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?
 A: You could be interested to notice that quite simple manipulations of the sines of multiple angles lead to  $$x_n=\sum_{i=1}^n\frac{\sin (i)}{2^i}= \frac{2^{-n} \left(\sin (n)-2 \sin (n+1)+2^{n+1} \sin (1)\right)}{5-4 \cos (1)}$$ Similarly $$x_n=\sum_{i=1}^n\frac{\sin (i)}{3^i}=\frac{3^{-n} \left(\sin (n)-3 \sin (n+1)+3^{n+1} \sin (1)\right)}{10 - 6 \cos (1)}$$ $$x_n=\sum_{i=1}^n\frac{\sin (i)}{4^i}=\frac{4^{-n} \left(\sin (n)-4 \sin (n+1)+4^{n+1} \sin (1)\right)}{17-8 \cos (1)}$$ 
If fact, for any value of $k$ 
$$x_n(k)=\sum_{i=1}^n\frac{\sin (i)}{k^i}=\frac{k^{-n} \left(\sin (n)-k \sin (n+1)+k^{n+1} \sin (1)\right)}{1+k^2-2 k \cos (1)}$$ 
A: 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!!!
A: Consider $\displaystyle|x_n-x_m|=|\sum^{n}_{k=m+1} \frac{\sin k}{2^k}|\le |\sum^{n}_{k=m} \frac{1}{2^k}|$, but the last term can be made less than $\epsilon$ since it is seqeunce of partial sums of convergent series $\displaystyle\sum^{\infty}_{k=1} \frac{1}{2^k}$
