How to calculate $\sum _{n=1}^{\infty }\:\left(\frac{2}{3}\right)^n\sin\left(\frac{\pi }{3}n\right)$ I believe that Complex numbers should be used in order to calculate this.
Let $z = \frac{2}{3}e^{\frac{i\pi }{3}}$,
So,
$$\sum _{n=1}^{\infty }\left(\frac{2}{3}\right)^n\sin\left(\frac{\pi }{3}n\right)=\sum _{n=1}^{\infty }\text{Im}\left(\frac{2}{3}e^{\frac{i\pi}{3}}\right)^n=\text{Im}\left[\sum _{n=1}^{\infty }\left(\frac{2}{3}e^{\frac{i\pi}{3}}\right)^n\right]$$
Is this correct?
How do you go about solving it further? I believe this might be a infinite geometric series where,
$a_1 = z, q = z$?
 A: It is also possible to do it directly in the real domain.
Since the series is absolutely convergent, we can rearrange terms to group the same values of the sinus.

*

*$n\equiv 0,3\pmod 6\implies \sin(\frac{n\pi}3)=0$

*$n\equiv 1,2\pmod 6\implies \sin(\frac{n\pi}3)=\frac 12\sqrt{3}$

*$n\equiv 4,5\pmod 6\implies \sin(\frac{n\pi}3)=-\frac 12\sqrt{3}$
Let $a=\frac 23$ then :
$\displaystyle\begin{align}\frac{2S}{\sqrt{3}}
&=\sum\limits_{n=0}^\infty\left(\frac 23\right)^{6n+1}+\sum\limits_{n=0}^\infty\left(\frac 23\right)^{6n+2}-\sum\limits_{n=0}^\infty\left(\frac 23\right)^{6n+4}-\sum\limits_{n=0}^\infty\left(\frac 23\right)^{6n+5}\\\\
&=(a+a^2-a^4-a^5)\sum\limits_{n=0}^\infty {(a^6)}^n\\\\
&=(a+a^2-a^4-a^5)\,\frac{1}{1-a^6}\\\\
&=\frac{a}{a^2-a+1}=\frac 67
\end{align}$

Therefore $S=\dfrac{3\sqrt{3}}{7}$
A: Indeed so:
$$\sum_{n=1}^\infty \left( \frac 2 3 z \right)^n
= \frac{2z/3}{1-2z/3} = \frac{2z}{3-2z}$$
You can justify this as $|2/3|<1,|e^{i \pi /3}| = 1$, so their product satisfies $|2z/3| < 1$ and you may use the infinite geometric series result.
Note that you can express $z = e^{i \pi /3}$ by Euler's formula as
$$z = \cos \frac \pi 3 + i \sin \frac \pi 3 = \frac{\sqrt 3}{2} + \frac 1 2 i$$
Substitute into the series' result, and rationalize to get a more standard form of answer (one in the style of $a+bi$ for $a,b$ real): then taking the imaginary part should be easy.
