A closed form of the series $ \sum_{n=1}^{\infty} q^n \sin(n\alpha) $

Question

I am having problems with the following series:

$$ \sum_{n=1}^{\infty} q^n \sin(n\alpha), \quad|q| < 1. $$

No restrictions on $\alpha$. I need to find out whether it converges and if yes, evaluate its sum.

I can see that it's convergent using the comparison test. But I fail to find its sum. So far I tried grouping subsequent terms and using trigonometric formulas, but it didn't help me much.

Where should I start when I see trigonometric functions in a series? In general, I have no idea where to take off in such situations.

Thanks in advance.

Answer 1

Hint. We assume $\alpha\in \mathbb{R}$ and $-1<q<1$. Then one may write $$ \sum_{n=1}^{\infty} q^n \sin(n\alpha)=\Im \sum_{n=1}^{\infty} (qe^{i\alpha})^n =\Im\: \frac{qe^{i\alpha}}{1-qe^{i\alpha}} $$ where we have used the standard evaluation of a geometric series.

Answer 2

I'd like to expand Olivier Oloa's hint:

$$ \sum _{ n=1 }^{ \infty } q^{ n }\sin (n\alpha )=q\sin { \alpha +{ q }^{ 2 }\sin { 2\alpha +...+{ q }^{ n }\sin { n\alpha +...\quad \quad \quad \quad \quad \quad \quad \quad \left( 1 \right) } } } \quad \\ \sum _{ n=1 }^{ \infty } q^{ n }\cos { \left( n\alpha \right) } =q\cos { \alpha +{ q }^{ 2 }\cos { 2\alpha } +...+{ q }^{ n }\cos { n\alpha +... } } ,\left| q \right| <1\quad \quad \left( 2 \right) $$

let denote partial sums of $(1)$ and $(2)$as follows:

$${ u }_{ n }=\sum _{ n=1 }^{ \infty } q^{ n }\sin (n\alpha )=q\sin { \alpha +{ q }^{ 2 }\sin { 2\alpha +...+{ q }^{ n }\sin { n\alpha } } } \quad \quad \\ { v }_{ n }=\sum _{ n=1 }^{ \infty } q^{ n }\cos { \left( n\alpha \right) } =q\cos { \alpha +{ q }^{ 2 }\cos { 2\alpha } +...+{ q }^{ n }\cos { n\alpha } } $$

by using Euler's formula ${ e }^{ i\varphi }=\cos { \varphi +i\sin { \varphi } } $ we get

$${ u }_{ n }+i{ v }_{ n }=q\left( \sin { \alpha } +i\cos { \alpha } \right) +{ q }^{ 2 }\left( \sin { 2\alpha } +\cos { 2\alpha } \right) +...+{ q }^{ n }\left( \sin { n\alpha +i\cos { n\alpha } } \right) =\\ =\frac { q{ e }^{ i\alpha }-{ q }^{ n+1 }{ e }^{ i\left( n+1 \right) \alpha } }{ 1-q{ e }^{ i\alpha } } $$

since $\left| q \right| <1\Rightarrow \left| q{ e }^{ i\alpha } \right| <1$ we have

$$\\ \lim _{ n\rightarrow \infty }{ \left( { q }^{ n+1 }{ e }^{ i\left( n+1 \right) \alpha } \right) =0 } $$

finally we get

$$u+iv=\lim _{ n\rightarrow \infty }{ \left( { u }_{ n }+i{ v }_{ n } \right) =\frac { q{ e }^{ i\alpha } }{ 1-q{ e }^{ i\alpha } } =q\left( \frac { \cos { \alpha -q } }{ 1-2q\cos { \alpha +{ q }^{ 2 } } } +i\frac { \sin { \alpha } }{ 1-2q\cos { \alpha +{ q }^{ 2 } } } \right) } $$ where $$u=q\frac { \cos { \alpha -q } }{ 1-2q\cos { \alpha +{ q }^{ 2 } } } ,v=\frac { q\sin { \alpha } }{ 1-2q\cos { \alpha +{ q }^{ 2 } } } $$

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To be more clearly understant the last part $$\frac { q{ e }^{ i\alpha } }{ 1-q{ e }^{ i\alpha } } =q\frac { \cos { \alpha +i\sin { \alpha } } }{ 1-q\cos { \alpha -iq\sin { \alpha } } } =q\frac { \cos { \alpha +i\sin { \alpha } } }{ \left( 1-q\cos { \alpha } \right) -iq\sin { \alpha } } =q\frac { \left( \cos { \alpha +i\sin { \alpha } } \right) \left( \left( 1-q\cos { \alpha } \right) +iq\sin { \alpha } \right) }{ \left( \left( 1-q\cos { \alpha } \right) -iq\sin { \alpha } \right) \left( \left( 1-q\cos { \alpha } \right) +iq\sin { \alpha } \right) } =\\ =q\frac { \cos { \alpha -q\cos ^{ 2 }{ \alpha +iq\cos { \alpha } \sin { \alpha +i\sin { \alpha } -iq\sin { \alpha } \cos { \alpha -q\sin ^{ 2 }{ \alpha } } } } } }{ 1-2q\cos { \alpha +{ q }^{ 2 }\cos ^{ 2 }{ \alpha +{ q }^{ 2 }\sin ^{ 2 }{ \alpha } } } } =q\frac { \cos { \alpha -q+i\sin { \alpha } } }{ 1-2q\cos { \alpha +{ q }^{ 2 } } } =\\ =q\left( \frac { \cos { \alpha -q } }{ 1-2q\cos { \alpha +{ q }^{ 2 } } } +i\frac { \sin { \alpha } }{ 1-2q\cos { \alpha +{ q }^{ 2 } } } \right) $$