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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.

pisoir
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    If you can choose any rational coefficients you like, you should be able to make that series converge to whatever you want it to. – bof Jun 20 '21 at 08:06
  • Did you mean to sum over all square-free integers in the second sum? – TonyK Jun 20 '21 at 08:08
  • @TonyK I updated the question. $c_k \neq 0$. In the second sum I assumed all integers. I simply thought that in that case we could split the sum into sum of non-square-free integers and square-free, thus having rational number + irrational number = irrational number. Is that a wrong thinking? – pisoir Jun 20 '21 at 08:17

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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 $\sum_{k=1}^n c_kx_k\in I_n$.

TonyK
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  • Nice argument. Interesting how finite and infinite sum make difference. So also the second summation ($a_k=\sqrt{k}$ can converge to whatever I want... – pisoir Jun 20 '21 at 08:39