1
$\begingroup$

This is a follow up on a previous question of mine.

Out of curiosity, I am wondering more generally when a closed form exists for

$$\sum_{n=1}^{\infty} \frac{P(n)}{Q(n)}$$

where $P$ and $Q$ are both polynomials. Obviously for convergence $\deg(Q)\ge \deg(P)+2$, but I don't know much from there.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Have a look at the book $A=B$ math.upenn.edu/~wilf/AeqB.html Such sums tend to be expressed as values of hypergeometric functions, not that it helps too much. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 16, 2015 at 18:17
  • 2
    $\begingroup$ More natural sum $\sum_{n=-\infty}^{\infty} \frac{P(n)}{Q(n)}$ can be expressed via residues of $P/Q$. $\endgroup$
    – Andrew
    Commented Oct 16, 2015 at 18:36

2 Answers 2

Reset to default
8
$\begingroup$

As GH from MO says, it depends what you think is a closed form.

Any function of the form $P(n)/Q(n)$ with $\deg P \leq \deg Q -2$ can be written as a linear combination of $1/(n+1) - 1/(n+\alpha)$ and $1/(n+\alpha)^k$, for various $\alpha$ and various $k>1$. (Use partial fraction decomposition. The bound on $\deg P$ ensures that the coefficients of $1/(n+1)$ will cancel out.)

The sum $\sum \left( \frac{1}{n+1} - \frac{1}{n+\alpha} \right)$ is $\gamma+\psi(\alpha)$, where $\psi$ is the digamma function. The sum $\sum \frac{1}{(n+\alpha)^k}$ is the polygamma function $\psi^k(z)$. In both cases, it is worth noting the recurrence identities and reflection relations in the linked Wikipedia articles, which let you relate values at $\alpha$ to values at $m \pm \alpha$, for $m$ an integer.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Your answer is certainly more constructive (in many ways) than mine! $\endgroup$
    – GH from MO
    Commented Oct 16, 2015 at 19:17
3
$\begingroup$

Closed form is a subjective term. At any rate, our understanding of these sums is quite limited, e.g. it is not known if $\sum_{n=1}^\infty n^{-5}$ is an irrational number (same for any smaller negative odd exponent).

Improve this answer
$\endgroup$
2
  • $\begingroup$ "Unknown" meaning unproven, yes, but it's morally certain that $\zeta(5) \notin \bf Q$... $\endgroup$
    – Noam D. Elkies
    Commented Oct 16, 2015 at 18:57
  • $\begingroup$ @NoamD.Elkies: Yes, I meant what is proven. Probably the odd zeta values are even algebraically independent, which shows in particular that the OP's question is too broad (in addition to being too vague). $\endgroup$
    – GH from MO
    Commented Oct 16, 2015 at 19:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .