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Determine whether the series convergent or divergent by expressinglescoping sum (as in Example 8).n(nconvergentdivergentconvergent, find its sum (If the quantity di...

Question

Determine whether the series convergent or divergent by expressinglescoping sum (as in Example 8).n(nconvergentdivergentconvergent, find its sum (If the quantity diverges_ enter DIVERGES )Need Help?Suqmllt AnswerPractice Anothor Version1/1 POINTSPREVIOUS ANSWERSSCALCETBM 11.2.052.Mi_

Determine whether the series convergent or divergent by expressing lescoping sum (as in Example 8). n(n convergent divergent convergent, find its sum (If the quantity diverges_ enter DIVERGES ) Need Help? Suqmllt Answer Practice Anothor Version 1/1 POINTS PREVIOUS ANSWERS SCALCETBM 11.2.052.Mi_



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Determine whether the series is convergent or divergent by expressing $ s_n $ as a telescoping sum (as in Examples 8). If it is convergent, find its sum.
$ \displaystyle \sum_{n = 2}^{\infty} \frac {1}{n^3 - n} $

Let's determine whether the Siri's converges or diverges by raiding the SN. This is recalled the end partial some In our case, since we're starting at two, this would just be a too all the way up to an So we'LL write. This thing is a telescoping some as they shone the examples in the sex book. And then, if it's conversion, will actually go ahead and find the sum. So here, before we do anything regarding the sum, let's just look at the A n here. So we have to Let's go ahead and factor that denominator. And then here we would have to do partial fractions so we would have to solve this equation here for and be We can multiply both sides by the denominator on the left and then you get a and minus one B and plus one. So here you been solved This for A and B and you end up with negative one for a one for Bea. So I'll need more room here to evaluate the sum, so I'll need to go to the next page here. R Sum is from two to infinity and then we just used a partial fraction So we have won over and minus one minus one over and plus one. This is after the partial fraction decomp then and of course, the witnesses are in value. Then we could write. This is a limit of S K. So now this is where the telescoping will happen. So if we won't worry about the limit till the very end, let's go ahead and deal with everything inside the Red Prentice's So this is just a finite sum here. So let's write this over here. So we start off by plugging in and equals two. So this is an equals to here. Then you plug in and equals three plugin and equals four. Maybe we'LL plug in one more in this direction if you plug in and equals five and we would keep adding in that direction. And then we will also plug in the last few terms like the very, very last. Sir, will you plug in K? We see this will show up. So this is for an equals K right before that is when you plug in K minus one and then even before that. So it's best to have to really see all the cancellation that will ever happen. If you want to be more sure, it's best to write as many terms as you can the first few terms and also not just the very last term K, but also a few terms before that. So here I am writing to more terms before this, and this is just so that I can hope that I can figure out how much cancelation there is. So let's look over here. We see that there's what's left over when we started canceling, we have a one that doesn't cancel with anything, and then do you see that there's a positive on? Happier. But there's never a negative one half because these air these negatives start at one over three, then one of our four one over five. That denominator is going to keep growing, so there's no negative one half, so the one in the one half will both stay in there. However, this positive one three cancels with the negative. The positive one four cancels with the negative, and this negative one six will eventually cancel with a positive one over six. Now it's called the Otherside involving the case. So here we see that this minus one over Kaye plus one does not cancel with anything because all the positive denominators they never reach Cape Plus one. They they only go up to K minus one. Similarly, one over. Okay, this term over here for the same exact reasoning. This negative will never cancel. Because if you look at the positive denominators they only go up to came on this one, however, everything under that which means we started looking at K minus one. Daniel, sir, getting your cancellation again and then came on this two would even cancel with the term corresponding to n equals K minus three. So everything else will cancel. And when we take the limit is kay goes to infinity. One over K goes to zero one of our K plus one goes zero. So we just have one plus a half, which is three over to sew the Siri's converges. And we recall there was two parts to this problem. We determined that it converges. And then if it is, it does converge. We were supposed to find the sum, and we just did that. And we have three over two, and that's our final answer.

Let's determine whether the Siri's is conversion or diversion by using the telescoping song. So hear what we'LL do is rewrite this summation as a limit. So instead of looking at the whole sum all at once, let's on Lee Look at that's two s k. So remember, But this is by definition of the Siri's. It's just a limit of the partial sums and you let the index go to infinity. Now, in our case, we have Lim kay goes to infinity s k means the sum from our starting point, which is one all the way up to Kay and then we have Ellen and then end over and plus one in this case is actually going to be better if we want to telescope usual odd properties to rewrite. This is natural log of end minus Ellen of plus one. So this these are the terms that were adding up. So before we take the limit, let's ignore that for a moment and on Lee, focus on what's inside the apprentices. And then after we simplify, we'LL come back over here and take the limit. So me come down here, Lim Now let's go ahead and right this some oak. So you start with n equals one. We have Ellen one minus ln of one plus one. Ellen to minus ln three Ellen three minus Ellen for and so on. We see the pattern and then we keep going. And for example, when we plug it K minus one. And then finally you were plugging and equals K. And let's go ahead and try to simplify this. Well, I want of one. We know that zero and then we have natural log up to, but that cancels with positive. Eleanor too. Negative, Ellen on three Positive, Eleanor, three and so on. All the negatives will cancel out with a positive even all the way up into you. Getto negative, natural lot of kay. Here's positive, natural on cocaine And even this term right here. This would cancel with the term right before it, because the term right before it would be Ellen of K minus two, minus Ellen of K minus one. And then that would cancel with this. So everything cancels except this very last term here with the negative sign. So we have let lim Kay goes to infinity of negatives. Natural log of K plus one. But we know from the O. R picture of the natural log that as thie input is larger, the graph goes to infinity. We have a negative sign here, so this is going to a negative infinity. So the limit is not a real number. And since the limits not a real number, this Siri's is diversion. Any time the limit is infinity minus infinity or even if the limit doesn't exist in any of those cases, we say that the syriza's divergent and that's our final answer by using the telescoping method from example eight.

