Real Analysis – Convergence of an Alternating Series: ?_{n?1} (-1)^n|sin n|/n

Question

Study the convergence of $$\displaystyle \sum_{n\geq 1} \frac{(-1)^n|\sin n|}{n}.$$

I am stuck with this series, we need probably some measure of irrationally of $\pi$, unfortunately I am unfamiliar with this. So here is my attempt :

Let $f(x) = \sum \frac{|\sin{n}|}{n} x^n, |x| < 1$

It's not difficult to compute the Fourier series of $|\sin(x)|$ :

$$
\displaystyle|\sin(x)|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{4n^2-1}
$$

Then Fubini's theorem (Series Version) works very well (because the previous series converges absolutely at $x$ fixed ) and all calculations made, we find that for all $x\in( -1,1)$:

$$
\displaystyle f(x)=\frac{2}{\pi}\sum_{n=1}^{+\infty}\frac{x^n}{n}-\frac{4}{\pi}\sum_{p=1}^{+\infty}\frac{x^2-2x\cos(p)}{(4p^2-1)(x^2-2x\cos(p)+1)}
$$

However, the second sum I have not been able to show the convergence. I feel the series diverge because the following series
$$
\displaystyle\sum\frac{1}{p^2\sin^2\left(\frac{p}{2}\right)}
$$

diverge because $0$ is an accumulation point of $\displaystyle (n\sin(n))$ sequence.

Any ideas (for the original series) ?

Answer 1

The series converges. It is enough to show that the sequence of the following partial sums converges: \begin{align} s_N &= \sum_{n=1}^{N} \left(\frac{ |\sin 2n|}{2n}-\frac{|\sin (2n+1)|}{2n+1}\right)=\sum_{n=1}^N \left(\frac{(2n+1)|\sin 2n|- 2n|\sin (2n+1)| }{2n(2n+1)} \right)\\ &=\sum_{n=1}^N \left( \frac{|\sin 2n|-|\sin (2n+1)|}{2n+1}+\frac{|\sin 2n|}{2n(2n+1)}\right). \end{align} Thus, it is enough to show that the following converges: $$ S_N = \sum_{n=1}^N \frac{|\sin 2n|-|\sin (2n+1)|}{2n+1}. $$ We consider a partition of $\mathbb{N}$ into four disjoint sets $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$ defined by: $$ A_{1}=\{n\in\mathbb{N}: \sin 2n >0, \sin (2n+1)>0\}, \ \ A_{2}=\{n\in\mathbb{N}: \sin 2n >0, \sin (2n+1)<0\}, $$ $$ A_{3}=\{n\in\mathbb{N}: \sin 2n <0, \sin (2n+1)>0\}, \ \ A_{4}=\{n\in\mathbb{N}: \sin 2n <0, \sin (2n+1)<0\}. $$ Note that $$ A_{1}=\{n\in\mathbb{N}: 2n \ \mathrm{mod} \ 2\pi \in (0,\pi-1)\}, \ \ A_{2}=\{n\in\mathbb{N}: 2n \ \mathrm{mod} \ 2\pi \in (\pi-1,\pi)\},$$ $$ A_{3}=\{n\in\mathbb{N}: 2n \ \mathrm{mod} \ 2\pi \in (-1,0)\}, \ \ A_{4}=\{n\in\mathbb{N}: 2n \ \mathrm{mod} \ 2\pi \in (-\pi, -1)\}.$$ By trigonometric identities, $$ n\in A_{1} \Longrightarrow |\sin 2n|-|\sin (2n+1)| = \sin 2n - \sin(2n+1) = -2\cos(2n+\frac12)\sin \frac12, $$ $$ n\in A_{2} \Longrightarrow |\sin 2n|-|\sin (2n+1)| = \sin 2n + \sin(2n+1) = 2\sin(2n+\frac12)\cos \frac12, $$ $$ n\in A_{3} \Longrightarrow |\sin 2n|-|\sin (2n+1)| = -\sin 2n - \sin(2n+1) = -2\sin(2n+\frac12)\cos\frac12, $$ $$ n\in A_{4} \Longrightarrow |\sin 2n|-|\sin (2n+1)| = -\sin 2n + \sin(2n+1) = 2\cos(2n+\frac12)\sin \frac12.$$ We define $$ f_1(x)=-I_{(0,\pi-1)}(x)2\cos(x+\frac12)\sin\frac12, \ \ f_2(x)=I_{(\pi-1,\pi)}(x)2\sin(x+\frac12)\cos\frac12,$$ $$ f_3(x)=-I_{(-1,0)}(x)2\sin(x+\frac12)\cos\frac12, \ \ f_4(x)=I_{(-\pi,-1)}(x)2\cos(x+\frac12)\sin\frac12 $$ where $I_A$ is the characteristic function of $A$. Note that these functions $f_i(x)$ are of bounded variation in $[-\pi,\pi]$. Thus, $f=f_1+f_2+f_3+f_4$ is of a bounded variation.

