Find the Sum $\sum_{n=1}^∞\frac{\sin(2nπ/3)}{2^n}$

Question

I found this converges but how can I find the sum of this series: $$\sum_{n=1}^∞\frac{\sin(2nπ/3)}{2^n}$$

Answer 1

Hint:

$$\dfrac{\sin\dfrac{2n\pi}3}{2^n}$$ is the imaginary part of $$\left(\dfrac{e^{i2\pi/3}}2\right)^n$$

$$\sum_{n=0}^\infty\left(\dfrac{e^{i2\pi/3}}2\right)^n$$

$=-1+\sum_{n=0}^\infty\left(\dfrac{e^{i2\pi/3}}2\right)^n$$

$$=-1+\dfrac1{1-\dfrac{e^{i2\pi/3}}2}$$

$$=-1+\dfrac2{2-\cos\dfrac{2\pi}3-i\sin\dfrac{2\pi}3}$$

Rationalize the denominator

$$=-1+\dfrac4{5-i\sqrt3}$$

Answer 2

$\sin(2n\pi/3) =(0, a, -a) $ repeated, where $a=\sqrt{3}/2$.

Therefore the sum is $a\sum_{m=0}^{\infty}(1/2^{3m+1}-1/2^{3m+2}) =a\sum_{m=0}^{\infty}(1/2^{1}-1/2^{2})/2^{3m} =(a/4)\sum_{m=0}^{\infty}1/8^{m} =(a/4)/(1-1/8) =2a/7 =\sqrt{3}/7 $.