Prove that $\sum_{n=1}^\infty\frac{\sin(nz)}{2^n}$ is analytic on $\{z\in\mathbb{C}:|\operatorname{Im}(z)|<\log(2)\}$ 
Prove that $f(z)=\sum_{n=1}^\infty\frac{\sin(nz)}{2^n}$ is analytic on $A=\{z\in\mathbb{C}:|\operatorname{Im}(z)|<\log(2)\}$ 

I tried expanding $\sin(nz)$ in terms of $e^{inz}$ but that did not help me unless I am doing something wrong.  I know Weierstrass's M-test comes in to play.

Using $\sin nz=\frac{1}{2i}(e^{inz}-e^{-inz})$ works. The constant is irrelevant for the convergence. 
Deal with the two exponentials separately. Let $z=x+iy$. Then $|e^{inz}|=e^{-ny}$, and  $|e^{i((n+1)z}|=e^{-(n+1)y}$. 
Thus, remembering about the $2^n$ in the denominator, we see that the norm of the ratio of two consecutive terms is $\frac{e^{-y}}{2}$.  This norm is $\lt 1$ precisely if $e^{-y} \lt 2$, that is, if $y\gt -\log 2$.
In the same way, for the term in $e^{-iz}$, the norm of the ratio of two consecutive terms is $\lt 1$ precisely if $y \lt \log 2$.  Thus, by the Weierstrass $M$-test, we have analyticity if $-\log 2\lt y\lt \log 2$.
