(T)\big)^s}=\frac{2^s-1}{2^s+1}\,B(s)\,.$$