4
$\begingroup$

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).

$\endgroup$
  • 2
    $\begingroup$ 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 at 18:49
  • $\begingroup$ Perfect, many thanks @GHfromMO $\endgroup$ – user142929 Aug 28 at 5:25
8
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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 at 17:55
  • 2
    $\begingroup$ @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. $\endgroup$ – Gerhard Paseman Jul 29 at 18:15
  • 4
    $\begingroup$ Shouldn't it be $L(s,\chi_{-4})^{-1}$, not $L(s,\chi_{-4})$? $\endgroup$ – Stopple Jul 29 at 18:44
  • $\begingroup$ @Stopple oops, you're totally right, edited. $\endgroup$ – Greg Martin Jul 30 at 6:07
  • 1
    $\begingroup$ @user142929 Thank you—that's the beauty of Stack Exchange, sometimes a question hits a person right in their specialty :) $\endgroup$ – Greg Martin Jul 30 at 6:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.