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JH
Numerade Educator

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Problem 31 Medium Difficulty

Determine whether the series converges or diverges.
$ \displaystyle \sum_{n = 1}^{\infty} \sin \left( \frac {1}{n} \right) $

Answer

DIVERGES

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Video Transcript

let's use the Lim comparison test to determine whether this given series convergence. So if I will use this test, I should call this a M. And then I need some other term being to compare with Ann. And here, just by a little luck, maybe by looking inside of the princess's we just take the end to be won over it. But maybe you're not so lucky. Waited to see why this is a good choice for being, and it gets really, really big sign of one over end and one over and really close together. This is just using the facts that sign and extra approximately equal when x a small, and this could be shown more rigorously. So here, let's look at the limit of a and over bian. That's just the limit as n goes to infinity of sign of one over it over one over. And and instead of simplifying this, let's just create another letter here. Let's call them to the one over, and that I can replace this notice that as n goes to infinity, that's a global in tow. M going to zero because one over infinity a zero and then I could write this a sign over and and this limit, which you might remember from your taking derivatives and limits You could use local tiles rule here, but sine X over X goes toe one in the limit. Therefore, we can use the Lim comparison test. So instead of looking at our Siri's, we just have to look at this series here, and we know the serious that verges. This is the well known harmonic series, or you could just say that it diverges because it's a piece. Areas with P equals one. So by the pee test, this is from eleven point three with p equals one, therefore by Lim comparison, which we're allowed to use because we have a positive number. That's so first of all, it's bigger than zero and two. It's not infinite part of me. So by Lim comparison, our Siri's also converges deceives me. Also diverges, diverges, and that's our final answer. Why limit comparison