# Prove $\sum_{k=1}^{\infty} \frac{\sin(kx)}{k}$ converges

How to prove $$\sum_{k=1}^{\infty} \frac{\sin(kx)}{k}$$ converges without using integral test?

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This follows from Dirchlet’s Test. We identify $a_n = \frac{1}{n}$ and $b_n = \sin(nx)$ as in the linked theorem.

We are left to check the three properties

1. $(a_n)$ is monotonic decreasing, and is bounded away from $0$.

2. $a_n \to 0$ as $n \to \infty$.

3. $\left | \sum_{n < N} b_n \right | \leq M$ for each $N$.

I think (1) and (2) are straightforward to see. For (3), notice that $$\sum_{n = 1}^{N} 2\sin(nx)\sin \left ( \frac{x}{2} \right ) = \sum_{n = 1}^N \cos \left ( n-\frac{1}{2} \right) x – \cos\left (n + \frac{1}{2} \right)x.$$

The sum on the right is telescoping, so dividing both sides by $2\sin \left ( \frac{x}{2} \right )$ gives $$\sum_{n = 1}^{N}\sin(nx) = \frac{\cos \frac{x}{2}- \cos\left (N + \frac{1}{2} \right)x.}{2\sin \left ( \frac{x}{2} \right )}$$

I will let you draw the rest of the conclusions.