A convergence problem going around Twitter

Ten days ago, Fermat’s library posted a tweet saying that it is unknown whether the sum

\sum_{n=1}^\infty \frac{1}{n^3 \sin(n)^2}

converges or diverges, stirring up a lot of discussion. Sam Walters has been part of this discussion and pointed to a paper that says this is known as the Flint Hills series.

My first thought was to replace the sine term with a random variable and see what happens, because the values of n mod 2π act a lot like a random sequence. To be precise, the series is ergodic.

If X is a uniform random variable on the interval [0, 2π], then the random variable Y = 1/sin(X)² is fat tailed, so fat that it has no finite expected value. If Y had a finite expected value E[Y], then one might expect the Flint Hills sum to be roughly E[Y] ζ(3), i.e. the Flint Hills sum with the sine terms replaced by E[Y]. But since E[Y] diverges, this suggests that the Flint Hills series diverges.

Of course this doesn’t settle the question because the values of n mod 2π are not actually random. They simply act as if they were random in some contexts. For example, if you wanted to use them as if they were random values in order to do Monte Carlo integration, they would work just fine.

The question is whether the values act sufficiently like random values in the context of the Flint Hills problem. This is not clear, and is a problem in number theory rather than in probability. (Though number theory and probability are surprisingly interconnected.)

Sam Walters suggested considering a variation on the Flint Hills problem where we replace sin(n) with sin(2πθn) for some constant θ, the original problem corresponding to θ = 1/2π. I suspect the series diverges for some (almost all?) values of θ. That is, unless you pick θ very carefully, number theory won’t rescue the series from divergence.

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4 thoughts on “A convergence problem going around Twitter

  1. This seems to be closely connected to rational approximations to pi, which I know John has posted about a few times. That is, if m/n is close to pi, then sin(m) is approximately m – n pi, or n times the error in the rational approximation. With this in mind, I conjecture that it’s possible to formulate this problem in terms of the continued fraction representation of pi (or theta more generally). It might be easier to show that the series diverges for almost all theta than to show that it diverges for a particular theta, much like showing that a number is transcendental.

  2. The series converges to 30.3145… and it might exist a relationship between it and the Bernoulli numbers, specially B4=-1/30 and B8=-1/30. The method to solve it is based on an old formula of the 90’s that deal with a particular summation of the Riemann zeta function and is based on a logarithm, the acceptance of the method depends of a current revision. But yes, I can achieve the expected value of 30.3145 and there is no any other jump after that.
    For the Cookson Hills convergence is 42.99…
    I need to wait for the approval of the paper but the method is clear although discovering the criterion has been the most difficult part,
    regards
    Carlos L

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