Convergence of 1 over sine serie, a hard one I want to find the radius of this series, I tried my best I proved that r<1.
$$ \sum _ { n = 1 } ^ \infty \frac { x ^ n } { \sin ( n \theta ) } $$
is convergent for $ \theta = t \pi $ such that $ t $ is irrational, e.g.  $ \theta = \sqrt 3 \pi $.
It is a really interesting question because it will tell something about how the $ \sin \left( \sqrt 3 \pi n \right)$ goes to zero for some sequence goes to infinity (since $ \sqrt 3 $ is irrational sin will have a dense image in $ [ 0 , 1 ] $).P
 A: tl;dr
$r=1$ for $t=\sqrt3$ and pretty much any other number you've ever heard of.Then there are just as many "bad" numbers for which $r<1$ and can be anything, all the way down to $0$.
Now come the details.
Roth's theorem basically says that an algebraic number $\alpha$ can't be approximated by the rationals "too well" - that is, $\left|\alpha-{\frac {p}{q}}\right|>\frac1{q^{2+\varepsilon}}$ for all but finitely many $\frac pq$. With that in mind, we have:
$$\left|{1\over\sin n\theta}\right|=\left|{1\over\sin(n\theta-p\pi)}\right| \leqslant{\pi/2\over|n\theta-p\pi|}={n\cdot const\over\left|{\theta\over\pi}-{p\over n}\right|}\leqslant n^{3+\varepsilon}\cdot const$$
and of course the series $\sum x^nn^4$ (here I chose $\varepsilon=1$) has the radius of convergence equal to $1$.
The same goes for all other numbers with finite irrationality measure.
Now the "bad" part. One can construct numbers with infinite irrationality measure, and they (though not even all of them) can make our radius a bit shorter. Say we have $x={1\over2}$; to break the convergence, we need $\left|{1\over\sin n\theta}\right|>2^n$ for infinitely many $n$. Can that be done? Why, that's easy. We know that for the continued fraction convergents of $\alpha$, $\left|\alpha-{p_n\over q_n}\right|<{1\over q_nq_{n+1}}$. Well, just make that $q_{n+1}$ bigger. How big can it be made? As big as you wish and even bigger. Can it be greater than $2^{q_n}$? Sure, and the extra topping is on the house.
So it goes.
