For $\sum_{n=1}^\infty z^n \frac{P(n)}{Q(n)}$ where $Q(n)$ and $P(n)$ are polynomials, does di(con)vergence only depends on $z$?

Write $d = \deg Q - \deg P$. Then we may write

$$ \frac{P(z)}{Q(z)} = \frac{R(z)}{z^d} $$

for some rational function $R(z)$ such that $R(z) \to 1$ as $z \to \infty$. This gives a decisive answer for the behavior of the sum along the circle of convergence:

• If $z = 1$, then

  $$ \sum_{n=1}^{\infty} z^n \frac{P(n)}{Q(n)} = \sum_{n=1}^{\infty} \frac{R(n)}{n^d}, $$

  and so, the sum converges absolutely for $d > 1$ and diverges for $d \leq 1$ by the limit comparison test.

• If $|z| = 1$ but $z \neq 1$, then

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

  converges absolutely for $d > 1$, converges conditionally for $d = 1$ by the Dirichlet test, and diverges for $d < 1$.