With $\mu(n)$ the Möbius function and $\zeta(s)$ the Riemann zeta function, it is well-known that:
$$\frac 1{\zeta(s)}=\sum_{n=1}^\infty \frac{\mu(n)}{n^s}$$
that is proven to converge for $\Re(s)>1$ and might converge $\frac12 < \Re(s) <1$ (which is the RH).
In a similar fashion as the Dirichlet $\eta$-function, I started to experiment with:
$$\sum_{n=1}^\infty \left(\frac{\mu(2n-1)}{(2n-1)^s}-\frac{\mu(2n)}{(2n)^s}\right)$$
and found that the following also seems to hold true:
$$\frac 1{\zeta(s)}=\frac{2^s-1}{2^s+1}\cdot\sum_{n=1}^\infty \left(\frac{\mu(2n-1)}{(2n-1)^s}-\frac{\mu(2n)}{(2n)^s}\right)$$
where $\frac{2^s-1}{2^s+1}$ could also be written as $\tanh\left(\frac{s\ln(2)}{2}\,\right)$.
I thought this was easy to proof using some known $\eta(s)$ and $\zeta(s)$ connections, however I got stuck and could not find a further reference on the web. So, is this relationship true (for $\Re(s) >1$) ?
Thanks.
It is true. Let $E=\sum_{n\text{ even}}\frac{\mu(n)}{n^s}$ and $F=\sum_{n\text{ odd}}\frac{\mu(n)}{n^s}$. Then $1/\zeta(s)=E+F$ (for $\sigma>1$) and
$$E=\sum_{n\text{ odd}}\frac{\mu(2n)}{(2n)^s}=-2^{-s}F.$$
So $1/\zeta(s)=(1-2^{-s})F$, and $$\frac{2^s-1}{2^s+1}(F-E)=\frac{2^s-1}{2^s+1}(1+2^{-s})F=1/\zeta(s).$$
(All valid for $\sigma>1$.) Note that $F-E$ (combining terms) does not converge absolutely for $\sigma>1/2$ as the terms for even $n$ with $2n-1$ squarefree are of order $1/n$, while they (most likely, I guess it's known) occur with positive density. Maybe there's a more clever combination of $E$ and $F$ that does give an analytic continuation beyond the line $\Re e\,s=1$...