On a known inequality

Let $x$ be a real number and let $n \in \mathbb{N}$. Prove the inequality:

$$\left | \sum_{k=1}^{n} \frac{\sin kx}{k} \right |\leq 2 \sqrt{\pi}$$

Solution

We give Titu's solution on the problem plus two comments.

Since $| \sin x|$ is periodic of period $\pi$ , we may assume that $x$ is in the interval $(0, \pi)$. For a fixed $x$ with $0<x<\pi$ , let $m$ be the non negative integer such that $m \leq \frac{\sqrt{\pi}}{x} < m+1$. Thus:

$$\left | \sum_{k=1}^{n} \frac{\sin kx}{k} \right | \leq \left | \sum_{k=1}^{m} \frac{\sin kx}{k} \right | + \left |\sum_{k=m+1}^{n} \frac{\sin kx}{k}  \right |$$

We set the first summation on the RHS to be zero if $m=0$ and the first summation taken from $1$ to $n$ and the second to be $0$ if $m \geq n$. It suffices to show that:

\begin{equation} \left | \sum_{k=1}^{m} \frac{\sin kx}{k} \right | \leq \sqrt{\pi} \end{equation}

and

\begin{equation} \left | \sum_{k=m+1}^{n} \frac{\sin kx}{k} \right | \leq \sqrt{\pi} \end{equation}

Since $\sin x \leq x$ then by the definition of $m$ it follows that:

$$\left | \sum_{k=1}^{m} \frac{\sin kx}{k} \right |\leq \sum_{k=1}^{m} \frac{kx}{k} = \sum_{k=1}^{m} x = mx \leq \sqrt{\pi}$$

establishing $(1)$. Now, we have the well celebrated inequality

$$\left | \sum_{k=m+1}^{n} \frac{\sin kx}{k} \right |\leq \frac{1}{(m+1) \left | \sin \frac{x}{2} \right | }$$

and since $\sin x$ is concave for $0<x<\frac{\pi}{2}$ then the tangent of the graph lies above it. Thus:

$$\left | \sum_{k=m+1}^{n} \frac{\sin kx}{k} \right |\leq \frac{1}{(m+1) \left | \sin \frac{x}{2} \right |} \leq \frac{1}{m+1} \cdot \frac{x}{\pi} \leq \frac{\sqrt{\pi}}{x} \cdot \frac{x}{\pi} \leq \sqrt{\pi}$$

proving $(2)$. Adding them up we get the wanted result.