Prove or disprove the convergence of the series $\sum_{n=1}^{\infty}\frac{|\sin n|\sin n}{n}$.

Question

The motivation of this question is from the improper integral $$\int_{0}^{\infty}\frac{|\sin x|\sin x}{x}\ {\rm{d}}x.\tag{*}$$ By Dirichlet test, we know the improper integral $(*)$ is convergent, because $$\int_{0}^{x}|\sin x|\sin x\ {\rm{d}}x,\quad x\in[0,\infty)$$ is bounded and $\frac1x\to0$ as $x\to\infty.$ Also by Lobachevsky integral formula (for reference you can see Lobachevsky), we know $$\int_{0}^{\infty}\frac{|\sin x|\sin x}{x}\ {\rm{d}}x=\int_0^\frac\pi2|\sin x|\ {\rm{d}}x=1.$$

So we consider the discrete case,i.e., the series $$\sum_{n=1}^{\infty}\frac{|\sin n|\sin n}{n}.$$ Let $a_n=|\sin n|\sin n$ and $b_n=\frac1n\to0$. If we can prove that the partial sum $$\sum_{k=1}^{n}a_k=\sum_{k=1}^{n}|\sin k|\sin k$$ is bounded, then, by Dirichlet test, we can get $$\sum_{n=1}^{\infty}\frac{|\sin n|\sin n}{n}$$ is convergent (it is easy to see that the convergence should be conditional convergent).

I'm stuck by the boundness of the partial sum $$\sum_{k=1}^{n}|\sin k|\sin k.$$

Any comments and hints will welcome! Or other methods to deal with this problem will welcome.

Answer 1

Step 1. From the Fourier series

$$ \left| \sin x\right| = \frac{2}{\pi} \biggl[ 1 - \sum_{p=1}^{\infty} \frac{2}{4p^2-1} \cos(2px) \biggr], $$

we get

\begin{align*} \left| \sin x \right| \sin x &= \frac{2}{\pi} \biggl[ \sin x - \sum_{p=1}^{\infty} \frac{2}{4p^2-1} \cos(2px)\sin x \biggr] \\ &= \frac{2}{\pi} \biggl[ \sin x - \sum_{p=1}^{\infty} \frac{\sin(2p+1)x - \sin(2p-1)x}{4p^2-1} \biggr] \\ &= \frac{2}{\pi} \sum_{p=1}^{\infty} \biggl( \frac{1}{4p^2-1} - \frac{1}{4(p-1)^2-1} \biggr) \sin(2p-1)x \end{align*}

Hence, for any $|z| < 1$ we get

\begin{align*} \sum_{k=1}^{\infty} \frac{\left| \sin k \right| \sin k}{k} z^k &= \frac{2}{\pi} \sum_{p=1}^{\infty} \biggl( \frac{1}{4p^2-1} - \frac{1}{4(p-1)^2-1} \biggr) \sum_{k=1}^{\infty} \frac{\sin(2p-1)k}{k} z^k. \tag{1} \end{align*}

(Check that Fubini's Theorem is indeed applicable!)

Step 2. Now let

$$ f(x, z) = \sum_{k=1}^{\infty} \frac{\sin(kx)}{k} z^k. $$

Then we can check that, for $-1 < z < 1$,

$$ f(x, z) = \arctan \left(\frac{z \sin x}{1-z \cos x}\right). $$

One important consequence is that $z \mapsto f(x, z)$ is monotone in $z \in [0, 1)$ for any $x \in \mathbb{R}$. In particular,

$$ \lim_{z \to 1^-} f(x, z) $$

exists for each $x$ and the limit lies between $[-\frac{\pi}{2}, \frac{\pi}{2}]$. And although it is not necessary, we can also prove that

$$ \lim_{z \to 1^-} f(x, z) = \begin{cases} \frac{\pi - (x \text{ mod } 2\pi)}{2}, & x \notin 2\pi \mathbb{Z}, \\ 0, & x \in 2\pi \mathbb{Z} \end{cases} $$

Step 3. The above discussion, combined with the representation $\text{(1)}$, tells that

$$ \lim_{z \to 1^-} \sum_{k=1}^{\infty} \frac{\left| \sin k \right| \sin k}{k} z^k $$

exists and is finite. Now the convergence of the original series follows from the following version of Tauberian Theorem proved by Littlewood:

Theorem (Littlewood). Suppose the sequence $(n a_n)$ is bounded, and $\sum_{n=0}^{\infty} a_n$ is Abel summable. Then $\sum_{n=0}^{\infty} a_n$ is summable in the ordinary sense.

(See this for instance.)