Sum of series $\sin x + \sin 2x + \sin 3x + \cdots $ Please help me compute the sum of the series:
 $$\sin(x)+\sin(2x)+\sin(3x)+\cdots$$
The series does not converge for all $x$. There are some $x$, for instance $x=0$ or $x=\pi$, for which the series converges to $0$, however if we consider $x=\frac \pi 2$ we find that our series is $1+0+-1+0+1+\cdots$ which does not converge.

If you think of the unit circle, imagine a line whose angle from the positive x-axis is the value of $x$. Then the x-coordinate of this point on the unit circle is the value of $\sin x $. Doubling the angle yields a point whose x-coordinate is $\sin 2x$. Tripling it yields a point whose x-coordinate is $\sin 3x$. Continuing this, you can see that the pattern will continue with varied positive and negative values, not approaching any particular limit unless the line representing an angle of $x$ was aligned with the positive or negative x-axis. (This is not rigorous, but can be made to be so.)

More technically, we have that $\lim_{n\to\infty} \sin nx =0$ iff $x=k\pi$ for some $k\in\Bbb Z$, so the series trivially converges to zero for such $x$ and diverges for all other $x$.
There is much more easier way to prove that the series does not converge when $x \ne k\pi$.

$\liminf\limits_{n\to\infty}\sin nx=-1$

The corresponding subsequence is $n_k=-\frac{\pi -4\pi k}{2x}$

$\limsup\limits_{n\to\infty}\sin nx=1$

The corresponding subsequence is $n_k=\frac{\pi + 4\pi k}{2x}$

Therefore there is no limit of $\sin nx$ and hence a necessary condition of convergence is violated.