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.

  • 2
    $\begingroup$ 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. $\endgroup$ Sep 16 '19 at 20:11
  • 2
    $\begingroup$ 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
    Sep 16 '19 at 20:15
  • 1
    $\begingroup$ 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. $\endgroup$ 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 $\frac{m}{\pi}$

  • $\begingroup$ 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
    Sep 17 '19 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.