# Sums of infinite algebraic numbers that are $\mathbb{Q}$-linearly independent

While working with infinite sums, I thought of the following problem

Consider the sequence $$\{a_n\}_{n=1}^{\infty}$$ of algebraic numbers that are $$\mathbb{Q}$$-linearly independent. Is it possible that $$\sum_{i=1}^{\infty}a_i\in\mathbb{Q}$$provided that the infinite sum converges.

Clearly, the above statement wont work if all $$a_i$$ were transcendental. Because in that case we would have consider $$a_i$$ to be the coefficient of $$x^{1+2i}$$ in the Taylor expansion of $$\sin(\pi/2)$$. Since in this case, all $$a_i$$ would be transcendental and $$\mathbb{Q}$$-linearly independent (because $$a_i$$ would be of the form $$\mathbb{Q}\pi^{1+2i}$$). So what can we say about the sequence of algebraic numbers? Is the statement written in yellow box has some counterexample?

Pick $$(b_n)_{n\geq 0}$$ a sequence of $$\mathbb{Q}$$-linearly independent algebraic numbers. Note that for $$0 \neq c_n\in \mathbb{Q}$$ we still have that $$(c_n \cdot b_n)_{n\geq 0}$$ is a sequence of $$\mathbb{Q}$$-linearly independent algebraic numbers. Hence, for any rational number $$q$$ we can find a sequence $$(a_n)_{n\geq 0}= (c_n \cdot b_n)_{n\geq 0}$$ of $$\mathbb{Q}$$-linearly independent algebraic numbers such that $$\sum_{n\geq 0} a_n = q.$$