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I'm very confident that $$\sum_{p \ \text{prime}} \sin p $$

diverges. Of course, it suffices to show that there are arbitrarily large primes which are not in the set $\bigcup_{n \geq 1} (\pi n - \epsilon, \pi n + \epsilon)$ for sufficiently small $\epsilon$. More strongly, it seems that $\sin p$ for prime $p$ is dense in $[-1,1]$.

This problem doesn't seem that hard though. Here's something that (to me) seems harder.

If $p_n$ is the nth prime, what is $$\limsup_{n \to +\infty} \sum_{p \ \text{prime} \leq p_n} \sin p?$$ What is $$\sup_{n \in \mathbb{N}} \sum_{p \ \text{prime} \leq p_n} \sin p? $$

Of course, we can ask analogous questions for $\inf$.

I'm happy with partial answers or ideas. For example, merely an upper bound.

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    $\begingroup$ $\sum_{p < x} e^{2 i \pi\alpha p}$ has been well studied, try searching for 'exponential sums over primes' (for example this) $\endgroup$
    – reuns
    May 8, 2017 at 22:29
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    $\begingroup$ Finding supremum or infimum of the partial sums will not be easy. However, it is true that $\{\sin p \}$ is dense in $[-1,1]$. This is because $\{p \ \mathrm{mod} \ 2\pi \}$ is equidistributed in $[-\pi,\pi]$. $\endgroup$
    – Sungjin Kim
    May 9, 2017 at 1:16
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    $\begingroup$ More work on this can prove that $\sum_p \frac{\sin p}{p}$ converges. $\endgroup$
    – Sungjin Kim
    May 9, 2017 at 2:39
  • $\begingroup$ @i707107 That's interesting. How is it shown? Is this something you've proven independently, or is it a known result? $\endgroup$
    – MathematicsStudent1122
    May 10, 2017 at 6:50
  • $\begingroup$ @MathematicsStudent1122 I am not sure if it is known result. I found it while I was trying your question. But, it is shown from a well-known result by Vinogradov. $\endgroup$
    – Sungjin Kim
    May 10, 2017 at 7:13

1 Answer 1

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My answer here only includes partial results. First, we use Vinogradov's inequality:

Let $\alpha$ be a real number. If integers $a$ and $q$ satisfies $(a,q)=1$ and $$ \left| \alpha - \frac aq \right| \leq \frac 1{q^2}, $$ then $$ \sum_{n\leq N} \Lambda(n) e^{2\pi i \alpha n} = O\left( (Nq^{-1/2} +N^{4/5} + N^{1/2}q^{1/2} ) (\log N)^4 \right) $$

With an error of $O(N^{1/2+\epsilon})$, we obtain the same upper bound for $\sum_{p\leq N} (\log p) \cdot e^{2\pi i \alpha p}$.

Since we have finiteness of irrationality measure of $\pi$ : see this, we may use the continued fraction convergents $a/q$ for $\alpha = 1/(2\pi)$. Thus, it is possible to find a denominator $q$ of the continued fraction convergent of $1/(2\pi)$ such that $N^{1/7}<q<N^{99/100}$. Then Vinogradov inequality yields that there is $\delta>0$ such that $$ \sum_{p\leq N} (\log p)e^{i p} = O(N^{1-\delta}). $$

Now, partial summation gives for some $\delta>0$, $$ \sum_{p\leq N} e^{ip} = O(N^{1-\delta}). $$

Therefore, by taking imaginary parts, $$ \left|\sum_{p\leq N} \sin p \right| = O(N^{1-\delta}). $$ With this result and partial summation, we obtain that $$ \sum_{p \ \mathrm{prime} } \frac{\sin p}p $$ converges.

It will be possible to find the best $\delta>0$ in the upper bound: $$ \left|\sum_{p\leq N} \sin p \right| = O(N^{1-\delta}) $$ by using Vinogradov's inequality more efficiently.

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  • $\begingroup$ Just a comment for which isn't required a response. Maybe is interesting (I don't know what are potentially feasible since to prove that there are infinitely many twin primes is an unsolved problem) to state problems similar than OP's questions or conjectures to get/propose inequalities,bounds or asymptotics involving the sine function and the sequence of twin primes invoking Brun's theorem, or say us for example assuming the well-known conjecture for twin primes (the section Further results of the Wikipedia's article Brun's theorem). $\endgroup$
    – user243301
    Sep 7, 2018 at 12:12

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