# Is there a $f: \mathbb{N} \to \mathbb{N}$ such that $\sum_{n=1}^{\infty} \frac{1}{n^2f(n)} \in \mathbb{Q}$?

Take by convention $$0 \not \in \mathbb{N}$$, and let $$f: \mathbb{N} \to \mathbb{N}$$. Define the real number $$N(f)$$ by

$$N(f) = \sum_{n=1}^{\infty} \frac{1}{n^2f(n)}.$$

$$N(f)$$ is well-defined because, for example, each summand is a positive number less than $$\frac{1}{n^2}$$, so $$N(f) \leq \frac{\pi^2}{6}$$, so we have a bounded above sequence of positive numbers.

My question: is $$N(f)$$ ever rational?

Note that if it is the case that $$N(f)$$ is never rational, we're almost certainly not going to be able to prove it in this thread. By taking $$f(n)=n$$ we obtain $$\zeta(3)$$, which was only proven to be irrational in the 1970s, and apparently "at least one of $$\zeta(5), \zeta(7), \zeta(11),\zeta(13)$$ must be irrational" (see link), so I assume the irrationality of these numbers individually is unknown.

But I'm wondering if I'm missing some $$f$$ for which it is not too hard show is a counterexample to the conjecture "$$N(f)$$ is never rational."

What I've tried thinking about is defining $$f$$ recursively, but not gotten far. If we say $$a_k, b_k$$ are defined such that $$\frac{a_k}{b_k} = \sum_{n=1}^{k} \frac{1}{n^2f(n)}$$ is in lowest terms, then we know

$$\frac{a_{k+1}}{b_{k+1}} = \frac{a_k}{b_k} + \frac{1}{(k+1)^2f(k+1)} = \frac{a_k (k+1)^2f(k+1) + b_k}{b_k (k+1)^2f(k+1)},$$

so in particular $$da_{k+1} = a_k (k+1)^2f(k+1) + b_k$$, $$db_{k+1} = b_k (k+1)^2f(k+1)$$, where $$d$$ is the gcd of the numerator and denominator. If we can pick $$f(k+1)$$ in such a way to make $$d$$ large, we can have the denominators of the partial sums grow slowly, which may at least be a starting point for looking for such an $$f$$, I'm not sure.

• Not super helpful, but I'm sharing my failed attempt: An easy way to force the result to be rational would be to make your $b_k$ be a constant $b$. However, this implies $k$ divides $b$ for all $k$, which is a contradiction. Consequently, you need $b_k \rightarrow \infty$ (else you could choose a suitable constant $b$). Commented May 29 at 11:28
• Is there some $f$ such that $N(f)$ is algebraic? Commented May 29 at 11:56
• @G.Fougeron I tried something like that too - like yours, my first attempt was to see if there was some way to bound $b_k$, but as you say that can’t be done. I’m not entirely sure that my suggestion to try give $b_k$ a slow growth rate would work anyways, as for example, the best approximants of $\varphi$ grow notoriously slow, and the denominators in the partial sums of $1 + \frac{1}{2} + \frac{1}{4} + \dots$ grow exponentially, yet this converges to an integer. A look at $a_k$, $b_k$ could possibly help, but I’m not sure what about them precisely is worth analysing. Commented May 29 at 12:22
• If we relax the condition to $f:\mathbb{N}\to \mathbb{Q}$ then there are examples such as $f(n)=\frac{(n+2)^2}{n+1}$, maybe even simpler. To the original problem, I have no solution but perhaps recursively choosing $f(k)$ such that some sort of condition like $\frac{3}{2}-\frac{1}{k}< \sum_{n=1}^{k} \frac{1}{n^2f(n)} < \frac{3}{2}+\frac{1}{k}$ would guarantee sum equal to $\frac{3}{2}$. Question of course is whether such choice is possible... (or in general $|\sum_{n=1}^{k} \frac{1}{n^2f(n)} - \frac{p}{q}|<1/g(k)$ for $\frac{p}{q}\in (1,\frac{\pi^2}{6})$ and some increasing function $g$)
– Sil
Commented May 29 at 13:36
• Another example is Bessel function $I_0(2)$. Proved transcendental by Siegel, 1929. (Irrationality is easier.) Commented May 29 at 13:48

Yes, there is an $$f$$ such that $$N(f)$$ is rational. Using $$\sum_{n=1}^\infty\frac1{n^2(n+1)}=\frac{\pi^2}6-1, \\ \sum_{n=1}^\infty\frac1{(2n+1)^2(2n-1)}=\frac34-\frac{\pi^2}{16}$$ (both are computed using partial fractions), we have a chance to cancel the $$\pi^2$$ stuff. Say, take $$f(n)=n+2$$ for $$n$$ even, $$f(n)=3n-6$$ for $$n>1$$ odd, and $$f(1)$$ arbitrary.