Assume $p, q \in \mathbb{R}[x_1,\ldots,x_k]$ and let $ \vec{0} \not\in V(q) := \{\vec{x} \in \mathbb{R}^k \mid q(\vec{x}) = 0 \}$ such that $r_q := \inf_{\vec{x}\in V(q)} |\!|\vec{x}|\!|_\infty < \infty$ exists. We consider $\frac{p}{q} = p \cdot \sum_{\alpha \in \mathbb{N}^k} a_\alpha \vec{x}^\alpha \in \mathbb{R}[\![\vec{x}]\!]$. Is the following true: $$ \forall \vec{b} \in B_{\vec{0}}(r_q) := \{\vec{x} \in \mathbb{R}^n \mid |\!|\vec{x}|\!|_\infty < r_q\}. \quad \frac{p(\vec{b})}{q(\vec{b})} ~=~ p(\vec{b}) \cdot \sum_{\alpha \in \mathbb{N}^k} a_\alpha \vec{b}^\alpha $$
Phrased in words: I have two polynomials $p$ and $q$ such that $q$ is invertible in the ring of formal power series. I choose $r_q$ as the minimal distance of $q$'s zeroes to the origin (w.r.t. the $\infty$-norm). Does evaluating the "closed-form" $p/q$ at all points $\vec{b}$ inside the Ball with radius $r_q$ coincide with computing the actual infinite sum at $\vec{b}$? Does this depend on the $\infty$-norm or would $|\!|\cdot|\!|_2$, or $|\!|\cdot|\!|_1$ also always work?
Example: \begin{align*} \frac{p}{q} &= \frac{1}{1-x-y} = \sum_{n \in \mathbb{N}} (x+y)^n\\ V(q) &= \{(x,y) \mid y = 1-x,~ x\in \mathbb{R}\}\\ \inf_{(x,y) \in V(q)} |\!|\binom{x}{y}|\!|_\infty &= \frac{1}{2} \end{align*} From Calculus 101 we know the geometric series $\sum_{n \in \mathbb{N}} (x+y)^n$ converges if and only if $|x+y| < 1$. Indeed, $B_\vec{0}(1/2) = \{(x,y) \mid |x| < 1/2, ~ |y| < 1/2\}$ satisfies the condition $|x+y| < 1$. However this result seems to be very conservative. If we would have taken the $|\!|\cdot|\!|_1$ norm instead, $\inf_{(x,y) \in V(q)} |\!|\binom{x}{y}|\!|_1$ becomes 1 and hence $B_\vec{0}(1)$ is the set of all points $(x,y)$ where $|x| + |y| < 1$ which is by triangle inequality at least as big as $|x + y|$ hence again satisfying the condition, but this time less conservative.
So in both cases we would be fine evaluating $\frac{1}{1-0.4-0.3} = \frac{10}{3} = \sum_{n \in \mathbb{N}} (0.7)^n = \sum_{n \in \mathbb{N}} (0.4 + 0.3)^n$.
