# $\sum_{k=1}^{n} \arcsin(\sin(k))= a_n+b_n \pi$, for $a_n,b_n\in\mathbb{Z}$; what can be said about the sequence $(a_n,b_n )$?

We have $$\sum_{k=1}^{n} \arcsin(\sin(k))= a_n+b_n \pi$$ for some integers $$a_n,b_n$$. I have several questions about the behavior of these numbers:

• For which $$n$$ does $$b_n=0$$ and $$a_n\ne 0$$, i.e. the sum is a non-zero integer?
• For which $$n$$ does $$(a_n,b_n)=(0,0)$$, i.e. the sum vanishes?
• Aside from these subsequences, based on numerical evidence I conjecture that $$\lim_{n\to\infty} a_n/b_n=-\pi$$, but I don't know how to prove it.

Recall the convergents of $$\pi$$ are $$3, 22/7, 333/106, 355/113, 103993/33102$$, etc. I computed $$(a_n,b_n)$$ for $$1\le n\le 120000$$ and here's what I found:

• As expected, I saw 'increased aberrations' for $$n$$ near the numerator of the numerators of the convergents. For example, $$b_n=0$$ and $$a_n\ne 0$$ when $$n=\{1,8,16,52,60,96,104,140,148,184,192,228,236,272,280,316,324,360,103632,103668,103676,103712,103720,103756,103764,103800,103808,103844,103852,103888,103896,103932,103940,103976,103984\}$$ and $$(a_n,b_n)=(0,0)$$ when $$n=\{24,44,68,88,112,132,156,176,200,220,244,264,288,308,332,352,103640,103660,103684,103704,10372 8,103748,103772,103792,103816,103836,103860,103880,103904,103924,103948,103968,103992\}.$$ In particular, when $$b_n=0$$ and $$a_n \ne0$$, aside from $$n=1$$ I found that $$a_n=2$$.
• Aside from these values, the ratio is quite close to $$\pi$$. We have $$a_{120000}/b_{120000} =-\frac{248162}{78993}$$, which differs from $$-\pi$$ by about $$0.0000231474$$. Here is a plot:

Any information or insight would be much appreciated. I've heard of several continued fractions for $$\pi$$ and was wondering if perhaps this has to do with that.

Update: I checked oeis.org for both the numerator and denominator and found nothing. Since $$\arcsin(\sin(k))=x_k+y_k \pi$$, $$x_k,y_k\in\mathbb{Z}$$, is essentially the signed distance between $$k$$ and its nearest multiple of $$\pi$$, it is is clear that $$x_k/y_k\approx -\pi$$. Then perhaps using the CLT or another probabilistic argument, one could examine the convergence of the random variable $$a_n/b_n$$.

• have you tried restricting to the subsequence $\{b_n\neq 0\}$ and then feeding $\{b_n\}$ to the inverse sequence encyclopedia? I feel like any problem involving some sort of irrationality measure or convergents of $\pi$ either end up having a trivial recurrence or completely unsolved. Very nice question though. Commented Dec 5, 2020 at 17:33
• What range for arcsin are you using? Is this basically $\sum_k \left(k\bmod 2\pi\right)$ ? Commented Dec 5, 2020 at 19:39
• @StevenStadnicki, I'm using $-\pi/2 \le \arcsin(x)\le \pi/2$. So for instance, $\arcsin(\sin(10))=3\pi -10$ and $\arcsin(\sin(13))=13-4\pi$. In particular, it's not as you suggest because for $n=24$ the sum vanishes but your version would give $300-72\pi$. Commented Dec 5, 2020 at 19:42
• Probably this can be useful: sequence $(k/(2\pi))$ is equidistributed modulo 1 (en.wikipedia.org/wiki/Equidistribution_theorem) and it follows that for any Riemann-integrable function the sum $\frac{1}{N}\sum_{k=1}^{N}f(\{k/2\pi\})$ tends to the $\int_{0}^{1}f(t)dt$. I guess this can help to approximate this sum, but not sure. Commented Dec 6, 2020 at 19:53
• Indeed, we can simply take $f(t)=\arcsin(\sin(2\pi t))$ and since $\int_{0}^{1}f(t)dt=0$ we have $(a_n+b_n\pi)/n\to 0$. It remains to prove that $a_n,b_n\sim n$ (up to constant). Commented Dec 6, 2020 at 20:00

This may help: $$\arcsin(\sin(x)) = |((x-\pi/2) \bmod (2\pi))-\pi|-\pi/2$$ or equivalently $$\arcsin(\sin(x)) = \left|x-3\pi/2 -2\pi\left\lfloor\frac{x-\pi/2}{2\pi}\right\rfloor\right|-\pi/2$$

If we write $$\arcsin(\sin(k)) = x_k + y_k \pi$$ with $$x_k, y_k \in \mathbb Z$$ we have $$x_k =\operatorname{sgn}(\cos(k))\,k$$ and $$y_k=-\operatorname{sgn}(\cos(k))\left(\frac{3}{2}+2\left\lfloor\frac{k-\pi/2}{2\pi}\right\rfloor\right)-\frac{1}{2}$$

• Thanks for getting a better hold of the terms; I updated the question with your notation. Commented Dec 6, 2020 at 20:33
• Did you find $y_k$ by trail and error method? @jjagmath Commented Nov 20, 2023 at 11:34
• @BobDobbs No, I got $y_k$ as $(\arcsin(\sin(k))- x_k)/\pi$ Commented Nov 20, 2023 at 11:53

This is very interesting !

For the fun of it, I computed the ratio's $$R_k=-\frac{a_{10^k}}{b_{10^k}}$$ and obtained the following sequence $$\left\{1,\frac{17}{6},3,\frac{22}{7},\frac{22}{7},\frac{931}{296},\frac{10559}{3361},\frac{1093611}{348107},\cdots\right\}$$ For the last one $$\frac{1093611}{348107}-\pi=1.74 \times 10^{-6}$$

• Not blazing fast convergence, but not terrible either :) Commented Dec 6, 2020 at 20:33