# The function $\sum_{n=0}^\infty\frac{(-1)^n\mu(2n+1)}{(2n+1)^s}$: reference request or particular values at integers and abscissa of convergence

We denote for integers $$m\geq 1$$ the Möbius function as $$\mu(m)$$. With the help of a CAS, Wolfram Alpha online calculator, I was calculating certain values of $$\sum_{n=0}^\infty\frac{(-1)^n\mu(2n+1)}{(2n+1)^k}\tag{1}$$ at integer arguments, for example $$k=2,3$$ and $$k=4$$. It is from the Möbius inversion of the Taylor series for the arctangent function and calculating integrals $$\int_0^1\frac{\arctan(z^{n+1})}{z}(\log z)^{k-2}dz$$ for the mentioned integers $$k$$.

Example. For example, it seems that $$\sum_{n=0}^\infty\frac{(-1)^n\mu(2n+1)}{(2n+1)^4}=\frac{1536}{\psi^{(3)}(\frac{1}{4})-\psi^{(3)}(\frac{3}{4})}$$, where $$\psi^{(3)}(z)$$ is the polygamma function of order $$3$$ (it is the notation from the Wikipedia Polygamma function).

Question 1. I am curious to know if these particular values ​​are in the literature and how to calculate them. Please if it is in the literature refers it and I try to search and read it from the literature, in other case can you express in terms of particular values of well-known functions and constants for example the case $$k=5$$ or $$k=6$$ of $$(1)$$? Many thanks.

Let $$s=x+iy$$ the complex variable, and we consider $$\eta(s):=\sum_{n=0}^\infty\frac{(-1)^n\mu(2n+1)}{(2n+1)^s}.$$

Then it is easy to see that $$\eta(s)$$ converges absolutely for $$\Re s >1$$, since by comparison with the harmonic series, $$|\sum_{n=0}^\infty\frac{(-1)^n\mu(2n+1)}{(2n+1)^s}|\leq \sum_{n=0}^\infty\frac{(-1)^n\mu(2n+1)}{(2n+1)^{\Re s}}.$$

Question 2. Can you provide useful hints or ideas to know what about the abscissa of convegence of our series $$\eta(s)$$? Please if it is in the literature refers it and I try to search and read it from the literature. Many thanks.

By the methods of analytic number theory for Dirichlet series, I think that should be useful calculate the asympotitc behaviour of $$\sum_{1\leq n\leq x}(-1)^{n}\mu(2n+1),$$ but I don't know nothing about it.

• I missed many of your questions (including this one), because you did not use the highest level tag "nt.number-theory". Please use this tag in the future. Aug 27, 2019 at 18:49
• Perfect, many thanks @GHfromMO Aug 28, 2019 at 5:25
• I recently came across this paper that at first sight seems it could be relevant to your question: degruyter.com/view/j/math.2019.17.issue-1/math-2019-0006/… Dec 26, 2019 at 0:55
• Many thanks for your help @Valerio I'm going to take a look Dec 26, 2019 at 13:32

This series can be written as a sum over all integers using a Dirchlet character of conductor $$4$$: $$\eta(s) = \sum_{n=0}^\infty\frac{(-1)^n\mu(2n+1)}{(2n+1)^s} = \sum_{m=1}^\infty \frac{\mu(m)\chi_{-4}(m)}{m^s},$$ where $$\chi_{-4}(m) = \begin{cases} 1, &\text{if } m\equiv1\pmod 4, \\ -1, &\text{if } m\equiv3\pmod 4, \\ 0, &\text{if m is even.} \end{cases}$$ Therefore this series is actually the reciprocal of the Dirichlet $$L$$-function $$L(s,\chi_{-4})$$, which factors into an infinite Euler product $$\eta(s) = L(s,\chi_{-4})^{-1} = \sum_{m=1}^\infty \frac{\mu(m)\chi_{-4}(m)}{m^s} = \prod_p \bigg( 1 - \frac{\chi_{-4}(p)}{p^s} \bigg)$$ and has a functional equation $$L(s,\chi_{-4}) = L(1-s,\chi_{-4}) \Gamma(1-s) (\tfrac\pi2)^{s-1} \cos(\tfrac \pi2s).$$ Finally, there is a known formula for the values of (all) Dirichlet $$L$$-functions at negative integers; for this function we have $$L(1-m,\chi_{-4}) = - \frac{4^{m-1}}{m} \big( B_{m}(\tfrac14) - B_{m}(\tfrac34) \big).$$ Here $$B_m(x)$$ is the usual Bernoulli polynomial, defined by $$\frac{t e^{Xt}}{e^t-1} = \sum_{n=0}^\infty B_n(X) \frac{t^n}{n!};$$ presumably its values at $$\frac14$$ and $$\frac34$$ are related to the polygamma values you discovered.
This information is collectively enough to evaluate $$L(s,\chi_{-4})$$ exactly at positive odd integers. (At positive even integers the functional equation gives merely $$0=0$$, and so those values are a mystery exactly like the values of the Riemann zeta function at positive odd integers.)
• Shouldn't it be $L(s,\chi_{-4})^{-1}$, not $L(s,\chi_{-4})$? Jul 29, 2019 at 18:44