An engineer has told me they need to evaluate $$\sum_{k=1}^{n}a^k\sin(k \frac{2\pi}{l})$$ as part of their problem. I am unable to provide any further motivation for this question at the moment.
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Using complex numbers might be a way to do it. – Sep 30 '13 at 05:09
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3This sum is the imaginary part of the sum $\sum_{k=1}^n a^k \exp(2\pi i k/L) = \sum_{k=1}^n (a\exp(2\pi i /L))^k$ which is a simple geometric sum. – Alexander Vlasev Sep 30 '13 at 05:15
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You could use $\sin \theta= \frac{e^{i \theta}-e^{-i \theta}}{2i}$, where $i =\sqrt{-1}$. – Shobhit Sep 30 '13 at 05:20
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@AleksVlasev : Now I see it, thanks – jimjim Sep 30 '13 at 07:18
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I will assume that $a$ and $\ell$ are real in what follows. Since $\sin x$ gives the imaginary part of $e^{ix}$, we may rewrite your sum as $$\sum_{k=1}^n a^k \sin(2\pi k/\ell)= \mathrm{Im}\left(\sum_{k=1}^n a^k e^{2\pi i k/\ell}\right)= \mathrm{Im}\left(\sum_{k=1}^n e^{k(2\pi i/\ell+\log a)}\right).$$ Let $x= 2\pi i/\ell + \log a$. As a geometric series, we have $$\mathrm{Im}\left(\sum_{k=1}^n e^{kx}\right)= \mathrm{Im}\left(\frac{e^{x}(e^{nx}-1)}{e^x-1}\right)=\mathrm{Im}\left(\frac{e^{2\pi i/\ell + \log a}(e^{n(2\pi i/\ell + \log a)}-1)}{e^{2\pi i/\ell + \log a}-1}\right).$$ At this point, we need only write what remains in `$a+bi$' form, and take the imaginary part. After multiplying numerator and denominator by $(e^{-x}-1)$, we find (after much simplification) $$\frac{a \left(a^n \sin ^3\left(2\pi/\ell\right) \left(-\cos \left(2\pi n/\ell\right)\right)+a^n \left(a-\cos \left(2\pi/\ell\right)\right) \sin \left(2\pi n/\ell\right)+\sin \left(2\pi/\ell\right)\right)}{a^2-2 a \cos \left(2\pi/\ell\right)+1}.$$ (These expressions tend to be easier to read in complex exponential form.)
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