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

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?

• I can't understand the downvote: the question is nicely and well redacted put. The OP shows a serious self effort and work....so what's the problem? Perhaps someone thinks the OP must be right otherwise he must be downvoted?! But then what to ask for? – Timbuc Nov 25 '14 at 14:21
• @Timbuc: After flagging a dozen questions questions per day from people who do nothing more than cut-and-paste their homework problems, I'm happy to upvote anyone who puts in a bit of effort. – anomaly Nov 25 '14 at 14:55

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

and the series converges absolutely and thus converges.

• Oh sure, the absolute value is what's been missing. – Shemafied Nov 25 '14 at 14:10
• And if $n = 3,$ or and multiple of $3$, we have $\cos 2\pi = 1$. I think you want to bound the numerator to get your result. The $n$th root will still evaluate to $1$, but as written, the first equality is incorrect. – Namaste Nov 25 '14 at 14:11
• @amWhy I guess that's what the absolute value brackets are for :) – Shemafied Nov 25 '14 at 14:14
• @amWhy is right, and if I were applying the original test, I should be taking the lim sup of that...and the outcome is the same, of course. Thanks. – Timbuc Nov 25 '14 at 14:16
• Absolute value signs here were assumed to make the \cos always evaluate to positive $1/2$, which is incorrect. Of course, the absolute value sign restricts $1/2 \leq |\cos(2n\pi/3)|\leq 1$. You're welcome Timbuc! – Namaste Nov 25 '14 at 14:17

Hint

Use the comparison with geometric series

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