# Is $\sum_{k=0}^n (|\sin(k)|-2/\pi)$ bounded by a constant $M$?

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.

• 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. – Iosif Pinelis Sep 16 '19 at 20:11
• 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. – Wojowu Sep 16 '19 at 20:15
• 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. – Iosif Pinelis Sep 16 '19 at 20:21

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 $$\frac{m}{\pi}$$

• 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/… ) – ueir Sep 17 '19 at 4:19