Alternating Dirichlet series involving the Möbius function.

Question

It is well known that:

$$\sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)} \qquad \Re(s) > 1$$

with $\mu(n)$ the Möbius function and $\zeta(s)$ the Riemann Zeta function.

Numerical evidence strongly suggests that the alternating Dirichlet series:

EDIT: as pointed out in the answer and the comment, I got the domain of absolute convergence wrong. To avoid any confusion, below are the corrected values.

$$\sum_{n=1}^\infty (-1)^{n+1}\frac{\mu(n)}{n^s} = \frac{2^s+1}{(2^s-1)\,\zeta(s)} \qquad \Re(s) > \require{cancel}\cancel{0} \color{ForestGreen}1$$

i.e. the series on the LHS should now also induce a pole at the non-trivial zeros in the critical strip. I did find that this series has been mentioned in this MSE-question, which didn't get much traction and there is no proof provided.

Q: Is the relation above true?

Assuming it is true, after a small manipulation we obtain:

$$ \coth\left(\frac{s}{2}\,\log(2)\right) = \zeta(s)\,\sum_{n=1}^\infty (-1)^{n+1}\frac{\mu(n)}{n^s} \qquad \Re(s) > \require{cancel}\cancel{0} \color{ForestGreen}1$$

so that also:

$$s = \zeta\left(\frac{\log(s+1)-\log(s-1)}{\log(2)}\right) \,\sum_{n=1}^\infty (-1)^{n+1}\frac{\mu(n)}{n^{\frac{\log(s+1)-\log(s-1)}{\log(2)}}} \qquad \Re(s) > \require{cancel}\cancel{0} \color{ForestGreen}1$$

which yields, with $\zeta(s)$ in terms of the Dirichlet $\eta$-function, the following relation:

$$\frac{s\,(3-s)}{s+1} = \left(\sum_{k=1}^\infty (-1)^{k+1}\frac{1}{k^{\frac{\log(s+1)-\log(s-1)}{\log(2)}}} \right) \cdot\left(\sum_{n=1}^\infty (-1)^{n+1}\frac{\mu(n)}{n^{\frac{\log(s+1)-\log(s-1)}{\log(2)}}}\right) \qquad \Re(s) >\require{cancel}\cancel{0} \color{ForestGreen}1$$

which for $s=3$ becomes zero. The first series becomes $\log(2)$, hence the second series must be equal to zero at this value.

Answer 1

It is indeed true. The function $(-1)^{n+1}$ is a multiplicative function (really! confirm this), and therefore the Dirichlet series can be expanded into an Euler product as usual (in its half-plane of absolute convergence, which is $\sigma>1$ in this case): \begin{align*} \sum_{n=1}^\infty (-1)^{n+1} \frac{\mu(n)}{n^s} &= \prod_p \sum_{k=0}^\infty (-1)^{p^k+1} \frac{\mu(p^k)}{(p^k)^s} \\ &= \prod_p \biggl( 1 + (-1)^{p+1} \frac{-1}{p^s} + \sum_{k=2}^\infty 0 \biggr) \\ &= \biggl( 1+\frac1{2^s}\biggr) \prod_{p\ge3} \biggl( 1 - \frac{1}{p^s} \biggr) \\ &= \biggl( 1+\frac1{2^s}\biggr) \biggl( 1-\frac1{2^s}\biggr)^{-1} \prod_p \biggl( 1 - \frac{1}{p^s} \biggr) = \frac{2^s+1}{2^s-1} \frac1{\zeta(s)} \end{align*} as claimed.