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.

Footnote (See comments). Greg Martin anticipated a follow up question ( I thought to ask about the possibility of a functional equation and what about the zeros).

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    $\begingroup$ 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. $\endgroup$
    – GH from MO
    Aug 27, 2019 at 18:49
  • $\begingroup$ Perfect, many thanks @GHfromMO $\endgroup$
    – user142929
    Aug 28, 2019 at 5:25
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    $\begingroup$ 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/… $\endgroup$
    – Valerio
    Dec 26, 2019 at 0:55
  • $\begingroup$ Many thanks for your help @Valerio I'm going to take a look $\endgroup$
    – user142929
    Dec 26, 2019 at 13:32

1 Answer 1


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.)

  • $\begingroup$ You are such a good mathematician that you've spoiled another post that I wanted to write asking about a possible functional equation and zeroes :) $\endgroup$
    – user142929
    Jul 29, 2019 at 17:55
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    $\begingroup$ @user142929 : it is fine to add a footnote to your current question, observing that Greg Martin anticipated a follow up question. Your footnote (when properly phrased and read) would aid future readers who might try forming a similar question after reading yours. Gerhard "Is Anticipating Anticipating Future Anticipation" Paseman, 2019.07.29. $\endgroup$ Jul 29, 2019 at 18:15
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    $\begingroup$ Shouldn't it be $L(s,\chi_{-4})^{-1}$, not $L(s,\chi_{-4})$? $\endgroup$
    – Stopple
    Jul 29, 2019 at 18:44
  • $\begingroup$ @Stopple oops, you're totally right, edited. $\endgroup$ Jul 30, 2019 at 6:07
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    $\begingroup$ @user142929 Thank you—that's the beauty of Stack Exchange, sometimes a question hits a person right in their specialty :) $\endgroup$ Jul 30, 2019 at 6:09

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