Express $\sum_{n=1}^{\infty} \dfrac1{\prod_{k=1}^m (n+a_k)}$ as a finite sum.

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 $$-\sum_{k=1}^mc_k\ln(a_k)$$.

• Do you mean that $a_k$ are distinct positive integers? – user Nov 3 '18 at 11:43
• Yes I do. 6 more – marty cohen Nov 3 '18 at 20:37

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 $$\psi(z)=\frac{d}{dz}\log\Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)}=H_{z-1}-\gamma$$.
• 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. – marty cohen Nov 1 '18 at 20:45
• @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. – Jack D'Aurizio Nov 1 '18 at 20:47
• @ Jack D'Aurizio: I want an explicit formula for arbitrary $m$. – marty cohen Nov 1 '18 at 20:53
• @martycohen: you already have one, $$-\sum_{k=1}^{m}\frac{H_{a_k}}{\prod_{j\neq k}(a_j-a_k)}.$$ – Jack D'Aurizio Nov 1 '18 at 20:55
• I was hoping for one similar in form to your equation (2). Maybe I could modify mine so the denominator is $\prod_{1 \le j < k \le m}(a_j-a_k)$. – marty cohen Nov 2 '18 at 0:08