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I imagine that, due to the fact that the average value of $\sin x$ in the interval $[0,\pi]$ is $\frac{1}{\pi-0}\displaystyle\int_{0}^{\pi} \sin x dx = \frac{2}{\pi},$ that the sequence

$$ a_n:= \sum_{k=1}^{n} \vert \sin k \vert - \left(\frac{2}{\pi}\right) n$$

is bounded, and that this can be proved by equidistributional theorems, and/or by utilising Riemann integration. I think the true statement, $\left\{n\bmod \pi: n\in\mathbb{N} \right\}$ is equidistributed modulo $\pi,$ is not strong enough to help in the proof (although I might be wrong about this).

Also, note that $\frac{a_n}{n}\to0,$ which may (or may not) be the case, but even if true, it $ \not\implies a_n$ is bounded.

So, is the sequence $(a_n)$ bounded, and if so, how can it be proved formally?

Maybe Birkhoff means? Although I am not that familiar with this.

I have calculated the first $11$ terms of $a_n$ and the absolute value of all terms are less than $\frac{1}{2},$ with no illuminating patterns yet. Edit: Calvin Khor has kindly posted an image of the graph of $a_n$ against $n$ in the comments. Indeed, it appears to be bounded.

The generalised version of this question:

If $f:[0,1]\to\mathbb{R}$ is continuous and has average value $M:=\frac{1}{1-0}\displaystyle\int_{0}^{1} f(x) dx,\ $ and $\alpha$ is irrational, then is the sequence

$$ a_n:= \sum_{k=1}^{n} f(\alpha k\pmod 1) - M n$$

bounded?

Obviously if you can prove the generalised version, then that would affirm the above instance where $f=\sin.$

Adam Rubinson
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    Nitpicking: For the special case you would choose $f(x) = \sin (\pi x)$, not $f=\sin$. – Try bmod instead of pmod for a better display. – Martin R Nov 06 '24 at 23:08
  • @MartinR ok done. Thanks – Adam Rubinson Nov 06 '24 at 23:10
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    looks bounded (N=1000) https://i.sstatic.net/26q5qDcM.png – Calvin Khor Nov 06 '24 at 23:16
  • The wavy hair comb graph. – Adam Rubinson Nov 06 '24 at 23:19
  • Maybe the Fourier series of $|\sin|$ can be useful: $$ |\sin (k)|={2\over\pi}-{4\over\pi}\sum_{m=1}^\infty{\cos(2m k)\over 4m^2-1} $$ and thus $$ \sum_{k=1}^{n} \vert \sin (k) \vert - \left(\frac{2}{\pi}\right) n = -{4\over\pi}\sum_{k=1}^{n}\sum_{m=1}^\infty{\cos(2m k)\over 4m^2-1} = -{4\over\pi}\sum_{m=1}^\infty{ \cos\left({(n+1)m}\right))\over 4m^2-1}{\sin(nm) \over \sin(m)}. $$ – user23571113 Nov 07 '24 at 02:43
  • A simily question for $\cos$ was asked by @ Ethan Splaver in the old post https://math.stackexchange.com/questions/650403/which-methods-different-than-the-natural-one-can-one-devise-to-confirm-that-the?noredirect=1 – Riemann Nov 07 '24 at 03:17
  • Here is a proof given by @Jack D'Aurizio of $$\sum_{k=1}^{n}\left|\cos k\right|=\frac{2n}{\pi}+O(1).$$ – Riemann Nov 07 '24 at 03:59
  • Up to $n=13$, $|a_n|<0.5$. But for $n=14$, $a_{14}=0.524176$. It seems that an upper bound is $0.684$ – Claude Leibovici Nov 07 '24 at 06:48
  • @Riemann I think you forgot to put the link to the proof by Jack D'Aurizio in your second comment. But thanks for the link in your first comment. – Adam Rubinson Nov 07 '24 at 12:33
  • @Riemann I assume you know but I am just making it clear for readers: Ethan's proof only manages to show $\sum_{k=1}^n |\cos k| = \frac{2n}{\pi} + o(n)$ – Calvin Khor Nov 07 '24 at 21:09
  • Oh, so your second comment was continuing from your first. – Adam Rubinson Nov 07 '24 at 22:00
  • @ Adam Rubinson. Yes, I forgot to put the link ,Here is the link https://math.stackexchange.com/questions/2930375/proving-sum-k-1n-vert-cosk-vert-ge-fracn4-for-all-n0?noredirect=1 – Riemann Nov 07 '24 at 23:35
  • @Riemann the proof linked is flawed since it implicitly assumes the irrationality degree of pi smaller than $3$ which is unknown as $(m^2d(m,\pi \mathbb Z))^{-1}$ doesn't converge to zero if the irrationality degree of pi is at least $3+\epsilon$ so the respective series in the answer cannot converge as claimed; note that the argument involving the sparseness of such $m$ and their growth works only if the general term goes to zero faster than some power of $m$ – Conrad Nov 07 '24 at 23:59

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