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.
Yes.
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.
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.