I am interested in the following problem. For $c_k\in \mathbb{Q}, c_k\neq 0$, and $a_k$ being distinct square-free integers, is the following (convergent) sum $$S=\sum_{k=1}^\infty c_k\sqrt{a_k}$$ an irrational number?
I found the following nice answer The sum of square roots of non-perfect squares is never integer answering my question for finite sum. My question can thus, in other words, be, does the argument there extend to the infinite series? I don't quite understand the proof there, but since it is done by induction, it should extend to $n\to\infty$, shouldn't it?
As a special case, is the following convergent sum ($c_k\in \mathbb{Q}, c_k \neq 0$) $$S=\sum_{k=1}^\infty c_k\sqrt{k}$$ an irrational number?
Any reference is also appreciated.
Given any $\xi\in\Bbb R$, and any sequence $(x_k)$ of non-zero real numbers (in your case, square-free integers), we can choose rational numbers $c_k$ so that $\sum_{k=1}^\infty c_kx_k=\xi$.
To see this, define the sequence of intervals $I_n=(\xi-\frac{1}{n},\xi+\frac{1}{n})$. Choose $c_1$ so that $c_1x_1\in I_1$, $c_2$ so that $c_1x_1+c_2x_2\in I_2$ etc. This is always possible if the $x_k$ are non-zero.
Then $\sum_{k=1}^n c_kx_k\in I_n$ for all $n$, so $|\sum_{k=1}^n c_kx_k-\xi|<\frac{1}{n}$, and so $\sum_{k=1}^\infty c_kx_k=\xi$.
Updated to add: Your updated question specifies that the $c_k$ must be non-zero. This changes nothing; we can always choose non-zero $c_n$ so that $\sum_{k=1}^n c_kx_k\in I_n$.