I am trying to show that the sequence $$f_n(x)=\sum_{m=1}^n sin(mx)$$ is uniformly bounded on $[a,2\pi-a]$ with $0<a\le\pi$. From playing around with the graph of $f$ its pretty clear that the maximum and minimum of $(f_n)$ on $[0,2\pi]$ are approaching positive and negative infinity, and as they grow in absolute value they are getting closer to $0$ and $2\pi$ respectively. So it is definitely going to be uniformly bounded I just can't figure out how to show this.
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You could simply compute $f_n(x)$: $$f_n(x)=\text{Im}\left[\sum_{m=1}^n e^{imx}\right] =\text{Im}\left[\frac{e^{i(n+1)x}-e^{ix}}{e^{ix}-1}\right] =\text{Im}\left[\frac{e^{i(n+1/2)x}-e^{ix/2}}{2i\sin (x/2)}\right] $$ (I multiplied numerator and denominator by $e^{-ix/2}$ to get last expression) Then, you can brutally bound the expression. Assuming $0< x<2\pi$, so that $\sin(x/2)>0$, $$|f_n(x)|\le \left|\frac{e^{i(n+1/2)x}-e^{ix/2}}{2i\sin (x/2)}\right|\le \frac1{\sin(x/2)}$$ This is your uniform bound.
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