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For $n\geq 1$, let:

$$ a_n = \text{min} \lbrace{|\sin(k)|: 1\leq k\leq n} \rbrace $$

So that $a_1=\sin(1)$,$a_2=\sin(1)$,$a_3=\sin(3)$,$a_4=\sin(3)$, $a_5=\sin(3)$ and so on.

And let: $$ S_n = \sum_{k=1}^n a_k $$

The questions:

1-Does $a_n$ converge? (yes, proven in a comment by Xander Henderson)

2-What is the limit of $a_n$ as $n$ goes to infinity? (proven to be zero by Matt F. in his answer)

3-Does $S_n$ converge? (Still open)

I believe this can be related to the precision of rational approximations of $\pi$ because for some integer $a$ there exists $b\in]0,\pi[$ such that:

$$ n= a\pi+b $$ Then: $$ |\sin(n)|=|\sin(a\pi+b)| = |\sin(b)| $$ And: $$ \pi = \frac{n}{a}-\frac{b}{a} $$

So $n/a$ is a rational approximation of $\pi$ with error smaller than $\pi/a$ in absolute value. But since $b$ is in the interval ]0,\pi[ (cannot be zero because $\pi$ is irrational), then the sequence is basically the value of the smallest $b$ found for $k\leq n$.

My guess is that the sequence converges to zero, even if it never reaches zero (just like a geometric progression). I would also believe the series is convergent, but these would depend on how fast the accuracy of the rational approximations to $\pi$ grows with respect to their denominator.

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    $\begingroup$ $a_n$ is monotonically decreasing and bounded below by zero. It must converge. Indeed, it converges to zero, though it takes some work to show this. $\endgroup$
    – Xander Henderson
    Commented Mar 23, 2018 at 17:32
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    $\begingroup$ Your question 3 is almost certainly unsolved, since it relates closely to the (unknown) irrationality measure of $\pi$. A proof that $S_n$ was divergent would also prove that $\pi$ had irrationality measure $2$, so we certainly don't have that. A proof that $S_n$ was convergent technically wouldn't imply anything about the irrationality measure of $\pi$, but still seems unlikely to exist to me. $\endgroup$
    – Micah
    Commented Mar 23, 2018 at 17:51
  • $\begingroup$ As others have said, the answer to 1 and 2 are fairly simple. As for 3, I think an easier question would be to ask about the sum of the unique values of $a_n$ (of which the first four are $\sin(1),\sin(2),\sin(3)$ and then $|\sin(22)|$) $\endgroup$
    – Ian
    Commented Mar 26, 2018 at 12:30
  • $\begingroup$ @Ian that's an interesting suggestion. But if your proposed sum were to diverge, then so would $S_n$, however, if your proposed sum were to converge, nothing could be said about $S_n$, it adds information, but is not the same the question. $\endgroup$
    – Mefitico
    Commented Mar 26, 2018 at 12:37
  • $\begingroup$ Yes, my suggestion is not equivalent to this one. $\endgroup$
    – Ian
    Commented Mar 26, 2018 at 13:02

1 Answer 1

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  1. Yes.

  2. The $a_n$ converge to $0$. Proof: Let $a = \liminf a_n$. If $a>0$ then the integers mod $\pi$ would have density $0$ in the interval $(0, \arcsin(a))$, contradicting equidistribution.

  3. The $S_n$ probably do not converge, using an analogy that analytic number theorists can probably make rigorous. Consider $X_i$ equidistributed on $[0, \pi]$, which is the same as the limiting distribution of the integers mod $\pi$. Let $$a’_n= \min(\{|\sin(X_i)|: 1\le i \le n\}),$$ $$E[a’_n] \sim \sin\left(\frac\pi{2(n+1)}\right),$$ since the expected quantile of the minimum is $1/(n+1)$. The expected sum of these $a’_n$ does not converge, and by analogy, $S_n$ should not converge either.

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  • $\begingroup$ Dear @Matt F. That's an interesting approach, which does add some light to the problem, though I guess you know it is not conclusive. I'd like to ask for a clarification on point 2: Equidistribution says the probability density of $n \pi \, \text{mod} \, 1$ is constant (and uniform). But the sequence I propose speaks about $n \,\text{mod} \, \pi$. How do link those two cases? $\endgroup$
    – Mefitico
    Commented Mar 26, 2018 at 18:16
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    $\begingroup$ @Mefitico Equidistribution says $nb \mod 1$ is equidistributed for any irrational $b$. So take $b=1/\pi$; then $n/\pi \mod 1$ is equidistributed, and this is just a rescaling of $n \mod \pi$. $\endgroup$
    – user210229
    Commented Mar 26, 2018 at 18:32
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    $\begingroup$ I've awarded the bonus to this answer which indeed has valuable insight, although the original problem still lacks a definitive solution. $\endgroup$
    – Mefitico
    Commented Apr 2, 2018 at 18:50

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