Let $\mu(n)$ the Möbius function, see its definition for example from this MathWorld, and we denote with $s$ the complex variable.

I'm curious to know if some case of the series $$\sum_{n=1}^\infty\frac{(-1)^n\mu(n)}{n^s}$$ for $\Re s\geq \frac{1}{2}$, were in the literature.

I did simple experiments with Wolfram Alpha and from those my belief is that one can calculate a closed-form for the case $s=2+0\cdot i=2$, and write an identity in terms of the constant $\frac{1}{\zeta(3)}$ for the case $s=3$.

Question. Was in the literature the formal series (or complex function defined on a domain of the complex plane) $$\sum_{n=1}^\infty\frac{(-1)^n\mu(n)}{n^s}\,?\tag{1}$$ Then refer the literature and I try to find and read those known facts about the complex function $(1)$. Many thanks.

  • $\begingroup$ I doubt it's in the literature, it's simple, but useless. $\endgroup$ – Professor Vector Dec 27 '17 at 10:34
  • $\begingroup$ Do you know how calculate $\sum_{n=1}^{\infty}\frac{(-1)^n\mu(n)}{n^2}$? Or properties of our function $(1)$ as the convergence abscissa @ProfessorVector ? $\endgroup$ – user243301 Dec 27 '17 at 10:35
  • $\begingroup$ It's trivial to derive from the known result without the $(-1)^n$. Give us a few own thoughts for a change, please! $\endgroup$ – Professor Vector Dec 27 '17 at 10:39
  • $\begingroup$ Many thanks Professor Vector. $\endgroup$ – user243301 Dec 27 '17 at 11:38
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    $\begingroup$ $(-1)^{n+1}$ is multiplicative not $(-1)^n$. Dirichlet series with multiplicative coefficients $\implies$ Euler product, this is the 1st page of every book on analytic number theory. Can you write this Euler product ? $\endgroup$ – reuns Dec 27 '17 at 23:27

Assume for a moment that $\operatorname{Re}(s) > 1$. Then using the fact that $\mu$ is multiplicative, we have

$$ \sum_{n\text{ even}} \frac{\mu(n)}{n^s} = \sum_{k=1}^{\infty} \frac{\mu(2k)}{(2k)^s} = - \frac{1}{2^s} \sum_{k\text{ odd}} \frac{\mu(k)}{k^s}. $$

So if we write $D(s) = \sum_{k\text{ odd}} \frac{\mu(k)}{k^s}$, then

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

and hence

\begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^n \mu(n)}{n^s} &= \sum_{n\text{ even}} \frac{\mu(n)}{n^s} - \sum_{n\text{ odd}} \frac{\mu(n)}{n^s} \\ &= -(1+2^{-s})D(s) = - \frac{2^s+1}{2^s-1} \cdot \frac{1}{\zeta(s)}. \end{align*}

  • $\begingroup$ Many thanks I would like to study more your answer, but it seems that agree with my examples. $\endgroup$ – user243301 Dec 27 '17 at 10:56

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