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JH
Numerade Educator

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Problem 14 Medium Difficulty

Find at least 10 partial sums of the series. Graph both the sequence of terms and the sequence of partial sums on the same screen. Does it appear that the series is convergent or divergent? If it is convergent, find the sum. If it is divergent, explain why.
$ \displaystyle \sum_{n = 1}^{\infty} \left( \sin \frac {1}{n} - \sin \frac {1}{n + 1} \right) $

Answer

The series converges to sin 1

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Video Transcript

let's find these ten partial sums for the Siri's. So, for example, we could do s one. As to all the wayto s ten, these will be the first ten partial sums. For example, we can do those ten. So here are all approximate these ten values by just putting them in a graphing calculator and then labeling the co ordinates. So here you could see in the graphing calculator Dez Moe's I have my a N terms sign one over it, minus sign one over and plus one just like we have in the formula and then here and read. I have the first ten partial sums The force co ordinate tells you the number of the partial some and then the right coordinate is the sum. So, for example, as one is point three six two approximately and then s ten is approximately point seven five o seven. So if you would like to write these down, you could just go ahead and pause the video and record these values. But that does take care of the first part of this problem. That's the first ten values of the Siri's and then we're already halfway through the second part well, graph both the sequence of terms. That's the sequence of a ends where AM is given by the sine formula over here. And we'LL also graft the sequence of partial sums which we've already done So in red Still, we still have our terms. Let me go ahead and remove the labels there so we could just see the graph. And then now I'LL go ahead and graph and purple the ends So a one a two all the way down to a ten I'll temporarily label these numbers. If you like to record those for the purple sequence the ends. So let me remove that graph that the label Excuse me. So the question is, is whether the red sequence converges. We see the purple sequence does go to zero so we can apply the diversions test. So maybe here before we make a guest, let's go ahead and increase the number of terms in the sequence. So if, for example, I could goto thirty or I could even go, for example to fifty, and it looks like this red sequence of partial sums will converge to something less than one, but bigger than point five now Let's go ahead and verify this. So based on the graph, my guess is that it? Convergence? Yeah. Now I'LL go ahead and evaluate the sum because the question here the last part is if it's conversion, find the sum So now my since I'm claiming you converges I need to back this up by evaluating the sun and we should get out a real number So what I'LL do here is basically set myself up for telescoping Siri's. So let me write This is the limit of partial sums Since then of writing the whole limit the the whole sum I just ready It is a limit of partial Then I have signed one over and minus sign one over in plus one So what I'LL do here is before taking the limit, I'Ll just evaluate this term in the apprentices by using the telescope in method. So here, So that's equal to this. And then now we're going down here to save some room. The limit is kay goes to infinity Now we could start evaluating the sun, so we have signed one minus sign too. Excuse me. Sign one half and then minus sign one has minus sign Wonder. So that should be a plus there, getting a little careless here. Sorry about that. The next term sign one of the three minus. Sign one over four. And I will keep going in this direction and let me plug in a few more terms. There. The end. So let me write down three terms here as well. So the very last term, it will be sine of one over. Okay, minus sign one over. Kaye plus one. Yeah, the term right before that sign. One overcame, minus one and then minus. Sign one over. Okay. And one more term in this direction. Sign one over. K minus two minus. Sign one. Overcame minus one. Now let's see how much we could cancel. First of all, I noticed that this sign one will not cancel with any term, because there's not a negative sign one there. So when I go ahead and write the next line here, which I will have to go to the next page, I will write sign of one. Still, because it never canceled. However, sign of one half is negative here, cancels with a positive negative sign. Wonder positive sign. One third and about her right? In the next term, I would get a positive sign one over for it'LL cancel with his negative sign longer before and so on. Now going to the other extreme You notice this very last term has the cape plus one in the denominator There's not another term to cancel with it. So that's Herm will remain in the final answer. And don't forget the negative out in the front, However, for all the other denominators So here one over k cancels with the negatives one over K minus one cancels with the negative. And if I want to write another term in this direction and I would see that the sign of one over came on, his two were also cancel. So this is I've done the telescoping. That's what this method down here is the cancel, all the successive terms. And now I'm left over with the limit and I have signed one minus side won over Kaye plus one, and that I'm taking the limit as Kay goes to infinity. Now we're finally ready to deal with the limit. This first term here doesn't happen. Okay, so when I take the limit is still sign of one. However, the seconds from here since sinus continuous. If you want, you could go out and you want to show the extra work here. So the reason I'm allowed to do that essence sign is continuous loops. Seiple there since sign is continuous, that's what allows me to rewrite the limit inside of the science. I pushed it on through inside, and now I see that this limit, the denominator gets big this close to zero. So I have signed one minus sign zero. However, I know that sign of zero is zero. So I get my final answer of sign one. So the Siri's converges. And after using the telescope method, we showed that a converges to sign one, and that's our final answer.