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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.

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    $\begingroup$ 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 $b$). $\endgroup$ Commented May 29 at 11:28
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    $\begingroup$ Is there some $f$ such that $N(f)$ is algebraic? $\endgroup$
    – psl2Z
    Commented May 29 at 11:56
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    $\begingroup$ @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. $\endgroup$
    – Robin
    Commented May 29 at 12:22
  • $\begingroup$ 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 $g$) $\endgroup$
    – Sil
    Commented May 29 at 13:36
  • $\begingroup$ Another example is Bessel function $I_0(2)$. Proved transcendental by Siegel, 1929. (Irrationality is easier.) $\endgroup$
    – GEdgar
    Commented May 29 at 13:48

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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.

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