Is the sequence $x_n = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} +\dots + \frac{\sin n}{2^n}$ Cauchy? $$x_{n} = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} + ... + \frac{\sin n}{2^n}$$
I came across this sequence while studying, and while it is convergent, I'm curious as to whether or not it is a Cauchy sequence. If so, what's the proof?

For $n > m > 0$, we have
$\vert x_n - x_m \vert = \vert \sum_{m + 1}^n \dfrac{\sin i}{2^i} \vert \le \sum_{m + 1}^n \dfrac{\vert \sin i \vert}{2^i} \le \sum_{m + 1}^n \dfrac{1}{2^i}, \tag{1}$
since $\vert \sin i \vert \le 1$.  But
$\sum_{m + 1}^n \dfrac{1}{2^i} = \dfrac{1}{2^{m + 1}} \sum_0^{n - m -1}\dfrac{1}{2^i} \le 2\dfrac{1}{2^{m + 1}} = \dfrac{1}{2^m}, \tag{2}$ 
since $\sum_0^\infty \dfrac{1}{2^i} = 2$. Thus by taking $m$ sufficiently large, we may make $\vert x_n - x_m \vert$ arbitrarily small; the sequence is Cauchy.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
