Show that the series $\sum \frac{\sin \left(\frac{\left( 3-4n \right)\pi }{6}\right) }{2^{n}}$ converges?

Question

Using the addition formula for the sine function I have managed to reduce this to a simpler form: $$\sum \frac{\cos \frac{2n\pi }{3}}{2^{n}}$$ It is obvious here that it passes the n-th term convergence test. But what next? I have applied Cauchy's root test, this is the result: $$\lim_{x\rightarrow \infty }\sqrt[n]{\frac{\cos \frac{2n\pi }{3}}{2^{n}}}$$ For the numerator being a "constant", I have gotten that the limit is $\frac{1}{2}$, which in turn means that the series is convergent. Is my reasoning behind this correct?

Answer 1

Going your way:

$$\lim\sup_{n\to\infty}\sqrt[n]{\left|\frac{\cos\frac{2n\pi}3}{2^n}\right|}=\lim_{n\to\infty}\frac{\sqrt[n]1}2=\frac12<1$$

and the series converges absolutely and thus converges.

Answer 2

Hint

Use the comparison with geometric series

$$ \frac{|\sin \frac{\left( 3-4n \right)\pi }{6}| }{2^{n}}\le\left(\frac12\right)^n$$