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Given the following series

$$\sum_{k=0}^\infty \frac{\sin 2k}{1+2^k}$$

I'm supposed to determine whether it converges or diverges. Am I supposed to use the comparison test for this? My guess would be to compare it to $\frac{1}{2^k}$ and since that is a geometric series that converges, my original series would converge as well. I'm not all too familiar with comparing series that have trig functions in them. Hope I'm going in the right direction

Thanks

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You have the right idea, but you need to do a little more, since some of the terms are negative. Use your idea and the fact that $|\sin x|\le 1$ for all $x$ to show that

$$\sum_{k\ge 0}\frac{\sin 2k}{1+2^k}$$

is absolutely convergent, i.e., that

$$\sum_{k\ge 0}\left|\frac{\sin 2k}{1+2^k}\right|$$

converges.

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  • $\begingroup$ Thanks for the response. So to clarify, I should do comparison test with $$\sum_{k\ge 0}\left|\frac{\sin 2k}{1+2^k}\right|$$ first and since that is convergent, my original is convergent as well? $\endgroup$ – trungnt Jul 6 '15 at 5:21
  • $\begingroup$ @trungnt: Yes, exactly. $\endgroup$ – Brian M. Scott Jul 6 '15 at 5:24
  • $\begingroup$ @trungnt: You’re welcome! $\endgroup$ – Brian M. Scott Jul 6 '15 at 5:28
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Yes, you can compare with geometric series. Thus you will show that your serie converge absolutely.

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$\sum_{k\geq 0} \left| \frac{\sin(2k)}{1+2^k} \right| \leq \sum_{k\geq 0} \left| \frac{1}{2^k} \right| = 1$

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Use asymptotic analysis: $$\Biggl\lvert\frac{\sin 2k}{1+2^k}\Biggr\rvert=_{\infty}O\Bigl(\frac1{2^k}\Bigr)$$ hence the series is absolutely convergent.

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