Explicit expression for a meromorphic function

Question

What is an explicit expression for the meromorphic function $$ \sum_{n\ge 1}\frac{(-1)^n}{z+n} ? $$

Answer 1

We have $$ \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right)=\gamma+\psi(1+z)\tag{1}$$ where $\psi(z)=\frac{d}{dz}\log\Gamma(z)$, by the Weierstrass product for the $\Gamma$ function. It follows that

$$ \sum_{n\geq 1}\frac{(-1)^n}{n+z}=\sum_{n\geq 1}\left(\frac{1}{2n+z}-\frac{1}{2n-1+z}\right)=\color{red}{\frac{\psi\left(\frac{1+z}{2}\right)-\psi\left(1+\frac{z}{2}\right)}{2}}.\tag{2}$$

The RHS of $(2)$ can be written in terms of $\pi$, algebraic numbers and logarithms of algebraic numbers for any $z\in\mathbb{Q}$, due to Gauss' digamma theorem. For instance, at $z=\frac{1}{2}$ the RHS of $(2)$ equals $\frac{\pi-4}{2}$ and at $z=\frac{1}{3}$ the RHS of $(2)$ equals $\frac{\pi\sqrt{3}-9+3\log 2}{3}$.

Answer 2

$$\begin{align} \sum_{n=1}^\infty\frac{(-1)^n}{z+n} & =(-1)^{-z}\sum_{n=z+1}^\infty\frac{(-1)^n}{n}\tag1 \\ & =e^{-z\pi i}\left(\sum_{n=1}^\infty\frac{(-1)^n}{n}-\sum_{n=1}^z\frac{(-1)^n}{n}\right)\tag2 \\ & =e^{-z\pi i}\left(\ln(2)+\int_{-1}^0\frac{1-r^z}{1-r}dr\right)\tag3 \\ \end{align}$$

$\diamondsuit\ 1:$ Reindexing the sum.

$\color{red}\diamondsuit\ 2:$ Changing the $(-1)^{-z}$ to its complex exponential form and split the sum into two sums.

$\diamondsuit\ 3:$ First sum is the alternating harmonic series and second sum has the form $\sum_{n=1}^z\frac{r^n}n=\int_0^r\frac{1-x^z}{1-x}dx$.

$\color{red}\diamondsuit$ remarks : I don't think any further simplifications are possible.