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

With the convention $0 \not\in \mathbb{N}$, does there exist a bijection $f: \mathbb{N} \to \mathbb{N}$ such that $$\sum_{n = 1}^\infty \frac{1}{n f(n)} \in \mathbb{Q}?$$ I have tried various telescoping sums (e.g. $f(n) = \begin{cases} n + 1 && \text{if $n$ odd}\\ n - 1 && \text{if $n$ even} \end{cases}$), and get irrational values which are difficult to cancel out with other series and maintain bijectivity.

Relaxing the constraints to injections gives the simple $f(n) = n + 1$ yielding a telescoping sum: $$\sum_{n = 1}^\infty \frac{1}{n(n+1)} = 1$$ If instead we just wanted a surjection, we could go for $f(1) = 1, f(n) = n - 1$ for $n \geq 2$, which yields $$1 + \sum_{n = 2}^\infty \frac{1}{n(n-1)} = 2$$ Unfortunately you can’t just cut the difference between the two to get a bijection here.

Sharky Kesa
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    In this context it may be interesting that for any bijection $f: \Bbb N \to \Bbb N$ the series $\sum_{n=1}^\infty \frac{1}{n f(n)}$ is convergent, see for example https://math.stackexchange.com/q/622604/42969 . – Martin R Jan 04 '25 at 08:12
  • @MartinR Indeed, you can prove that by rearrangement inequality on the finite sums and taking to infinity – Sharky Kesa Jan 04 '25 at 08:46
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    I believe some algorithm like the following should produce a series summing to $1$ (for example): define $f(n)$ recursively by greedily letting $f(n)$ be the smallest “valid” positive integer not in ${f(1),\dots,f(n-1)}$ if $n$ is odd, or just the smallest positive integer not in ${f(1),\dots,f(n-1)}$ if $n$ is even. Here “valid” when $n$ is odd means that $1-\sum_{j=1}^n 1/jf(j)$ is strictly greater than the largest possible value of $\sum_{i=(n+1)/2}^\infty 1/(2i\cdot f(2i))$ that remains possible at that moment. (continued...) – Greg Martin Jan 04 '25 at 09:16
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    (...continued) This strategy separates the need to have the series have a particular sum (which is borne by the values of $f(n)$ with $n$ odd) from the need to have $f$ be a bijection (which is borne by the values of $f(n)$ with $n$ even). It clearly produces a bijection $f$ and a series $\sum_{n=1}^\infty 1/nf(n)$ that converges to a value that is at most $1$; the remaining part is to prove that the greedy assignments for $n$ odd force the value to be exactly $1$. – Greg Martin Jan 04 '25 at 09:17

1 Answers1

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I believe the comments by Greg Martin give us a way to converge our sum to any number in some range, rational or not, but let's think the other way. Simple questions need simple answers.

A simple answer would be

$$f(n)=\begin{cases} 2, & \text{for }n=1 \\ n+2,& \text{for even } n\\ n-2,& \text{for odd } n>1\\ \end{cases}$$

Motivation: there are two nice telescoping sums mentioned in the question itself, which are, however, one off from being a bijection. Well, then why don't we split $\mathbb N$ in two halves and use different almost-bijections on both, so they can fix each other's problems?

The sum would obviously be $$\sum_{n=1}^\infty{1\over n\cdot f(n)}=\frac12+\sum_{i=1}^\infty{1\over\underbrace{2i\strut}_{\small\text{that's }n}\underbrace{(2i+2)}_{\small\text{that's }f(n)}}+\sum_{i=1}^\infty{1\over\underbrace{(2i+1)}_{\small\text{that's }n}\underbrace{(2i-1)}_{\small\text{that's }f(n)}} = \\ =\frac12+\frac12\left(\frac12-\frac14+\frac14-\frac16+\dots\right)+\frac12\left(1-\frac13+\frac13-\frac15+\frac15-\frac17+\dots\right) = \\ =\color{red}{5\over4}$$

So it goes.

Ivan Neretin
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