real analysis - Partial Sums of a Trignometric Series

I'm attempting to show that the partial sums of

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

are uniformly bounded on the interval $(-\pi,\pi).$ This expression

$$S_N(x)=\sum_{k=1}^N\frac{(e^{ix})^k}{k}$$

comes up where $S_N(x)$ is not the partial sum of $f(x),$ just a piece of it, after using Euler's formula. Is there a closed form expression of $S_N(x)$ similar in spirit to Dirichlet's kernel?

Answer

We have

$$\left|\sum_{n=1}^m (-1)^n\frac{\sin nx}{n} \right| = \left|\sum_{n=1}^m \frac{\sin n(x+ \pi)}{n} \right|.$$

The partial sums for $\sin n(x+\pi)$ are bounded as
$$|S_n| = \left|\sum_{k=1}^n \sin k(x+\pi)\right| = \frac{|\sin\left(n(x +\pi)/2\right)||\sin((n+1)(x+\pi)/2)|}{|\,\sin[(x + \pi)/2]\,|} \leqslant B,$$

and the bound is uniform on an interval $[-\pi + \delta,\pi-\delta]$ with $\delta > 0$:

$$|S_n| \leqslant B = \frac{1}{\sin(\delta/2)}$$

Summing by parts we get,

$$\left|\sum_{n=1}^m \frac{\sin n(x+ \pi)}{n} \right| \leqslant \frac{|S_m|}{m} + \sum_{n=1}^{m-1} |S_n|(1/n - 1/(n+1)) \\ \leqslant B/m + B (1 - 1/m) \\ = B$$