# Why does $\left|\frac{\sin(n+1)}{2^{n+1}}+…+\frac{\sin(n+p)}{2^{n+p}}\right|\leq\frac{|\sin(n+1)|}{2^{n+1}}+…+\frac{|\sin(n+p)|}{2^{n+p}}$ hold?

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) ?

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.

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 $$N\to +\infty$$, $$\sum_{n\geq 1}\frac{\sin n}{2^n}=\text{Im}\left(\frac{e^i}{2-e^i}\right)=\text{Im}\left(\frac{e^i(2-e^{-i})}{(2-e^i)(2-e^{-i})}\right)=\frac{2\sin(1)}{5-4\cos(1)}.$$

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

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