For each $n=1,2,{\dots}$ and $x\in(0,{\pi})$, prove that the series $$S_n(x)=\sum_{k=1}^{n} \frac{\sin(kx)}{k}>0$$
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Incomplete thought: $S_n'(x) = \sum \cos(kx)$, and $\cos y = \frac12(e^{iy}+e^{-iy})$, so the derivative is just the sum of two geometric series and can be calculated in closed form. Does it help...? – Greg Martin Apr 05 '14 at 20:18
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And you can get a look here : http://math.stackexchange.com/questions/710694/positivness-of-the-sum-of-frac-sin2k-1x2k-1/710744#710744 – Apr 05 '14 at 20:24
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The expression $x\in(0,\pi)$ looked typographically odd: $x{\in}(0,\pi)$. It was coded as x{\in}(0,\pi) instead of x\in(0,\pi). Spacing conventions are built in to the software, but this circumvents them. This is useful for treating \sim as an object rather than as a binary relation, as in a quotient structure $R/{\sim}$, coded as R/{\sim}, but in this present posting it doesn't make sense and I changed it. – Michael Hardy Apr 05 '14 at 22:36