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) $.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.