Sum $\sum_{k=1}^{n}a^k\sin(k \frac{2\pi}{l})$ 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.
 A: 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.)
