3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.