Happy holidays!
So suppose that I have a finite order polynomial expressed as: $$f(x)=\sum_{i=0}^n a_ix^i$$ where $a_i \in \mathbb{R}$. The roots of $f(x)$ can be real or complex. Can $f(x)$ be represented as a ratio of polynomials of infinite order, i.e, $$f(x) = \frac{\sum_{i=0}^\infty c_ix^i}{\sum_{i=0}^\infty b_ix^i}$$ with $\{b_i,c_i\} \in \mathbb{R}$?
You are essentially asking that whether for a given a finite order polynomial $f(x)=\sum_{i=0}^n a_ix^i$ there exists infinite polynomial sums $(\sum_{i=0}^\infty b_ix^i) \neq 0$ and $(\sum_{i=0}^\infty c_ix^i)$ such that $$f(x)(\sum_{i=0}^\infty b_ix^i) = (\sum_{i=0}^n a_ix^i)(\sum_{i=0}^\infty b_ix^i) = \sum_{i=0}^\infty c_ix^i$$ or not, where $a_i, b_i,c_i \in \mathbb{R}$.
It obviously does. Take any non-zero infinite order polynomial $b(x) = \sum_{i=0}^\infty b_ix^i$. Then, $$f(x)(\sum_{i=0}^\infty b_ix^i) = (\sum_{i=0}^n a_ix^i)(\sum_{i=0}^\infty b_ix^i) \\ = a_0b_0 + (a_0b_1+a_1b_0)x + (a_0b_2+a_1b_1+a_2b_0)x^2 + \dots + (\sum_{k=0}^n a_kb_{n-k})x^n + \dots \\= \sum_{i=0}^\infty c_ix^i \ \ \text{where,} \ c_n = \sum_{k=0}^n a_kb_{n-k} \in R$$
$$\text{So,} \ \ \ \fbox{$f(x) = \frac{\sum_{i=0}^\infty c_ix^i}{\sum_{i=0}^\infty b_ix^i}$} \ \ \text{with} \ b_i,c_i \in \mathbb{R}.$$