Consider the set $A = \{a : \sum_{n=0}^m c_n a^n =0 \}$, for some m, Where the coefficients are any sequence of rational numbers (i.e the set of algebraic numbers). Does there exist a function, $f$ such that $$\sum_{a \in A} f(a) < \infty$$ Excluding trivial examples such as the 0 function.
I tried constructing one as such: Let $f(x)$ be defined by saying$$\begin{cases} x^2 & 0\leq x \leq 1(A^*) \\ \frac{1}{x^2} & 1\leq x (A^+) \end{cases} $$
So $$\sum_{a \in A} f(a) = \sum_{a \in A^*} a^2 + \sum_{a \in A^+} \frac{1}{a^2}$$ But surly $$\sum_{a \in A^*} a^2 + \sum_{a \in A^+} \frac{1}{a^2}<\int_{0}^{1} x^2 \,dx + \int_{1}^{\infty} \frac{1}{x^2} \,dx =\frac{4}{3}$$ which would mean the sum converges.
But $$\frac{4}{3} < \frac{\pi ^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2}$$
Which implies what is (to me) a paradoxial result $$\sum_{a \in A^+} \frac{1}{a^2} < \sum_{n=1}^{\infty} \frac{1}{n^2}.$$ Would this mean that $$\sum_{a \in A^+ - \mathbb{N}} \frac{1}{a^2} < 0?$$
Regardless, does anyone have any input?
NOTE: $A$ is countable.
$A$ is countable, so there exists a bijection $\,\varphi : A \to \mathbb{N}\,$. Let $g$ be any function such that the series $\displaystyle\sum_{n \ge 1} g(n)$ converges absolutely, and let $f(a) = g(\varphi(a))$, then $\,\displaystyle \sum_{a \in A} f(a)\,$ converges absolutely, and in fact $\,\displaystyle \sum_{a \in A} f(a) = \sum_{n \ge 1} g(n)\,$ because absolutely convergent series can be arbitrarily rearranged.