Can an infinite sum of irrational numbers be rational? Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational.
Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear combination of the other ?
By linear combination, we mean there exists some rational numbers $u,v$ such that $a_i = ua_j + v$.

Taking
$a_k = \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}}$
we have
$\sum_{k=1}^\infty a_k = (1-\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}})+(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}) +\dots = 1$

As per Mario Carneiro's suggestion in the comments, let us instead take
$a_k = \frac{1}{\sqrt{p_{k-1}}} - \frac{1}{\sqrt{p_{k}}}$, 
where
$p_0 = 1$ and $p_k$ for $k > 0$ is the k-th prime number ($p_1 = 2$, $p_2 = 3$, $p_3 = 5$, $p_4 = 7$ etc.), so
$\sum_{k=1}^\infty a_k = (1-\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}})+(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}) +(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}) +\dots = 1$
