<s>[INST] 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?
 [/INST] 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 $k$ 
$$x_n(k)=\sum_{i=1}^n\frac{\sin (i)}{k^i}=\frac{k^{-n} \left(\sin (n)-k \sin (n+1)+k^{n+1} \sin (1)\right)}{1+k^2-2 k \cos (1)}$$ 
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