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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.

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    I think this is the same Dirichlet series: http://math.stackexchange.com/q/607060/8530 But I can't tell for sure from your formula. – 2017-01-10
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    Come on... You need a reference for that ? It is trivial. – 2017-01-11

1 Answers 1

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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 $\Re e\,s=1$...

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    Nice and simple. Thanks Barto. I guess it doesn't say anything about convergence in the domain $\Re(s) < 1$, right? ;-) – 2017-01-10
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    I was just expanding om that :) – 2017-01-10
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    My guess about the positive density of square-free numbers of the form $4k+3$ is true; here's a related result: http://mathoverflow.net/questions/59741 Though probably the density of square-free numbers $\equiv a\pmod b$ with $\gcd(a,b)$ not a square is always the same for $b$ fixed. – 2017-01-10
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    Thanks for the expanded answer, Barto. I also found that for $s=2k, k=1,2,3...$, closed forms do exist for both $E$ and $F$ individually. One last question: could it be proven that both $E$ and $F$ converge to $0$ at $s=1$, or is even that out of reach? – 2017-01-10
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    $F$ is just $1/\zeta$ multiplied with $1/(1-2^{-s})$, which is analytic for $\sigma>0$, so yes, closed forms for $\zeta(2k)$ imply closed forms for $E(2k)$ and $F(2k)$ and $E$ and $F$ both tend to zero with order $\asymp (s-1)$ at $s=1$. – 2017-01-10
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    Got it. Many thanks for you help. – 2017-01-10