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Yesterday I came with a question: if rational numbers are countable, that means that all rational numbers between 0 and 1 can be listed in a sequence. Let be $Q(n)$ that sequence. It is pretty clear that $\sum_{n=1}^{\infty}Q(n) >\sum_{n=1}^{\infty}\frac{1}{n}$, it diverges. But what about $\sum_{n=1}^{\infty}Q(n)^2$? Does this serie converge? Is there even a way to define $Q(n)$ in a precise way?

Many thanks in advance!!

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    Hint: there are infinitely many such rationals which are greater than $\frac 12$. – 2017-02-05
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    Note: I doubt you mean "sum of reciprocals of all rationals...". Surely the word "reciprocals" doesn't belong there. – 2017-02-05
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    That sequence does not converge to $0$. Infinitely many rationals have a square larger than $1/4$ (all those larger than $1/2$). – 2017-02-05
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    Many thanks!! I've clearly seen what you're talking about now! Thanks!! – 2017-02-05

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I will prove a little more general statement:

For all $\varepsilon>0$, the sum of all squared rational numbers between $0$ and $\varepsilon$ diverges.

Let's fix $\varepsilon >0$.

Let's denote by $(Q_{\varepsilon}(n))$ an enumeration of the rationnals between $0$ and $\varepsilon$.

Since $[\varepsilon/2,\varepsilon]\cap \mathbb Q$ is infinite (because $\varepsilon>0$), we can extract an infinite sub-sequence $(Q'_{\varepsilon}(n))$ of $(Q_{\varepsilon}(n))$ by conserving only the $Q_{\varepsilon}(n)$ such that $Q'_{\varepsilon}(n)\geqslant \frac{\varepsilon}2$.

We then have:

$$\sum_{n\in \mathbb N} Q_{\varepsilon}(n)^2\geqslant\sum_{n\in \mathbb N} Q'_{\varepsilon}(n)^2\geqslant \sum_{n\in \mathbb N} \frac{\varepsilon^2}4=+\infty.$$

So the original series diverges, and you can deduce your result from the case $\varepsilon =1$.