Convergence of $\sum_{n=1}^\infty n^2 \sin(\frac{π}{2^n}) $ This is my series: $$\sum_{n=1}^\infty n^2 \sin(\frac{π}{2^n}) $$
WolframAlpha says it converges, but I have no idea how to get the answer. I have learned comparison test, ratio test, root test and integral test. I don't really know which one of those to use. So far the only decent option seems the regular comparison test.
$ \lim \limits_{x \to \infty} \frac{a_n}{b_n} = c, c \ne 0, c \ne \infty$ 
I tried something taking an geometric series for $b_k$ (like $\frac{1}{2^n}$ ) to get: $ \lim \limits_{x \to \infty} \frac{a_n}{b_n} = \lim \limits_{x \to \infty} n^2 \sin(\frac{π}{2^n}) 2^n = \lim \limits_{x \to \infty} n^2 \frac{\sin(\frac{π} {2^n})}{\frac{π} {2^n}} \frac{π} {2^n} 2^n = \lim \limits_{x \to \infty} n^2 \frac{\sin(\frac{π} {2^n})}{\frac{π} {2^n}} π$ 
But that still comes to infinity. If i use harmonic series to get rid of infinity (n^2), I can't get rid of the 0 from sinus and if I use geometric series it's vice-versa.

The following inequality holds:
$$
\sin x \le x\qquad(x\ge 0)
$$
Then
$$
0\le \sin\left(\frac{\pi}{2^n}\right)\le \frac{\pi}{2^n}.
$$
