Generalized FoxTrot Series $F(a,b,q,x) = \sum_{k=q}^{\infty} \dfrac {(-1)^{k+1} k^a}{k^b+x}$

The FoxTrot Series is defined as:

$$F = \sum_{k=1}^{\infty} \dfrac {(-1)^{k+1} k^2}{k^3+1}.$$

Using partial fraction decomposition we can show that

$$F = \frac 13 \left[ 1 - \ln2 + \pi\operatorname{sech}\left(\frac 12 \sqrt3 \, \pi\right) \right].$$

More details about the evaluation at FoxTros Series MathWorld article or in this math.se answer.

Note that we could write $F$ in term of digamma functions:

$$F = \frac 13 \left[ 1 - \ln2 - \frac 12 \psi_0\left( \frac 12 (-1)^{1/3} \right) - \frac 12 \psi_0\left( -\frac 12 (-1)^{2/3} \right) + \frac 12 \psi_0\left( \frac 12 \left( 1+ (-1)^{1/3} \right) \right) + \frac 12 \psi_0\left( \frac 12 \left(1 - 1(-1)^{2/3} \right) \right) \right].$$

Now we define the following parametric series:

$$F(a,b,q,x) = \sum_{k=q}^{\infty} \dfrac {(-1)^{k+1} k^a}{k^b+x},$$

where $a,b,q$ are nonnegative integers, $a<b$, and $x\in\mathbb{C}$.

Question. Is there a closed-form for $F(a,b,q,x)$?

Of course $F=F(2,3,1,1)$. We also know that $F(0,1,1,1)=1-\ln2$. I've evaluated with Maple $F(i,j,1,1)$ for all $(i,j)$ for $0 \leq i <j$, $0 < j \leq 4$. Maple could solve them in term of digamma functions, so I guess that there is a general closed-form. Beside the closed-form maybe we could get or use a nice digamma identity as well.

Solution 1:

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:

1. 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 $$\frac{1}{2}\left [ \psi^0\left (\frac{q+1}{2}-\frac{x}{2}\right )-\psi^0\left ( \frac{q}{2}-\frac{x}{2}\right ) \right ]=\\ -\frac{1}{2}\sum_{n=0}^\infty \left ( \frac{1}{n+1}-\frac{1}{n+(q/2-x/2)}\right )+\frac{1}{2}\sum_{n=0}^\infty \left ( \frac{1}{n+1}-\frac{1}{n+q/2+1/2-x/2}\right )\\ =\frac{1}{2}\sum_{n=0}^\infty\left (\frac{1}{n+q/2-x/2}-\frac{1}{n+q/2+1/2-x/2} \right ) \\ =\sum_{n=0}^\infty\left (\frac{1}{(2n+q)-x}-\frac{1}{(2n+q+1)-x} \right )\\ =\frac{1}{q-x}-\frac{1}{q+1-x}+\frac{1}{q+2-x}-\frac{1}{q+3-x}...\\ =\sum_{k=q}^\infty \frac{(-1)^{k-q}}{k-x} \Box.$$

2. $$k^b+x=\prod_{m=1}^{b}\left( k - \exp\left[\frac{1}{b}(2\pi i m+i\text{Arg}(-x)+\log|x|)\right] \right )\\ =\prod_{m=1}^{b}\left( k - c_m \right ),$$ 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.

3. 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.