Writting $S = \sum_{k=0}^{\infty} \frac{1}{(r_1+k+1)(r_2+k+1)(r_3+k+1)}$ as a rational function of $r_1,r_2$ and $r_3$

Question

I'll give an example of what I'm trying to say. Consider the equation, where $r_1>r_2>0$ and both integers:

\begin{align} \sum_{k=0}^{\infty} \frac{1}{(r_1+k+1)(r_2+k+1)} = \frac{1}{r_1-r_2}\left( \frac{1}{r_2+1}+ \frac{1}{r_2+2} + \cdots + \frac{1}{r_1} \tag{1} \right) \end{align}

I wanted something similar for \begin{align} S = \sum_{k=0}^{\infty} \frac{1}{(r_1+k+1)(r_2+k+1)(r_3+k+1)} \end{align}

With $r_1>r_2>r_3>0$ and integers. Where $S$ is written as a simple rational function like $(1)$. This way we shall avoid anything written as digamma functions or harmonic series.

I'm not sure which tags to use for this question. If someone can provide more suitable ones it would be appreciated.

Answer 1

According to Wolfy,

$\frac1{(a + k) (b + k) (c + k)} =-\frac1{(a - b) (b - c) (b + k)} - \frac1{(c-a) (b-c) (c + k)} - \frac1{(a - b) (c - a) (a + k)} $.

You can use this to split your sum into 3 sums of a single reciprocal.

The generalization to an arbitrary number of terms seems somewhat evident, but I am too lazy right now to work out the details. Probably something like

$\dfrac1{\prod_{j=1}^m (a_j+k)} =\sum_{j=1}^m \dfrac{(-1)^{\text{something like }j}}{(a_j+k)\prod_{i=1, i\ne j}^m(a_i-a_j)} $.

As is evident, I'm not sure what the sign of the expressions should be.

Answer 2

Given any set of $n$ distinct positive numbers $r_1 > r_2 > \cdots > r_n > 0$, let $S_n(r_1,\cdots,r_n)$ be the sum

$$S_n(r_1,\ldots,r_n) = \sum_{s=1}^\infty \prod_{k=1}^n \frac{1}{r_k+s}$$

The sum we want is $S_3(r_1,r_2,r_3)$ and we already know the formula for $S_2(r_1,r_2)$ for integers.

In general, we can express $S_n$ in terms of $S_{n-1}$. For any $n \ge 3$, we have

$$\begin{align}\prod_{k=1}^n \frac{1}{r_k +s} &= \frac{1}{r_1+s}\frac{1}{r_n+s} \prod_{k=2}^{n-1}\frac{1}{r_k+s}\\ &= \frac{1}{r_n-r_1}\left(\frac{1}{r_1+s} - \frac{1}{r_n+s}\right) \prod_{k=2}^{n-1}\frac{1}{r_k+s} \\ &= \frac{1}{r_n-r_1}\left(\prod_{k=1}^{n-1} \frac{1}{r_k+s} - \prod_{k=2}^n \frac{1}{r_k+s}\right) \end{align} $$ Summing over $s$ leads to

$$S_n(r_1,\ldots,r_n) = \frac{1}{r_n - r_1}[ S_{n-1}(r_1\ldots,r_{n-1}) - S_{n-1}(r_2,\ldots,r_n)]\tag{*1}$$ In particular, for $n = 3$, we have

$$S_3(r_1,r_2,r_3) = \frac{1}{r_3-r_1}\left[S_2(r_1,r_2) - S_2(r_2,r_3)\right]$$

For integers $r_1, r_2, r_3$, you can use this identity and the known formula for $S_2(r_1,r_2)$ and $S_2(r_2,r_3)$ to derive the closed form you want.

For general $n > 3$, we can repeatedly apply $(*1)$ and express $S_n(r_1,\ldots,r_n)$ as a linear combination of $S_2(r_1,r_2)$, $S_2(r_2,r_3), \ldots, S_2(r_{n-1},r_n)$ with rational functions in $r_1,\ldots,r_n$ as coefficients.

