# Show that the series converges absolutely $\sum \frac{\sin n\theta}{2^n}$

Show that the series converges absolutely $$\sum \frac{\sin n\theta}{2^n}$$

Considering $a_n = \frac{\sin n\theta}{2^n}$ , $\forall n \geq 0$. We have $|a_n| = \frac{|\sin n\theta|}{2^n}$ where $0 \leq \sin n\theta \leq 1$, .

Considering $b_n = \frac{1}{2^n}$, we have $\int_0^{\infty} 2^{-n} < \infty$. It follows that $\sum b_n < \infty$

As $|a_n|,b_n \geq 0$, and $|a_n| \leq b_n$, by the comparison test $\sum|a_n|<\infty$ since $\sum b_n< \infty$.

Hence, $\sum \frac{\sin n\theta}{2^n}$ is absolutely convergent.

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Is this the correct argumentation for absolute convergence? much appreciated for the input.

• Does it make sense to say that the integral is zero? – user296602 Aug 9 '17 at 6:39
• And how is the integral even related to your purpose? – Did Aug 9 '17 at 6:42

You can directly use the comparison test and the fact that the term $\frac{1}{2^n}$ generates a geometric series.
Let $a_n = \frac{\sin {n \theta}}{2^n}$ and $a_n = \frac{1}{2^n}$. Then, using the argument you have presented, it is clear that,
Hence, the comparison test for absolute convergence applies if the series generated by $b_n$ is convergent. But this is clear, since,
Hence $\sum_{n=1}^{\infty} \frac{\sin{n \theta}}{2^n}$ is convegenet.