enter image description here

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.


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

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

  2. 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.

  • $\begingroup$ It turns out that the function which I defined as $S(q,x)=\Phi(-1,1,q-x)$, where $\Phi(z,s,a)$ is the Hurwitz-Lerch transcendent. $\endgroup$ – SDiv Oct 7 '14 at 22:29
  • $\begingroup$ At Addendum I $c_p$ means $c_p(b,x)$, right? $\endgroup$ – user153012 Oct 9 '14 at 5:56
  • $\begingroup$ @user153012. Correct, $c_p=c_p(b,x)$ $\endgroup$ – SDiv Oct 9 '14 at 6:10
  • $\begingroup$ @user153012. I have re-formatted the result in the addendum to reflect this point. $\endgroup$ – SDiv Oct 9 '14 at 8:30
  • $\begingroup$ Really nice solution. Thank you. $\endgroup$ – user153012 Oct 10 '14 at 20:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.