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JH
Numerade Educator

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Problem 13 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 = 2}^{\infty} \frac {2}{n^2 - n} $

Answer

2

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Video Transcript

let's find at least ten partial sums of the Siri's since the SYRIZA starting into notice why it's not starting at one. Because here, if the denominator would be zero. So here, if we'd like to find ten sons, we can do as to as three all the way up to s eleven. That's ten partial zones. So let's go ahead and approximate these partial sums with a calculator. So here I am in decimals calculator. You could see my N two over and squared minus end. And then here I'm doing the partial sums from two all the way up to eleven. So here there we go. We see up to eleven here. So this is telling us that as two is about one point three three three three as threes, about one point five s, fours about one point six and going all the way up to S eleven, which is about one point a three three three. So that takes care of this first part. That's the first ten values of the partial sons. And then for the second part, which we're already halfway done with. Let's graft both the sequence of terms. So that's the ends So here. That's an two over and squared minus end. And then we'LL also graph as to obsess eleven on the same screen. So now let me remove these numbers. So the red graph is the graph of the partial sons, and then the following graph will be the sequence of terms. So the purple graph is the sequence of a one. Er excuse me, a two a three a four all the way up to eleven. Based on the picture, it looks like they ends. We're getting closer to zero, and it looks like the sums may start converging. If we really want to be sure, we could try to increase in and then maybe zoom out here a little bit. And it looks like the sums, maybe converging to around two. And here we can see that the limit of the and zero. So let's go ahead and check if the answers to So based on the graph, my guess is that it converges. And then let's actually show why in its conversion. So if it says that if it's convergent excuse me well, actually, go out and find the song. So here we have two over and square minus end. Let's re write this doing partial fraction. So we were basically going to set up a telescoping some. So we have two equals a and minus one plus B end. So a plus B and minus A. So we see a plus B equals zero negative A equals two. So that means a is minus two is minus two and then be equals negative. A equals positive too. So this becomes negative two over and plus two over and minus one. So let's go on to the next page. And then we're doing the sum from n equals two to infinity. So here, this looks like we can do the telescoping method that was introduced in eleven point two. So the method here is the first rewrite the summit as a limit. So here, let me do the limit, as is. Kay goes to infinity and then we do the partial some s k so that partial sums from to decay negative two over end, two over and minus one. So now we'LL ignore the limit for a moment. And let's just focus on this partial, some on the inside. So come down here to make some room. Now let's go ahead and evaluate this partial song So we'LL go ahead and start up by plugging in and equals two. That's the first term. Then we plug in and equals three, then four and so on and on the other end Well, we'Ll keep adding terms until we get to the very last one When we plug in and equals K so that'LL be negative to over King Plus two overcame minus one however, will see in a moment that it will be helpful to actually write a few more terms. So here I should do minus one. So that's a minus two down there and then let me do one more turn negative two over K minus two plus two overcame minus three So then we'LL go ahead and see how much cancellation we can do. So I see that this tour for one will not cancel. So I'll write this in my next line I'm gonna cross off a cz much as I can and I'Ll rewrite the things that don't cancel So there's two over one which is just too that'LL State, on the other hand, we see negative to over two will cancel with positive to over two negative to over three cancels with positive to over three and so on. And I'll also be able to cancel the negative two over four if I were to write the next term out. And similarly, if we see this and this can end denominator, there's no other time which k appears in the denominator So this negative to over k will not cancel with anything else. So I should also right in the next line. However, if K is less than excuse me if we look at any denominator less than K, for example came on this Want that cancels okay, minus two that cancels. And if I wanna write another the previous term came on his three were also cancel. So the on ly terms that remain or two over one and negative to over Okay, so we've evaluated the partial some we've done the telescoping to cancel all terms. Now we just go ahead and take this limit As kay goes to infinity. This number doesn't depend on K. So that's two and then we have minus zero because two over cables to zero. So we get a final answer of two. So the Siri's converges and it converges to two. That's our final answer