Showing $\sum_{n=1}^\infty \sin x \sin nx$ is uniformly bounded

Question

I need to show that for every $x$: $$\sum_{n=1}^\infty \sin x \sin nx \lt M$$

So the first thing came into my mind is applying a well-known trigonometric identity:

$$\sum_{n=1}^\infty \sin x \sin nx = \frac{1}{2} \sum_{n=1}^\infty \cos (x-nx) - \cos(x+nx)$$

For a second I thought I'd get a telescoping series but it isn't.

What should I do next?

EDIT
Basically I'm trying to use here Dirichlet's test to show uniform converges for the functions series:

$$f_n(x) = \sum_{n=1}^\infty \frac{\sin x \sin nx}{\sqrt {n+x^2}}$$

Answer 1

Since $\cos$ is an even function, you have in fact a telescoping series:

\begin{align} \sum_{n = 1}^N \sin x\sin (nx) &= \frac{1}{2}\sum_{n = 1}^N \bigl(\cos\bigl((n-1)x\bigr) - \cos \bigl((n+1)x\bigr)\bigr)\\ &= \frac{1}{2}\bigl( 1 + \cos x - \cos (Nx) - \cos \bigl((N+1)x\bigr)\bigr). \end{align}

Answer 2

In this answer, I showed that

$$\begin{align} \left|\sum_{n=1}^N \sin(nx)\right| \le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\csc\left(\frac{x}{2}\right)\right| \end{align}$$

Thus,

$$\begin{align} \left|\sum_{n=1}^N \sin x \sin(nx)\right| &\le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\sin x\csc\left(\frac{x}{2}\right)\right|\\\\ &=\left(1+\left|\cos (\frac{x}{2})\right|\right)\left|\cos\left(\frac{x}{2}\right)\right|\\\\ &\le2 \tag 1 \end{align}$$

For the Dirichlet test of $f_n(x)=\sum_{n=1}^{\infty}\frac{\sin x\sin nx}{\sqrt{n+x^2}}$, we only require the following two conditions:

Condition $(1)$

The sequence $\frac{1}{\sqrt{n+x^2}}$ decreases monotonically to zero.

Condition $(2)$

The partial sums $\sum_{n=1}^N \sin x \sin(nx)$ be bounded by a constant.

Condition $(1)$ is trivially confirmed while equation $(1)$ confirms Condition $(2)$.