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.

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).
 A: 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.)
