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This is a generalization (and answer to) 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 $r_3$.

Let $S =\sum_{n=1}^{\infty} \dfrac1{\prod_{k=1}^m (n+a_k)} $ where $m \ge 2$ and the $a_k$ are positive integers.

Find an expression for $S$ involving only finite sums.

I get $S =-\sum_{k=1}^mc_kH_{a_k} $ where $c_k= \dfrac1{\prod_{j=1, j\ne k}^m (a_j-a_k)} $ and $H_n = \sum_{j=1}^n \dfrac1{j}$.

It is interesting that a quite good approximation to $S$ is $-\sum_{k=1}^mc_k\ln(a_k) $.

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  • $\begingroup$ Do you mean that $a_k$ are distinct positive integers? $\endgroup$ – user Nov 3 '18 at 11:43
  • $\begingroup$ Yes I do. 6 more $\endgroup$ – marty cohen Nov 3 '18 at 20:37
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Provided that $\text{Re}(a),\text{Re}(b)>0$ we have $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b},\tag{1} $$ hence by partial fraction decomposition, assuming $\text{Re}(c)>0$,

$$\begin{eqnarray*} \sum_{n\geq 0}\frac{1}{(n+a)(n+b)(n+c)}&=&\frac{1}{c-a}\left[\frac{\psi(a)-\psi(b)}{a-b}-\frac{\psi(b)-\psi(c)}{b-c}\right]\\&=&\frac{(c-b)\psi(a)+(a-c)\psi(b)+(a-b)\psi(c)}{(a-c)(a-b)(b-c)}\end{eqnarray*}\tag{2} $$ with $\psi(z)=\frac{d}{dz}\log\Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)}=H_{z-1}-\gamma$.
If more variables are involved the approach is just the same.

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  • $\begingroup$ I am looking for an explicit formula in terms of the $a_k$. I am aware of the discussion in places such as en.wikipedia.org/wiki/Digamma_function#Series_formula, but would like it made explicit. $\endgroup$ – marty cohen Nov 1 '18 at 20:45
  • $\begingroup$ @martycohen: I do not understand your request. $(2)$ gives an explicit relation between the given series and $H_{a_1},\ldots,H_{a_n}$. Only finite sums are involved. $\endgroup$ – Jack D'Aurizio Nov 1 '18 at 20:47
  • $\begingroup$ @ Jack D'Aurizio: I want an explicit formula for arbitrary $m$. $\endgroup$ – marty cohen Nov 1 '18 at 20:53
  • $\begingroup$ @martycohen: you already have one, $$ -\sum_{k=1}^{m}\frac{H_{a_k}}{\prod_{j\neq k}(a_j-a_k)}.$$ $\endgroup$ – Jack D'Aurizio Nov 1 '18 at 20:55
  • $\begingroup$ I was hoping for one similar in form to your equation (2). Maybe I could modify mine so the denominator is $\prod_{1 \le j < k \le m}(a_j-a_k)$. $\endgroup$ – marty cohen Nov 2 '18 at 0:08

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