It's not hard to show that the radius of convergence of the series $\sum_{n=0}^{\infty} \sin(n)\,x^n$ is $1$, however I'm at a loss for finding a closed form of the series... although Mathematica seems to have no trouble whatsoever:
$$ \sum_{n=0}^{\infty} \sin(n)\,x^n = \frac{-ix(e^{2i}-1)}{2(x-e^{i})(e^{i}x-1)} \\ $$ $$ = \frac{x \sin(1)}{1 + x^2 - 2 x \cos(1)} $$
which oddly enough is real-valued, after simplifying a bit with Mathematica.
How would one go about finding this? I couldn't believe that Mathematica had such an easy time with the series, let alone that there was an actual closed form.
Hint $$\sum_{n=0}^{\infty} (e^{i}x)^n=\sum_{n=0}^{\infty} e^{in}\,x^n=\sum_{n=0}^{\infty} \cos(n)\,x^n+i\sum_{n=0}^{\infty} \sin(n)\,x^n$$