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.


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

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

\begin{align} |a_n| \leq |b_n| \quad \text{for all} \; n \in \mathbb{N}. \end{align}

Hence, the comparison test for absolute convergence applies if the series generated by $b_n$ is convergent. But this is clear, since,

\begin{align} \sum_{n}^{\infty} \frac{1}{2^n} = \frac{1}{1- \frac{1}{2}} = 2. \end{align}

Hence $\sum_{n=1}^{\infty} \frac{\sin{n \theta}}{2^n}$ is convegenet.


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