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) ?


By Dirichlet's convergence test, this series will converge if we can show that there exists a constant $C$ such that $$\left|\sum_{n\leq x}(-1)^{n}|\sin(n)|\right|\leq C$$ for all $x$.

Lets write $$\sum_{n\leq x}(-1)^{n}|\sin(n)|=\sum_{n\leq\frac{x}{2}}|\sin(2n)|-\sum_{n\leq\frac{x+1}{2}}|\sin(2n-1)|.$$ Then for $x=2N$, an even number, Euler Maclaurin summation yields $$\sum_{n\leq N}|\sin(2n)|=\int_{1}^{N}|\sin(2t)|dt+\sum_{k=1}^{K}\frac{(-1)^{k}}{k!}B_{k}\left(\frac{d^{k-1}}{dt^{k-1}}|\sin(2t)|\biggr|_{t=1}^{t=N}\right)$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\frac{(-1)^{K}}{K!}\int_{1}^{N}B_{K}(\{t\})\left(\frac{d^{k}}{dt^{k}}|\sin(2t)|\right)dt.$$ Note that $|\sin(x)|$ has infinitely many derivatives everywhere except at integer multiples of $\pi$, and so the above holds for any $K>0$. Since $$|B_{k}(\{x\})|\leq k!2^{1-k}\pi^{-k}\zeta(k),$$ and since the derivatives of $|\sin(t)|$ are bounded in absolute value by $1$, it follows that $$\left|\sum_{n\leq N}|\sin(2n)|-\int_{1}^{N}|\sin(2t)|dt\right|\leq4\sum_{k=1}^{K}\frac{\zeta(k)}{(2\pi)^{k}}+\frac{2\zeta(K)N}{(2\pi)^{K}}.$$ The series $\sum_{k=1}^{\infty}\frac{\zeta(k)}{(2\pi)^{k}}$ converges absolutely, so by taking $K=N$ we see that there exists a constant $C_{1}$ such that $$\left|\sum_{n\leq N}|\sin(2n)|-\int_{1}^{N}|\sin(2t)|dt\right|\leq C_{1}$$ for all $N$. Similarly, there exists a constant $C_{2}$ such that $$\left|\sum_{n\leq N}|\sin(2n-1)|-\int_{1}^{N}|\sin(2t-1)|dt\right|\leq C_{2}.$$ Thus by the triangle inequality, $$\left|\sum_{n\leq x}(-1)^{n}|\sin(n)|\right|\leq C_{1}+C_{2}+\left|\int_{1}^{N}|\sin(2t)|dt-\int_{1}^{N}|\sin(2t-1)|dt\right|$$

$$\leq C_{1}+C_{2}+\int_{N-1/2}^{N}|\sin(2t)|dt+\int_{1}^{3/2}|\sin(2t-1)|dt$$

$$\leq C_{3}$$ for some constant $C_{3}$. This implies the desired result.

  • 2
    $\begingroup$ Euler Maclaurin summation formula might not work for functions with discontinuous derivatives. $\endgroup$ – Sungjin Kim Apr 25 '17 at 1:23
  • $\begingroup$ See my answer to the same problem: math.stackexchange.com/questions/2202138/… This uses different idea, but it is quite a standard approach for such series. $\endgroup$ – Sungjin Kim Apr 25 '17 at 1:31
  • $\begingroup$ See en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula for the notations I use, and for the case $p=1$, $S=\sum_{m<j\leq n} f(j)$, $I=\int_m^n f(x) dx$, $$ S-I=\frac{B_1}{1!}(f(n)-f(m)) + R_1$$ $$=\frac{f(n)-f(m)}2 +\int_m^n \left( \{x\}-\frac12\right) df(x)$$ This is integration by parts for Riemann-Stiltjes integral. A bound for the remainder $$ |R_1|\leq \frac 2{2\pi} \int_m^n |f'(x)| dx,$$ in case $f'$ is continuous. But, it is $$ |R_1|\leq \frac 2{2\pi} \int_m^n |df(x)| $$ if $f'$ has discontinuity. The function $f(t)=|\sin t|$ has a discontinuous derivative. $\endgroup$ – Sungjin Kim Jan 9 '19 at 0:06

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.


