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!

  • 3
    $\begingroup$ 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

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.

  • $\begingroup$ thanks! now the error is fixed =) $\endgroup$ – math student Sep 25 '14 at 3:57
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.