To derive the linear combination, let $P(s) = \prod_{k=1}^n (r_k + s)$. Since all $r_k$ are distinct, it has following partial fraction decomposition:

$$\frac{1}{P(s)} = \sum_{k=1}^n \frac{\alpha_k}{r_k + s}\quad\text{ where }\quad \alpha_k = \frac{1}{P'(-r_k)}$$

As long as $n > 1$, we have $$\sum_{k=1}^n \alpha_k = \lim_{s\to\infty} \sum_{k=1}^n \frac{a_k s}{r_k+s} = \lim_{s\to\infty} \frac{s}{P(s)} = 0$$ This allow us to rewrite $\frac{1}{P(s)}$ as

$$\frac{1}{P(s)} = \sum_{k=1}^n \alpha_k\left(\frac{1}{r_k+s} - \frac{1}{s}\right)$$

Summing over $s$ give us

$$S_n(r_1,r_2,\ldots,r_n) = \sum_{s=1}^\infty \frac{1}{P(s)} = \sum_{k=1}^n \alpha_k S_1(r_k)\tag{*2}$$

where $S_1(r) = \sum\limits_{s=1}^\infty \left(\frac{1}{r+s} - \frac{1}{s}\right) = -\gamma - \psi(r+1)$ and $\psi(r)$ is the digamma function.

In particular for $n = 2$, we obtain

$$S_2(r_1,r_2) = \frac{1}{r_2-r_1}\left(S_1(r_1) - S_1(r_2)\right) = \frac{1}{r_1-r_2}\left(\psi(r_1+1)-\psi(r_2+1)\right) $$

With this choice of $S_1(r)$, $(*1)$ is also true for $n=2$. For intergers $r_1,r_2$, this reduces to the formula we already know

$$S_2(r_1,r_2) = \frac{1}{r_1-r_2} \sum_{r=r_2+1}^{r_1} \frac{1}{r}\tag{*3}$$

Let $A_k = \sum\limits_{j=1}^k \alpha_j$, we can rewrite $(*2)$ as $$\begin{align} S_n(r_1,\ldots,r_n) &= A_1 S_1(r_1) + \sum_{k=2}^n ( A_k - A_{k-1} )S_1(r_k)\\ &= A_n S_1(r_n) + \sum_{k=1}^{n-1} A_k (S_1(r_k) - S_1(r_{k+1}))\\ \end{align} $$

Since $A_n = 0$, we obtain following decomposition of $S_n(\cdots)$:

$$S_n(r_1,\ldots,r_n) = \sum_{k=1}^{n-1} B_k S_2(r_k,r_{k+1}) \quad\text{ where }\quad B_k = (r_{k+1}-r_k)\sum_{j=1}^k \frac{1}{P'(-r_j)}$$

As an example, when $n = 3$

$$\begin{align} B_1 &= (r_2-r_1)\frac{1}{P'(-r_1)} = \frac{1}{r_3 - r_1}\\ B_2 &= (r_3-r_2)\left(\frac{1}{P'(-r_1)} + \frac{1}{P'(-r_2)}\right) = \frac{1}{r_1-r_3} \end{align}$$ For integers $r_1 > r_2 > r_3 > 0$, we can use $(*3)$ to deduce

$$S_3(r_1,r_2,r_3) = -\frac{1}{r_1-r_3}\left[\frac{1}{r_1-r_2}\sum_{r=r_2+1}^{r_1} \frac{1}{r} - \frac{1}{r_2-r_3}\sum_{r=r_3+1}^{r_2}\frac1r\right]$$

For general $n$ and integers $r_1 > r_2 > \cdots > r_n > 0$, the corresponding formula is

$$S_n(r_1,\ldots,r_n) = -\sum_{k=1}^{n-1}\left[\left(\sum_{j=1}^k\prod_{i=1,\ne j}^{n}\frac{1}{r_i-r_j}\right)\sum_{r=r_{k+1}+1}^{r_k} \frac{1}{r}\right]$$