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Prove that the following sequence ($x_n$) is convergent:

$$ x_n = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} + ... + \frac{\sin n}{2^n} $$ I have tried to use to the sequence is contractive, but am unable to do so. Any help as to which direction I should head to? Thanks!

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    $\begingroup$ Since you're only interested in proving convergence, I'll confine this (well-known) result to a comment: $$\sum_{n=0}^\infty \frac{\sin n}{2^n}=\text{Im}\left[\sum_{n=0}^\infty \left(\frac{e^{i}}{2}\right)^{n}\right] =\text{Im}\left[\frac{1}{1-e^i/2}\right]=\frac{\sin 1}{5-4\cos 1}$$ which is to say, it's not only convergent but can be explicitly summed $\endgroup$ – Semiclassical Sep 25 '14 at 4:25
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We have $\frac{|\sin \ n|}{2^n} \leq \frac{1}{2^n}$ for all $n$ (because $|\sin x| \leq 1$ for all $x$). The series $\sum 1/2^n$ (it is a geometric series) is convergent. By the comparison test, the series $\sum \frac{|\sin n|}{2^n}$ converges. Then $\sum \frac{\sin \ n}{2^n}$ converges.

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  • $\begingroup$ thanks! now the error is fixed =) $\endgroup$ – math student Sep 25 '14 at 3:57
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    $\begingroup$ This is exactly what I would have said. Points for speed to post. 4 minutes, with all the right formulas. +1 $\endgroup$ – Asimov Sep 25 '14 at 3:57
  • $\begingroup$ Thanks so much for this! Really solve my problem :D $\endgroup$ – Jason Sep 25 '14 at 4:00
  • $\begingroup$ you are welcome $\endgroup$ – math student Sep 25 '14 at 4:14

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