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I know $\sum_{k=0}^{n} \sin(k)$ is bounded by a constant and $\sum_{k=0}^{n} \sin(k^2)$ is not bounded by a constant. Then, what about $\sum_{k=0}^{n} (|\sin(k)|-2/\pi)$?

From numerical calculation, $\max_{n=0...10^8}(\sum_{k=0}^{n} (|\sin(k)|-2/\pi))=0.0900478$ which is much small compared to $\max_{n=0...10^8}\sum_{k=0}^{n} \sin(k^2)=1882.86$.

So, I suppose $\sum_{k=0}^{n} (|\sin(k)|-2/\pi)$ can be bounded by a constant, but I don't know how to prove it.

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    $\begingroup$ For $s_n:=\sum_{k=0}^{n} (|\sin(k)|-2/\pi))$, I get $\max_{n\le10^6}s_n\approx0.0900475$ and $\min_{n\le10^6}s_n\approx-0.726668$. So, $(s_n)$ seems to be bounded from below as well. $\endgroup$
    – Iosif Pinelis
    Commented Sep 16, 2019 at 20:11
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    $\begingroup$ Since integers are equidistributed modulo $\pi$, Weyl's criterion shows that $\frac{1}{n}\sum_{k\leq n}|\sin(k)|$ converges to $\frac{1}{\pi}\int_0^\pi|\sin(x)|dx=\frac{2}{\pi}$. This gives that your sum is $o(n)$. Long short from it being bounded, but that's a start. $\endgroup$
    – Wojowu
    Commented Sep 16, 2019 at 20:15
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    $\begingroup$ Assuming $(s_n)$ is bounded, one may further ask if $\sup_n s_n$ and $\inf_n s_n$ are attained; my guess they are not. $\endgroup$
    – Iosif Pinelis
    Commented Sep 16, 2019 at 20:21

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As a partial answer, we can show that if $S_N=\sum_{k=0}^N (|\sin(k)|-2/\pi)$, then:

$S_N=O(1)-\frac{4}{\pi}\sum_{k=2}^{N}\sum_{m=1}^{[k \log^2 k]}\frac{\cos 2mk}{4m^2-1}=O(\sum_{m=1}^{[N\log^2 N]}{\frac{\min(N\log^2N, ||\frac{m}{\pi}||^{-1})}{m^2}})$,

where as usual $||x||$ represents the distance to the closest integer.

In particular if we know that $||\frac{m}{\pi}||$ is not too "small" (e.g it is $>>m^{-1+\epsilon}$) for enough $m$ we may be able to show boundness or at least some better estimate than the uniform distribution $o(N)$ noted by @Wojowu above. Conversely, if there are infinitely many $m$ with $||\frac{m}{\pi}|| < C\frac{1}{m^2}$, the estimate doesn't work at least as boundness goes, though a cleverer estimate of the double trigonometric sum above could work.

Proof: We use the Fourier series $|\sin k|=\frac{8}{\pi}\sum_{m=1}^{\infty}\frac{\sin^2 mk}{4m^2-1}$ which converges absolutely and we cut it at $[k\log^2 k]$ for $k \ge 2$ as the terms for $k=0,1$ in $S_N$ are absorbed in the $O(1)$ above, so we won't bother with them.

Then the tail of the Fourier series for $|\sin k|$ is obviously bounded by $C\sum_{m > k\log^2 k}\frac{1}{m^2}=O(\frac{1}{k\log^2 k})$, hence summing on $k \ge 2$ we get the tail $O(1)$.

For the rest, using $2\sin^2(mk)=1-\cos(2mk)$ and $2\frac{1}{4m^2-1}=\frac{1}{2m-1}-\frac{1}{2m+1}$, we get that $\frac{8}{\pi}\sum_{m=1}^{[k \log^2 k]}\frac{\sin^2 mk}{4m^2-1}=\frac{2}{\pi}+O(\frac{1}{k\log^2 k})-\frac{4}{\pi}\sum_{m=1}^{[k \log^2 k]}\frac{\cos 2mk}{4m^2-1}$

Putting the above together we get the first statement above, namely that:

$S_N=O(1)-\frac{4}{\pi}\sum_{k=2}^{N}\sum_{m=1}^{[k \log^2 k]}\frac{\cos 2mk}{4m^2-1}$

Rearranging the terms with the same $1 \le m \le [N\log^2 N]$ and using the well known (and easy to prove) estimate:

$\sum_{j=m}^{n}\cos(2\pi jx)=O(\min(n-m+1, ||x||^{-1}))$, we get our claimed estimate result:

$\sum_{k=2}^{N}\sum_{m=1}^{[k \log^2 k]}\frac{\cos 2mk}{4m^2-1}=O(\sum_{m=1}^{[N\log^2 N]}{\frac{\min(N\log^2N, ||\frac{m}{\pi}||^{-1})}{m^2}})$

since for fixed $m$, the sum in $k$ has at most $N \log^2N$ terms and the corresponding $x$ is of course $\frac{m}{\pi}$

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  • $\begingroup$ Thank you. By using irrationality measure $\mu$ of $\pi$, for all large $m$, $||\frac{m}{\pi}|| \gg \frac{1}{m^{\mu-1+\varepsilon}}$. ( mathoverflow.net/questions/282259/… ) $\endgroup$
    – ueir
    Commented Sep 17, 2019 at 4:19

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