Be patient and the nice digamma identity that you asked for, shown below, will pop up in my answer. Lets begin with a few notes:
- Claim: $$\sum_{k=q}^\infty \frac{(-1)^{k-q}}{k-x}=\frac{1}{2}\left [ \psi^0\left (\frac{q+1}{2}-\frac{x}{2}\right ) - \psi^0\left ( \frac{q}{2}-\frac{x}{2}\right )\right ], q\in\mathbb{Z}^\ast$$ where $\psi^0$ is the digamma function or the polygamma function of order zero.
Proof:
By Wikipedia we have $$\psi^0(z)=-\gamma+\sum_{n=0}^\infty \left ( \frac{1}{n+1}-\frac{1}{n+z}\right ),$$ where $\gamma$ is a famous constant.
Then $$\frac{1}{2}\left [ \psi^0\left (\frac{q+1}{2}-\frac{x}{2}\right )-\psi^0\left ( \frac{q}{2}-\frac{x}{2}\right ) \right ]=\\
-\frac{1}{2}\sum_{n=0}^\infty \left ( \frac{1}{n+1}-\frac{1}{n+(q/2-x/2)}\right )+\frac{1}{2}\sum_{n=0}^\infty \left ( \frac{1}{n+1}-\frac{1}{n+q/2+1/2-x/2}\right )\\
=\frac{1}{2}\sum_{n=0}^\infty\left (\frac{1}{n+q/2-x/2}-\frac{1}{n+q/2+1/2-x/2} \right ) \\
=\sum_{n=0}^\infty\left (\frac{1}{(2n+q)-x}-\frac{1}{(2n+q+1)-x} \right )\\
=\frac{1}{q-x}-\frac{1}{q+1-x}+\frac{1}{q+2-x}-\frac{1}{q+3-x}...\\
=\sum_{k=q}^\infty \frac{(-1)^{k-q}}{k-x} \Box.$$
$$k^b+x=\prod_{m=1}^{b}\left( k - \exp\left[\frac{1}{b}(2\pi i m+i\text{Arg}(-x)+\log|x|)\right] \right )\\
=\prod_{m=1}^{b}\left( k - c_m \right ),$$ where we have taken the product using the $b$th roots of $-x$ and defined $c_m(b,x):=\exp\left[\frac{1}{b}(2\pi i m+i\text{Arg}(-x)+\log|x|)\right]$, $1\leq m \leq b$. It is important to note that each $c_m$ is unique.
Now for an interesting identity concerning partial fractions.
Claim: $$\frac{k^a}{k^b+x}=\sum_{m=1}^b\frac{c_m^a}{(k-c_m)\prod_{n\neq m}(c_m-c_n)} $$
where each $c_m$ is unique and each $c_m\in\mathbb{C}$, $(a,b)\in\mathbb{Z}^\ast$, $b>a$. To be clear, the index $n$ runs from 1 to $b$ but skips $m$. We could also write $\prod_{n\neq m}(c_m-c_n)$ as $\prod_{n=1}^b(\delta_{mn}+[1-\delta_{mn}][c_m-c_n])$, where $\delta_{mn}$ is the Kronecker delta.
Proof:
Proof is given in post A, which uses results from note 2 of the current post and from post B.
Now we put it all together. We have the Generalized FoxTrot series $$F(a,b,q,x):=\sum_{k=q}^\infty\frac{(-1)^{k-q}k^a}{k^b+x},$$
with $(a,b,q)\in\mathbb{Z}^\ast$, $b>a$, $x\in\mathbb{C}$, $|x|>0$.
I define $S(q,x):=\frac{1}{2}\left [ \psi^0\left (\frac{q+1}{2}-\frac{x}{2}\right ) - \psi^0\left ( \frac{q}{2}-\frac{x}{2}\right )\right ].$
Using note 2, we have $$F(a,b,q,x)=\sum_{k=q}^\infty\frac{(-1)^{k-q}k^a}{\prod_{m=1}^b(k-c_m)}. $$
Using note 3, we do a little sum-flipping: $$F(a,b,q,x)=\sum_{k=q}^\infty\sum_{m=1}^b\frac{(-1)^{k-q}c_m^a}{(k-c_m)\prod_{n\neq m}(c_m-c_n)}\\
=\sum_{m=1}^b\left (\frac{c_m^a}{\prod_{n\neq m}(c_m-c_n)} \sum_{k=q}^\infty\frac{(-1)^{k-q}}{(k-c_m)} \right ). $$
Here comes the digamma identity from note 1:
$$F(a,b,q,x)=\sum_{m=1}^b\left (\frac{c_m^a(b,x)}{\prod_{n\neq m}(c_m(b,x)-c_n(b,x))} S(q,c_m[b,x]) \right )\\
= \sum_{m=1}^b\left (\frac{c_m^a(b,x)}{\prod_{n=1}^b(\delta_{mn}+[1-\delta_{mn}][c_m(b,x)-c_n(b,x)])} S(q,c_m[b,x]) \right ) $$
Thus, we have turned an infinite series into a finite series by using a nice digamma identity!
Addendum I
As suggested by a simpler partial fraction identity given in this post, we can instead write
$$F(a,b,q,x)=\sum_{k=q}^\infty \sum_{p=1}^b \frac{(-1)^{k-q+1}}{bx}\frac{c_p^{a+1}}{k-c_p}\\
= \sum_{p=1}^b \left ( \frac{-c_p^{a+1}}{bx} \sum_{k=q}^\infty \frac{(-1)^{k-q}}{k-c_p} \right ) \\
= \frac{-1}{bx}\sum_{p=1}^b c_p^{a+1}(b,x)S(q,c_p[b,x]),$$ which is much prettier.
