The sum of $\sum_{n=0}^{\infty}\frac{(-1)^n}{(zn+1)^2}$

Question

I need to find the value of the following series
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(zn+1)^2}$$
I know there is a nice closed form in terms of trigamma functions, as shown here, I simply can't get it.

What I know is:
$$\sum_{n=0}^{\infty}\frac{1}{(z+n)^2}=\psi_1(z)$$
and
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{zn+1}=\frac{1}{2z}\left[\psi_0\left(\frac{z+1}{2z}\right)-\psi_0\left(\frac{1}{2z}\right)\right]$$
In particular, differentiating this last series w.r.t $z$ doesn't help, since you get an $n$ in the numerator as well. How to proceed?

Answer 1

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{zn+1}=\frac{1}{2z}\left[\psi_0\left(\frac{z+1}{2z}\right)-\psi_0\left(\frac{1}{2z}\right)\right]$$

Take derivative works,

$$\begin{align}\frac{d}{dz}\sum_{n=0}^{\infty}\frac{(-1)^n}{zn+1}&=-\sum_{n=0}^{\infty}\frac{(-1)^nn}{(zn+1)^2}=-\sum_{n=0}^{\infty}\frac{(-1)^n(n+\frac1z-\frac1z)}{(zn+1)^2}\\ \\ &=-\frac1z\sum_{n=0}^{\infty}\frac{(-1)^n}{zn+1}+\frac1z\sum_{n=0}^{\infty}\frac{(-1)^n}{(zn+1)^2}\end{align}$$

Can you proceed from here?