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How to find $\sum\limits_{n=1}^\infty q^n \sin(nx)$, where $|q|<1$ and $x \in \mathbb{R}$? I was thinking about rewriting it as $\sum\limits_{n=1}^\infty (q(\Im(\cos x+i\sin x)))^n$. It is a geometric series with the first term $q \cdot \sin x$, but what is the quotient? I can find $\sum\limits_{n=1}^\infty \sin(nx)$, but how to deal with the imaginary part, when multiplied by $q$?

Thanks!

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    $\begingroup$ Euler to the rescue: $$\sin (nx) = \frac{1}{2i}\left(e^{inx} - e^{-inx}\right).$$ $\endgroup$ – Daniel Fischer Mar 16 '14 at 22:11
  • $\begingroup$ I think the variable names are a bit confused. Do you mean $\sum_{n=1}^\infty q^n \sin(nx)$? $\endgroup$ – Argon Mar 16 '14 at 22:21
  • $\begingroup$ @Argon I edited the question. $\endgroup$ – glebovg Mar 16 '14 at 22:40
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You are on the right track. Since $|e^{ix}| = 1$ and $|q| < 1$, $|qe^{ix}| < 1$, so $\sum_{n = 1}^\infty q^n e^{i n x}$ is a geometric series. In particular, $$\sum_{n = 1}^\infty q^n e^{i n x} = \sum_{n = 1}^\infty (qe^{ix})^n = qe^{ix} + (qe^{ix})^2 + (qe^{ix})^3 + \cdots = \frac{qe^{ix}}{1 - qe^{i x}}.$$ Now you simply need the imaginary part of the right-hand side, which, if I am not mistaken, is $$\frac{q \sin x}{1 + q^2 - 2q \cos x}.$$

Observe, also, that $$\sum_{n = 1}^\infty q^n \sin(n x) = \frac{q \sin x}{1 + q^2 - 2q \cos x}.$$ is almost the Poisson kernel.

Note: Since $$a + ar + ar^2 + \cdots = \frac{a}{1 - r}, \quad |r| < 1,$$ we have $$ar + ar^2 + ar^3 + \cdots = \frac{ar}{1 - r}, \quad |r| < 1.$$ In your case $r = qe^{ix}$ and $a = 1$.

Hope this helps.

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Hint: Consider:

$$\Im \sum_{n=1}^\infty q^n e^{inx}$$

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  • $\begingroup$ This is a good, though a rather faint, hint. It most definitely is neither an invalid answer nor a very low-quality one. +1 $\endgroup$ – DonAntonio Mar 17 '14 at 3:31

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