5

How can we find $$L=\limsup_{n\to\infty}\sum_{k=0}^n(-1)^k|\sin k|,$$ where $|\cdot|$ denotes the absolute value of $\cdot$?

According to this answer, we can see the limit does exists. Unfortunately, this answer is not constructive. It does not provide a method to find the value of the limit.
Since $(-1)^{2k+1}|\sin k|<0$, $$L=\limsup_{n\to\infty}\sum_{k=0}^{2n}(-1)^k|\sin k|\\ =\limsup_{n\to\infty}\sum_{k=1}^{n}\Big(|\sin (2k)|-|\sin (2k-1)|\Big)$$ I noticed that if we can estimate $\sum_{k=1}^{n}|\sin (xk)|$ to $o(1)$ term, the problem will be solved, but I have no idea how to reach it.

Kemono Chen
  • 8,629
  • $\sum_{k=1}^{n}\left|\sin(xk)\right|$ won't ever be $o(1)$ (unless $x\in\pi\mathbb{Z}$) since the average value of $\left|\sin z\right|$ is positive, and it is $\frac{2}{\pi}$. A more reasonable bound is $$\sum_{k=1}^{n}\left|\sin(xk)\right|=\frac{2n}{\pi}+O_x(1).$$ – Jack D'Aurizio Jan 02 '19 at 02:35
  • By estimate to $o(1)$ term I mean find a known function $f(n)$ such that $$\sum_{k=1}^n|\sin(xk)|=f(n)+o(1)$$ but it seems impossible. – Kemono Chen Jan 02 '19 at 09:02

2 Answers2

2

We have $$ \left|\sin x\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{m\geq 1}\frac{\cos(2mx)}{4m^2-1} $$ hence $$\begin{eqnarray*} \sum_{k=0}^{2n}(-1)^k\left|\sin k\right| &=& \frac{2}{\pi}-\frac{4}{\pi}\sum_{m\geq 1}\frac{1}{4m^2-1}\sum_{k=0}^{2n}(-1)^k \cos(2mk)\\&=&\frac{2}{\pi}-\frac{2}{\pi}\sum_{m\geq 1}\frac{1}{4m^2-1}\cdot\frac{\cos(m(4n+1))+\cos(mn)}{\cos m}\end{eqnarray*}$$ and if the series $\sum_{m\geq 1}\frac{1}{m^2\left|\cos m\right|}$ were convergent, your $\limsup$ would be finite. This is not the case, since $\left|\cos m\right|$ can be as small as $\frac{1}{m^2}$ if $m$ is constructed from the numerator of a convergent of $\pi$ (consider $m=573204,m=52174$ or just $m=11$ from the Archimedean approximation). And if $\cos m$ is very close to zero and $n$ is even then $$ \cos(m(4n+1))+\cos(mn) = T_{4n+1}(\cos m)+T_{n}(\cos m) $$ might be dangerously close to $+1$ or $-1$. Indeed the convergence of the involved series follows from a pretty heavy machinery, relying on the fact that the irrationality measure of $\pi$ is finite and we have the theorems of Denjoy-Koksma, Erdos-Turan and Van der Corput for exponential sums (see the brilliant answer of i707107 here. I agree with him that the application of the EMC formula to a non-differentiable function is very fishy. I am less skeptical about standard tools in harmonic analysis and maximal operators.) Besides that, the computation of the exact value of the wanted $\limsup$ is both extremely difficult and probably irrelevant. This brings to the table an Italian motto: $$\text{"Non ora, non noi"} $$ meaning that the exact computation will be carried out by not us, not now.

Jack D'Aurizio
  • 353,855
  • To counter-act my laziness, I have posted an answer to the original question only relying on the finite-ness of the irrationality measure of $\pi$ and on the identity $$\sum_{n\geq 1}\frac{(-1)^n}{n}\cos(2mn) = -\log 2-\log\left|\cos m\right|.$$ – Jack D'Aurizio Jan 02 '19 at 04:23
0

Hint :

Using Fourier Series for $|\sin x|$, it is :

$$|\sin x|=\sum_{m=0}^\infty c_m\cos (mx) $$

Rebellos
  • 21,324
  • This is a hint... towards what, exactly? What happens by computing such $c_m$ and exchanging the sums on $m$ and $k$? – Jack D'Aurizio Jan 02 '19 at 02:32
  • I am not even sure the wanted $\limsup$ is finite. We might have $$\max_{1\leq n \leq N}\left|\sum_{k=0}^{2n}(-1)^k\left|\sin k\right|\right|=\Omega(\log\log N)$$ like in this famous case – Jack D'Aurizio Jan 02 '19 at 03:01