Convergence for all $\theta$ of a sum with periodic function

Question

How can I show that:

$$ \sum_{n \geq 1} \dfrac{\sin(n\theta)}{n} $$

converges for all $\theta \in \mathbb{R}$?

Answer 1

If $\theta\in\pi\mathbb{Z}$ the series yields obviously zero, while assuming $\theta\not\in\pi\mathbb{Z}$ we have $\sin\frac{\theta}{2}\neq 0$ and: $$\sum_{n=1}^{N}\sin(n\theta) = \frac{\sin\left(\frac{N}{2}\theta\right)\sin\left(\frac{N+1}{2}\theta\right)}{\sin\frac{\theta}{2}},$$ so by setting $S_N=\sum_{n=1}^{N}\sin(n\theta)$ we have: $$|S_N|\leq\frac{1}{|\sin(\theta/2)|}=K.\tag{1}$$

Now we apply summation by parts:

$$\sum_{n=1}^{M}\frac{\sin n\theta}{n}=\frac{S_M}{M}+\sum_{n=1}^{M-1}\frac{S_n}{n(n+1)}\tag{2}.$$

Since the absolute value of any $S_j$ is bounded by $K$ by $(1)$, the last identity gives that the original series is a convergent one, converging towards: $$\sum_{n=1}^{+\infty}\frac{S_n}{n(n+1)}$$ whose absolute value is less than $K$. There is also a general criterion: if $\{a_n\}_{n\in\mathbb{N}}$ is a sequence with bounded partial sums and $\{b_n\}_{n\in\mathbb{N}}$ is an increasing sequence diverging to $+\infty$, the series: $$\sum_{n\geq 1}\frac{a_n}{b_n}$$ converges. The proof goes along the same lines.

The continuous analogue is the following: if $f,g$ are two Riemann-integrable functions over $\mathbb{R}^+$, $g\in C^{0}(\mathbb{R}^+)$ is a non-increasing function and $\int_{0}^{x}f(t)\,dt$ is bounded for any $x\in\mathbb{R}^+$, then $f\cdot g$ is a Riemann-integrable function over $\mathbb{R}^+$.

Sometimes, both criteria are known as "the Dirichlet criterion".

Another possible approach is to notice that your series is just the Fourier series of the sawtooth wave, yielding: $$\sum_{n=1}^{+\infty}\frac{\sin(n\theta)}{n}=\frac{\pi}{2}-\pi\left\{\frac{x}{2\pi}\right\}\tag{3}$$ for any $\theta\not\in\pi\mathbb{Z}$, where $\{y\}=y-\lfloor y\rfloor$.

Answer 2

Set $$ s_n(\vartheta)=\sin\vartheta+\sin(2\vartheta)+\cdots+\sin(n\vartheta)=\mathrm{Re}\left(\mathrm{e}^{i\vartheta}+\cdots+\mathrm{e}^{in\vartheta}\right)=\cdots=\frac{\sin(n\vartheta/2) \sin[(n+1)\vartheta/2]}{\sin(\vartheta/2)} $$ and $s_n(0)=0$. Hence $s_n(\vartheta)$ is bounded for every $\vartheta$, say by $M(\vartheta)$. Then $$ \sum_{k=1}^n \frac{\sin (k\vartheta)}{k}=\sum_{k=1}^n \frac{s_{k}(\vartheta)-s_{k-1}(\vartheta)}{k}=\sum_{k=1}^n \frac{s_{k}(\vartheta)}{k}-\sum_{k=1}^n \frac{s_{k-1}(\vartheta)}{k} \\ =\frac{s_n(\vartheta)}{n}+\sum_{k=1}^{n-1} \frac{s_{k}(\vartheta)}{k}-\sum_{k=1}^{n-1} \frac{s_{k}(\vartheta)}{k+1}=\frac{\sin(n\vartheta/2) \sin[(n+1)\vartheta/2]}{n\sin(\vartheta/2)} +\sum_{k=1}^{n-1} \frac{s_{k}(\vartheta)}{k(k+1)}. $$ Now the first term $\frac{\sin(n\vartheta/2) \sin[(n+1)\vartheta/2]}{n\sin(\vartheta/2)}$ tends to zero, for every $\vartheta$, while the second term $\sum_{k=1}^{n-1} \frac{s_{k}(\vartheta)}{k(k+1)}$ is bounded by $$ M(\vartheta)\sum_{k=1}^{n-1} \frac{1}{k(k+1)}, $$ which establishes summability (conditional of course) of our series.