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.

$\endgroup$
  • 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 Oct 16 '15 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 Oct 16 '15 at 18:36
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.

$\endgroup$
  • $\begingroup$ Your answer is certainly more constructive (in many ways) than mine! $\endgroup$ – GH from MO Oct 16 '15 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).

$\endgroup$
  • $\begingroup$ "Unknown" meaning unproven, yes, but it's morally certain that $\zeta(5) \notin \bf Q$... $\endgroup$ – Noam D. Elkies Oct 16 '15 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 Oct 16 '15 at 19:15

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.