# Convergence of Trigonometric Dirichlet series

Can it be proved that the following series converges for some integer value of $s$?

$$\sum_{n=1}^\infty\frac{1}{n^s|\sin(n)|}$$

If so what value(s) of $s$ would it converge for?

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## 1 Answer

It certainly converges for $s>7.61$. Salikhov proved that $|\pi - p/q| \geq q^{-7.61}$ for all integers $(p,q)$ with $q$ sufficiently large. So $|q \pi - p| \geq q^{-6.61}$ and, since $p$ and $q$ must be basically proportional for this to be at all close to true, $|p - q \pi| \geq p^{-6.61}$.

For any $p$, choose $q$ to be the integer so that $(q-1/2) \pi < p < (q+1/2) \pi$. Since $|\sin x| \geq |x-q \pi| (2/\pi)$ for $x \in ((q-1/2) \pi, (q+1/2) \pi)$ (draw a graph), this shows that $|\sin p| \geq (2/\pi) p^{-6.61}$. We have $$\left| \frac{1}{p^s \sin(p)} \right| \leq p^{-s+6.61}$$ so this converges for $s > 7.61$.

The general assumption is that the truly order of irrationality of $\pi$ is $2 + \epsilon$, in which case this argument shows that the sum converges for $s > 2$.

I would actually guess that the sum converges all the way up to $s>1$, but I don't have an argument right now.

This question isn't going to have a truly easy answer. Let $a_i$ be a very rapidly growing sequence of integers (I'll decide how fast later.) Let $n_i = a_1 a_2 \cdots a_i$. Let $\theta = \pi \sum \frac{1}{n_i}$. I claim that we can make $\sum \frac{1}{n^s \sin (\theta n)}$ diverge for all $s$.

Observe that $\theta n_i$ is extremely close to an integer multiple of $\pi$: The error is essentially $\pi/a_{i+1}$. So $1/|\sin(\theta n_i)|$ is as large as $a_{i+1}/\pi$. If we choose $a_i$ to grow so fast that $\lim_{i \to \infty} a_{i+1}/n_i^s = \infty$ for any $s$, then the terms of $\sum 1/(n^s \sin(\theta n))$ won't go to $0$, so the sum won't converge.

So, in order to prove your result, you need to know that $1/\pi$ isn't extremely closely approximated by rationals. This is essentially the same as Salikhov's result (although it would be good enough to use Mahler's result, which is the same bound with the exponent $20$.)

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Thanks a lot for this I upvoted, what I really need is a very simple proof that there exists such an integer. It doesn't matter how large the integer is, or how good the bound is either. Do you know where I could find that? If you want I could create another question and accept your answer here also. –  Ethan Jan 27 '14 at 0:30
Give me a little time to dig around in the archives. I saw a question like this before, with a very clever answer... –  David Speyer Jan 27 '14 at 0:35