For which $f: \mathbb{Q}\rightarrow\mathbb{Q}^+$ does the sum $\sum_{q\in\mathbb{Q}} f(q)$ converge?

Question

BACKGROUND:

This is not a homework question -- I am not even in school. I am purely interested in the question itself. I actually asked this question in my second semester of real analysis during my math major, but that was years ago. The professor found the question interesting but was unable to provide a satisfying answer.

I know the sum $\sum_{q\in\mathbb{Q}\cap[0,\varepsilon)}q$ diverges to $\infty$ (since infinitely many terms are larger than $\frac{\varepsilon}{2}$). But $\mathbb{Q}$ is countable, and so there must be some "sums over the rationals" which converge.

Of course we need to avoid the ambiguity of unordered infinite series with positive and negative terms (since rearrangement of terms does not preserve the sum). So I am only interested in sums of positive terms, i.e. $\sum_{q\in\mathbb{Q}}f(q)$ where $f:\mathbb{Q}\rightarrow\mathbb{Q}^+$.

Intuitively, the density of $\mathbb{Q}$ makes convergence a lot harder to achieve for an arbitrary sum. We won't have success with something like $\sum_{q\in\mathbb{Q}\backslash\{0\}}\frac{1}{q^2}$, since there will still be too many terms that are bigger than any $\varepsilon>0$, in this case for all $q\in \mathbb{Q}^{\neq0}\cap\big(-\frac{1}{\sqrt{\varepsilon}},\frac{1}{\sqrt{\varepsilon}}\big)$. This seems to suggest that functions that have a simple formula for $f(q)$, written in terms of $q$, will rarely (almost never?) have a convergent sum.

So to make my question a bit more specific and therefore easier to answer, I want to focus my question only on the following example.

MY ACTUAL QUESTION:

Any rational number can be uniquely described as a fraction in lowest terms. For which positive real numbers $a$ and $b$ does the following sum converge?

$$\sum_{\frac{m}{n} \in \mathbb{Q}} \frac{1}{|m|^a+|n|^b}$$

Intuitively, if we use powers of $1$, then this fraction might--like the harmonic series--not shrink to zero quickly enough for the sum to converge. To further build intuition, I will informally define the "${l}^a$ complexity" of rational numbers to be $$\bigg|\frac{m}{n}\bigg|_a=\big(|m|^a+|n|^a\big)^{\frac{1}{a}}$$

To illustrate, $\big|\frac{-3}{4}\big|_2=(3^2+4^2)^\frac{1}{2}=\sqrt{25}=5$, but $\big|\frac{-3}{4}\big|_1=(3^1+4^1)^\frac{1}{1}=7$. To avoid ambiguity, we will choose the canonical representation of $0$ to be the fraction $\frac{0}{1}$, so that $|0|_a=(0^a+1^a)^\frac{1}{a}=1$. Since $0=\frac{0}{1}$ is the rational number written with the smallest possible numerator and denominator, we see that $|q|_a \geq 1$ for any $q\in\mathbb{Q}$. In general, "more complicated" rational numbers will have larger $l^a$ complexity than "simpler" rational numbers: $\big|\frac{456}{1207}\big|_a>\big|\frac{1}{2}\big|_a$.

This leads to an even smaller and more restrictive question.

A SIMPLER QUESTION THAT MIGHT HELP ME MAKE PROGRESS:

For which $a>1$ does the following sum converge?

$$\sum_{q\in\mathbb{Q}} \frac{1}{|q|_a^a}$$

Answer 1

To the last question: The answer is that it converges iff $a>2$. It is easy to check that your question is equivalent to asking for which $a >1$ the following sum converges: $$ \sum_{m,n \in \mathbb{N} \atop\text{ relative prime }} \frac{1}{n^a+m^a}. $$ The cruical step it to get rid of the "relative prime" here, i.e. show that $$ \sum_{m,n \in \mathbb{N} } \frac{1}{n^a+m^a} < \infty \iff \sum_{m,n \in \mathbb{N} \atop\text{ relative prime }} \frac{1}{n^a+m^a} < \infty $$ The direct implication is trivial because the left sum is bigger than the right. For the reverse implication assume the right sum to be finite and denote $\text{gcd}(m,n)$ the greatest common divisor of $m$ and $n$. Note that \begin{align} \sum_{m,n \in \mathbb{N} } \frac{1}{n^a+m^a} &= \sum_{k=1}^\infty \sum_{m,n \in \mathbb{N} \atop\text{gcd}(m,n)=k} \frac{1}{n^a+m^a} = \sum_{k=1}^\infty \sum_{m,n \in \mathbb{N} \atop\text{ relative prime }} \frac{1}{(kn)^a+(km)^a}\\ &=\sum_{k=1}^\infty \frac{1}{k^a} \sum_{m,n \in \mathbb{N} \atop\text{ relative prime }} \frac{1}{n^a+m^a} < \infty. \end{align}

Now, it is easy to check that $\sum_{m,n \in \mathbb{N} } \frac{1}{n^a+m^a} < \infty $, iff the integral $$ \int_{\mathbb{R}^2\setminus B(0,1)} \frac{1}{||x||^a} dx $$ is finite. ($B(0,1)$ denotes the unit ball). This is known to be true, iff $a>2$.

Answer 2

You wrote: "This seems to suggest that functions that have a simple formula for $f(q)$, written in terms of $q$, will rarely (almost never?) have a convergent sum."

This is in some sense true, because one can prove the following rigorous statement:

Let $f : \mathbb{Q} \to [0,\infty)$ be continuous s.t. $\sum_{q \in \mathbb{Q}} f(q)$ is convergent. Then $f=0$.

Proof: Assume that $f$ is not constant zero. Then exitsis $q_0$ s.t. $f(q_0)>0$. By continuity there is some $\epsilon>0$ s.t. for all $q \in (q_0-\epsilon,q_0+\epsilon)$ we have $f(q)>f(q_0)/2$. Since $(q_0-\epsilon,q_0+\epsilon)$ contains infintely many elements, that implies that the sum diverges.

Remark: Clearly, there are discontinuous non-zero functions $f : \mathbb{Q} \to [0,\infty)$ s.t. $\sum_{q \in \mathbb{Q}} f(q)$ converges. For example: Let $\Phi: \mathbb{N} \to \mathbb{Q}$ a bijection. Let $f(q):= 2^{-\Phi^{-1}(q)}$, then $\sum_{q \in \mathbb{Q}} f(q) = \sum_{n \in \mathbb{N}}2^{-n}=1$.