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If $A\subset \mathbb{N}$ is large, that is, $\displaystyle\sum_{n\in A} \frac{1}{n}$ diverges, then does $\displaystyle\sum_{n\in A} \frac{\vert \sin n \vert }{n}$ diverge also?

I know that $\displaystyle\sum_{n\in \mathbb{N}} \frac{\vert \sin n \vert }{n}$ diverges. For example, see here. So this deals with the case $A=\mathbb{N}.$

I guess my question has to do with the irrationality measure of $\pi$ also.

Or is there some standard convergence test which can help here?

Adam Rubinson
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  • What does "large" mean? – Lee Mosher Nov 26 '23 at 16:38
  • @LeeMosher It is standard terminology which I also explained what it means in the blockquote part of the question. – Adam Rubinson Nov 26 '23 at 16:42
  • If $A$ is the set of primes, the series $$\sum_{p} \frac{\vert \sin p \vert }{p}$$ diverges as we can show by using the prime number theorem with a good remainder, so even for sparser large sets we get divergence (also clearly the series diverges if the set has positive density); on the other hand it may be possible to use the uniform distribution of $n/\pi$ modulo $1$ to construct large but sparse sets (zero density like the primes) where the series is not that large, though it may still diverge – Conrad Nov 26 '23 at 16:42
  • @Conrad ok but we can try for better $A$ than the set of primes. I don't see why that would be a good attempt at a counter-example to my question. – Adam Rubinson Nov 26 '23 at 16:44
  • Ah, got it. Your "that is" language was a bit abbreviated and threw me off. – Lee Mosher Nov 26 '23 at 16:51

1 Answers1

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I believe that there does exist $A\subset\mathbb{N}$ such that $\displaystyle\sum_{n\in A}\frac{1}{n}$ diverges but $\displaystyle\sum_{n\in A}\frac{|\sin(n)|}{n}$ converges. For example, consider the set $A=\{n:|\sin(n)|<\ln(n)^{-1},n\in\mathbb{N}\}$, then $A$ contains roughly $\frac{n}{\pi\ln(n)}$ elements $<n$. Thus, $\displaystyle\sum_{n\in A}\frac{1}{n}\approx\sum_{n=2}^\infty\frac{1}{\pi n\ln(n)}$ which diverges, then since we know that for all $n\in A$ we have $|\sin(n)|<\ln(n)^{-1}$, we can also say that $\displaystyle\sum_{n\in A}\frac{|\sin(n)|}{n}\lesssim\sum_{n\in A}\frac{1}{n\ln(n)}\approx\sum_{n=2}^\infty\frac{1}{n\ln(n)^2}$ which converges. I don't have a rigorous proof, in particular I'm not sure how to go about showing how frequently $|\sin(n)|<\ln(n)^{-1}$ occurs, but the idea seems pretty convincing.

Edit: The intuition for $A$ containing roughly $\frac{n}{\pi\ln(n)}$ elements is that $|\sin(n)|<\ln(n)^{-1}$ around where $|\sin(n)|$ is near $0$, or when $n$ is near a multiple of $\pi$, and since the first order of the series expansion of $|\sin(n)|$ around any of there zeros is $n$, the width of the interval for which $|\sin(n)|<\ln(n)^{-1}$ will be about $2\ln(n)^{-1}$ for small $\ln(n)^{-1}$. Then, there will be on average about one such interval in any interval of length $\pi$, so the percentage of reals in these ranges will be about $2(\pi\ln(n))^{-1}$. Now these intervals do not always contain an integer, but since $\pi$ is irrational, the centers of the intervals will get arbitrarily close to integers, and it can be shown using irrationality measure that intervals of width $\ln(n)^{-1}$ spaced at any constant irrational distance $\alpha$ apart will intersect with infinitely many integers. Additionally, since $\ln(n)^{-1}$ decreases so "slowly", the intervals will tend to contain integers at a rate of about $\ln(c)^{-1}$ for $c<n$ for a while, say until $n/c=e$, so showing this rigorously could be enough for the proof.

For now I'll consider this a work in progress, and I'll update with proofs if I figure them out. Any thoughts/critiques/ideas are appreciated.

David Chew
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