What is 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

We can find the answer by breaking n into even and odd numbers $$S=\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty(a_{2n}+a_{2n+1})$$ $$\sum_{n=0}^\infty\frac{(-1)^n}{(zn+1)^2}=\sum_{n=0}^\infty\left(\frac{(-1)^{(2n)}}{(z(2n)+1)^2}+\frac{(-1)^{(2n+1)}}{(z(2n+1)+1)^2)}\right)=\sum_{n=0}^\infty\left(\frac{1}{(2nz+1)^2}-\frac{1}{(2nz+z+1)^2}\right)=\sum_{n=0}^\infty\left(\frac{1}{(2z)^2}\frac{1}{(n+\frac{1}{2z})^2}-\frac{1}{(2z)^2}\frac{1}{(n+\frac{z+1}{2z})^2}\right)$$ The sum obviously can be evaluated using the trigamma function

Answer 2

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

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?

Answer 3

Here is an integral representation.

Let $|x|<1$. Write $$f_z(x)=\sum_{n\ge0}x^{zn}=\frac{1}{1-x^z}.$$ Then $$\int_0^tf_z(y)=\int_0^1 tf_z(ty)dy=\sum_{n\ge0}\frac{t^{zn+1}}{zn+1}.$$ Dividing by $t$ and integrating again, $$\int_0^x\int_0^1f_z(ty)dydt=\int_0^1\int_0^1xf_z(xty)dydt=\sum_{n\ge0}\frac{x^{zn+1}}{(zn+1)^2}.$$ Thus $$\int_0^1\int_0^1f_z((-1)^{1/z}yt)dydt=\int_0^1\int_0^1\frac{dydt}{1+(yt)^z}=\sum_{n\ge0}\frac{(-1)^n}{(zn+1)^2}.$$

Answer 4

Using only that $$ \sum_{n=0}^\infty\frac1{(z+n)^2}=\psi_1(z)\tag1 $$ we can derive $$ \begin{align} \sum_{n=0}^\infty\frac{(-1)^n}{(nz+1)^2} &=\sum_{n=0}^\infty\frac2{(2nz+1)^2}-\sum_{n=0}^\infty\frac1{(zn+1)^2}\tag{2a}\\ &=\frac1{2z^2}\sum_{n=0}^\infty\frac1{\left(n+\frac1{2z}\right)^2}-\frac1{z^2}\sum_{n=0}^\infty\frac1{\left(n+\frac1z\right)^2}\tag{2b}\\ &=\frac1{2z^2}\psi_1\!\left(\frac1{2z}\right)-\frac1{z^2}\psi_1\!\left(\frac1z\right)\tag{2c} \end{align} $$ Explanation:
$\text{(2a):}$ the alternating sum is twice the sum of the even terms
$\phantom{\text{(2a):}}$ minus the sum of all the terms
$\text{(2b):}$ factor common pieces out of each sum
$\text{(2c):}$ apply $(1)$