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
$$ \sum_{n\geq 0} a_n = q.$$
