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I trying to understand a proof (using Cauchy's general criterion of convergence) of why the series $\sum_{n=1}^{\infty }\frac{\sin (n)}{2^{n}}$ converges . At the beginning, the following inequality is expressed: $$\left | \frac{\sin (n+1)}{2^{n+1}}+...+\frac{\sin (n+p)}{2^{n+p}} \right |\leq \frac{|\sin (n+1)|}{2^{n+1}}+...+ \frac{|\sin (n+p)|}{2^{n+p}}$$ where $n,p$ are natural numbers.

Why does this hold? Is the triangle inequality with more than 2 terms on $\mathbb{R}$ a valid fact (from what seems to be the case here) ?

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Yes. You can deduce the three-term version by using the two-term version twice:

$$|x+y+z|\leq |x+y|+|z|\leq |x|+|y|+|z|$$

and you can similarly prove it for any number of terms by induction.

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You can directly use geometric sums: $$\sum_{n=1}^{N}\frac{\sin n}{2^n}=\text{Im}\sum_{n=1}^{N}\left(\frac{e^i}{2}\right)^n = \text{Im}\left[\frac{e^i}{2-e^i}\left(1-\frac{e^{Ni}}{2^N}\right)\right]$$ and since $\frac{e^{Ni}}{2^N}\to 0$ as $N\to +\infty$, $$\sum_{n\geq 1}\frac{\sin n}{2^n}=\text{Im}\left(\frac{e^i}{2-e^i}\right)=\text{Im}\left(\frac{e^i(2-e^{-i})}{(2-e^i)(2-e^{-i})}\right)=\frac{2\sin(1)}{5-4\cos(1)}.$$

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$\left | \frac{\sin n }{2^n} \right |\leq \left |\frac{1}{2^n} \right |$

We know $ \sum \frac{1}{2^n} $ is Geometric series with common ratio $\frac{1}{2}$ convergent series

By comparison the given series is also convergent

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