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: $$ \left|\sum_{n\leq N}f(x_n)\right|\leq CN^{1-\frac1{\mu-1}+\epsilon}. $$

  • 5
    $\begingroup$ Wow. You need a lot of sledgehammers to crack this nut. $\endgroup$ – marty cohen Mar 30 '17 at 0:39
  • 1
    $\begingroup$ @martycohen I will be interested to see if we can avoid them. But, I do not know any easier ways by now. $\endgroup$ – Sungjin Kim Mar 30 '17 at 0:54
  • 1
    $\begingroup$ @i707107 Your solution is absolutely brilliant. $\endgroup$ – Sanjoy Kundu Mar 31 '17 at 4:36
  • $\begingroup$ Yes, we can avoid using the bound on the irrationality measure of $\pi$ by splitting the sum in Erdős–Turán into the sum over the numbers $h$ that are denominators of the convergents of $1/\pi$ and the sum over the rest. $\endgroup$ – Jarek Kuben May 10 '17 at 21:32
  • $\begingroup$ @JarekKuben If we really can avoid using irrationality measure of $\pi$, please post your entire solution separately as an answer. $\endgroup$ – Sungjin Kim May 11 '17 at 0:24

$\color{red}{\text{Not an answer, just an idea needing more work}}$ :

I'd say $\sum_n \frac{(-1)^n}{n} b_n$ converges whenever $\Delta^k b(n) = O(w^{k}),w< 2$ where $\Delta^k b(n)$ is the $k$th forward difference, here $b(n) = |\sin n|$

If you sum by parts $k$ times, using that $\sum_{n=1}^N \frac{(-1)^n}{n} = \frac{(-1)^N}{2 N}+ O(\frac{1}{2 N^2})$ you'll get a main term $2^{-k} \sum_{n=1}^{N-k} \frac{(-1)^n}{n} \Delta^k b(n)$

use that $A(2M)=\sum_{n=M}^\infty \frac{1}{(2n+1)(2n+2)} = \sum_{n=m}^\infty \frac{1}{(2n+1)^2}-\frac{1}{(2n+1)^2(2n+2)}$ and approximate with $\int_{2M}^\infty \frac{dx}{x^2}-\int_{2M}^\infty \frac{dx}{x^3}$ to obtain

$$A(N) = \ln 2+\sum_{n=1}^N \frac{(-1)^n}{n}=\frac{(-1)^N}{2N}+O(\frac{1}{2 N^2})$$

Summing by parts $$\sum_{n=1}^N \frac{(-1)^n}{n} |\sin n| = A(N)|\sin(N)|+\sum_{n=1}^{N-1} A(n) (|\sin n| - |\sin (n +1)|)$$

The problematic term is $\sum_{n=1}^{N-1} \frac{(-1)^N}{2N} (|\sin n| - |\sin (n +1)|) $ that we can sum by parts again to get a new problematic term $$\sum_{n=1}^{N-2} \frac{(-1)^n}{4n} \Delta^2 b(n)$$ where $\Delta^2 b(n)=(|\sin n| - |\sin( n +1)|)-(|\sin( n+1)| - |\sin( n +2)|)$

summing by parts $k$ times we'll have $$\frac{1}{2^k}\sum_{n=1}^{N-k} \frac{(-1)^n}{n} \Delta^k b(n)$$ Where $\Delta^k b(n)$ is the $k$th forward difference of $b(n) = |\sin n|$


By using the Fourier series of $\left|\sin(n)\right|$ the problem of showing that $$ \sum_{n\geq 1}\frac{(-1)^n \left|\sin n\right|}{n} $$ is convergent boils down to the problem of showing that $$ \sum_{m\geq 1}\frac{\log\left|\cos m\right|}{4m^2-1}$$ is convergent. The only issue is given by the values of $m$ such that $m$ is close to an odd multiple of $\frac{\pi}{2}$.
On the other hand, since the irrationality measure of $\pi$ is finite, even if $\cos m$ is close to zero it cannot be closer than $\frac{1}{m^{10}}$ for any $m$ large enough. Since $$ \sum_{m\geq 1}\frac{10\log m}{4m^2-1} $$ is absolutely convergent, the original series is convergent as well.


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