Let $z=\cos\theta + i\sin\theta$. Then \begin{align} &\Big(\cos\theta + \frac{1}{2}\cdot\cos 2\theta + \frac{1}{4}\cdot\cos 3\theta\Big) + i\Big(\sin\theta + \frac{1}{2}\cdot\theta + \frac{1}{4}\cdot 3\theta + \cdots \Big)\\ =&\quad z + \frac{1}{2}\cdot z^2 + \frac{1}{4}\cdot z^3 + \cdots \\ =&\quad \frac{z}{1-\frac{1}{2}\cdot z}\\ =&\quad \frac{2z}{2-z}\\ =&\quad \frac{2(\cos\theta + i\sin\theta)}{(2-\cos\theta) -i\sin\theta}\\ =&\quad \frac{2\Big(\cos\theta + i\sin\theta\Big)\Big((2-\cos\theta) +i\sin\theta\Big)}{\Big((2-\cos\theta) -i\sin\theta\Big)\Big((2-\cos\theta) +i\sin\theta\Big)}\\ =&\quad \frac{(\dots)+i(4\sin\theta)}{5-4\cos\theta} \end{align} Where $(\cdots)$ is an expression of $\theta$ which does not involve $i$. Therefore we conclude $$\sin\theta + \frac{1}{2}\cdot\sin 2\theta + \frac{1}{4}\cdot\sin 3\theta + \cdots=\boxed{\frac{4\sin\theta}{5-4\cos\theta}}$$