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i need to show this equality: $$ \sum_{k=1}^n \frac{\sin(kx)}{k} = \frac{\pi - x}{2}$$

I should use that $\displaystyle\frac{\sin(kx)}{k} = \int_\pi ^x \cos(kt)\,\mathrm dt$.

I tried many times to solve this, but I just got stuck. Is there a trick or anything I don't see?

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  • $\begingroup$ Do you know what the sum of $\cos(kx)$ is with respect to $k$? $\endgroup$ – Chinny84 May 16 '17 at 20:53
  • $\begingroup$ That holds only for $x\in(0,2\pi)$ when you consider the limit as $n\to +\infty$. It is a standard exercise in Fourier series than can be simply tackled through $\sin(z)=\text{Im}\,e^{iz}$ and $\sum_{n\geq 1}\frac{z^n}{n}=\log(1-z)$ for any $z$ such that $|z|<1$. $\endgroup$ – Jack D'Aurizio May 16 '17 at 20:55
  • $\begingroup$ well ..i got as far as $\int_\pi ^x \frac{sin(nt/2)* cos(tn+t))/2)}{sin(t/2) }dt $ do you mean that? $\endgroup$ – wondering1123 May 16 '17 at 20:57
  • $\begingroup$ we dont do fourier analysis yet..:( $\endgroup$ – wondering1123 May 16 '17 at 20:58
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Hint:

$\displaystyle \sum_{k=1}^n \dfrac{\sin(kx)}{k} = \int_\pi ^x \sum_{k=1}^n \cos(kt)\,\mathrm dt$ by linearity of integral then write $\displaystyle\sum_{k=1}^n \cos(kt)$ as the real part of $\displaystyle\sum_{k=1}^n e^{ikt}$ which is a geometric sum.

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HINT

$$\sum\limits_{k=1}^n{\sin(kx)\over k} = -\sum\limits_{k=1}^n\int\limits_0^x\cos(kx)dx = -\Re\int\limits_0^x\left(\sum\limits_{k=1}^ne^{ikx}\right)dx = -\Re\int\limits_0^x{e^{ikx}\over1-e^{ikx}}dx$$ $$ = \Re\left({1\over ik}\int\limits_0^x{d\left(1-e^{ikx}\right)\over1-e^{ikx}}dx\right) = \Re\left({1\over ik}\log(1-ikx)\right) = {1\over k}\Im\log(1-ikx) = {1\over k}\left(\arg(1-ikx)+2\pi m\right) = {1\over k}\left(-\arctan kx+2\pi m\right)$$

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