We need Koksma's inequality p. 143, Theorem 5.1 of 'Uniform Distribution of Sequences' by Kuipers and Niederreiter:

Theorem [Koksma]

Let $f$ be a function on $I=[0,1]$ of bounded variation $V(f)$, and suppose we are given $N$ points $x_1, \ldots , x_N$ in $I$ with discrepancy $$ D_N:=\sup_{0\leq a\leq b\leq 1} \left|\frac1N \#\{1\leq n\leq N: x_n \in (a,b) \} -(b-a)\right|. $$ Then $$ \left|\frac1N \sum_{n\leq N} f(x_n) - \int_I f(x)dx \right|\leq V(f)D_N. $$

To control the discrepancy, we apply Erdos-Turan inequality p. 112, Theorem 2.5 of Kuipers and Niederreiter's book.

Theorem[Erdos-Turan]

Let $x_1, \ldots, x_N$ be $N$ points in $I=[0,1]$. Then there is an absolute constant $C>0$ such that for any positive integer $m$, $$ D_N\leq C \left( \frac1m+ \sum_{h=1}^m \frac1h \left| \frac1N\sum_{n=1}^N e^{2\pi i h x_n}\right|\right). $$

The sequence of our interest is $x_n = 2n$ mod $2\pi$. The above two inequalities together applied to $f=f_1+f_2+f_3+f_4$ yields the following: There is an absolute constant $C>0$ such that for any positive integer $m$, $$ \left|\frac1N \sum_{n\leq N} f(x_n) - \frac1{2\pi} \int_{-\pi}^{\pi} f(x)dx \right|\leq C \left( \frac1m+ \frac1N\sum_{h=1}^m \frac1{h\langle \frac h{\pi} \rangle}\right). $$ A result on the irrationality measure of $\pi$ by Salikhov implies that $$ \left| \frac1{\pi} - \frac pq \right| \geq \frac 1{q^{\mu+\epsilon}} $$ for all integers $p, q$ and $q$ is sufficiently large, and $\mu=7.60631$, $\epsilon>0$. This implies $$ h\left\langle \frac h{\pi} \right\rangle \geq h^{2-\mu-\epsilon} $$ for sufficiently large $h$. Then for some absolute constant $C>0$, $$ \left|\frac1N \sum_{n\leq N} f(x_n) - \frac1{2\pi} \int_{-\pi}^{\pi} f(x)dx \right|\leq C \left( \frac1m+ \frac1N m^{\mu-1+\epsilon}\right). $$ Taking $m=\lfloor N^{1/\mu}\rfloor$, we obtain $$ \left|\frac1N \sum_{n\leq N} f(x_n) - \frac1{2\pi} \int_{-\pi}^{\pi} f(x)dx \right|\leq C N^{-\frac1{\mu}+\epsilon}. $$ It is easy to see that $\int_{-\pi}^{\pi} f(x) dx = 0$. Therefore, $$ \left|\sum_{n\leq N}f(x_n)\right|\leq CN^{1-\frac1{\mu}+\epsilon}. $$ The convergence of the series now follows from Abel's summation formula.

Added on 12/1/2018

By the bound for the discrepancy here: https://en.wikipedia.org/wiki/Low-discrepancy_sequence#Additive_recurrence

We may have a better error term: $$ \left|\sum_{n\leq N}f(x_n)\right|\leq CN^{1-\frac1{\mu-1}+\epsilon}. $$