Explicit expression for a meromorphic function What is an explicit expression for the meromorphic function
$$
\sum_{n\ge 1}\frac{(-1)^n}{z+n} ?
$$
 A: 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 $\frac{\pi\sqrt{3}-9+3\log 2}{3}$.
A: $$\begin{align}
\sum_{n=1}^\infty\frac{(-1)^n}{z+n} & =(-1)^{-z}\sum_{n=z+1}^\infty\frac{(-1)^n}{n}\tag1 \\
& =e^{-z\pi i}\left(\sum_{n=1}^\infty\frac{(-1)^n}{n}-\sum_{n=1}^z\frac{(-1)^n}{n}\right)\tag2 \\
& =e^{-z\pi i}\left(\ln(2)+\int_{-1}^0\frac{1-r^z}{1-r}dr\right)\tag3 \\
\end{align}$$
$\diamondsuit\ 1:$ 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.