Let's determine whether the Siri's converges air diversions by expressing SN as a telescoping some. And then, if it converges, will go and find that summer's well. So, SN recalled, by definition is just the sum of the first and numbers where your a N is this thing that's being added up. So let's just go ahead and write. This is a limit. Kay goes to infinity, and then we have the sum from n equals one. Take a hee won over in minus E one over and plus one. Now this term here in the apprentices, this whole sum this is S K. That's the sum from one all the way of Tien, she says. It's defined. So let's telescope. Let's do a telescoping, some on ly inside of the prophecies. And then after that, we'LL come out here and take the limit, and that should answer our question. So let's write this. So now let's go ahead and start adding, so plug in and equals one first. That's our starting point. So eat to the one over one minus E to the one over two. This's the first term after plugging in and equals one and then next you increase and buy one. So and is now too. It's an equals two there, maybe one more term in this direction. So hee So now industry. So plug that in and you might see some cancellation happening there already, so we would keep going in this direction. So put some dots there to indicate that. And then eventually, at the very end, when we plug in and equals care would have ete to the one over K minus. Eat the one over Kaye plus one. But in order, Tio, maybe help you see the pattern here, it might be best to write a term before n equals K. So the term right before the last one came in this one and by writing that you could see some more cancellation over here, the e to the one over K will cancel with negative eats of the one of Okay, so now let's keep see how much we could cancel each of the one over one. This doesn't cancel with anything because all the remaining denominators are larger than one. Here we have negative eats of the one half that'LL cancel with the positive each other one half more cancelation there. If I wanna keep writing, I would see that e to the neck. Negative. Each of the one over four would cancel, and I would keep getting cancellation all the way until I get to eat to the one over. Okay, that will cancel. So this means this came on this one would also cancel with the term right before, however, is I go all the way to the end. I see that I have each of the one over Kaye plus one. And that won't cancel because all the other denominators are less. Think a plus one. So everything is canceled except eat of the one and then negative e to the one over. Kaye plus one saw me write that on the next page. And then we had inside the apprentices each of the one. What should just eat. However, as we take care to go to infinity this term here, that's just e. But this term over here becomes one over infinity, which is zero. And so in this case, if we go out and plug that in, we just have e minus one. And that's your final answer.

Let's determine whether the series is convergent or divergent by expressing the end partial some SN recall sn just the some from some starting point in this case, which is one. So we must start at one we go up to end and then three over N N Plus three. Well, express. This guy is a telescoping some and then determine whether it converges or diverges. So let's do that. So first two steps here, I'll do it once. The first step will be to write this as a limit of the partial sons. The next part will be to use partial fraction decomposition to rewrite this. So from our earlier chapter, we know that this can be written as like a over n plus B over n plus three. So one would have to do a bit of work here just to find A and B. It doesn't take too long, and it turns out that a is one and that be is minus one. So that gives me this expression here and now it's more clear that I'll be able to telescope. Let me switch up the color here. So now I'll rewrite the limit. The limit will just come along for the ride until the very last part of the problem. So right now, let's only focus on the inside part of the summation and let's expand the some. So go ahead and plug in and equals one and equals two. So here, because there's a gap between the denominators of size three, the more terms you right out here, the higher your chances will be to see the cancellation pattern and what terms remain, what terms get removed and so on. So let me write five terms in this direction so that it becomes more clear what will cancel. And with that said, I'll also right five terms in the other direction. So that would be after this line. So the very last one when you plug in the end or this case K excuse me one over K minus one over K plus three before that, men before that before that and one more time before that. So notice that the denominator being subtracted is always three larger than the positive denominator in all cases, you can see that that is a true statement. Now let's see what will cancel what will be left over in the next line, let me circle what will stay. That will stay because there's no negative one. The reason there's no negative one is because the smallest negative denominators is for for the same reason. 1/2 will stay. 1/3 will stay. But once I get to four, it will start canceling out with the negatives. So 1/4 cancels 1/5 cancels, and if I were to write one more term, I would have a positive 1/6 that would cancel with this negative 1/6. So the only numbers so far is just 11 half and one third. Now let's go to the other side to see how much cancellation we really have. So looking at the one over K plus three, I see that this will stay because the largest positive denominator is K for the same reason K plus two will stay and then negative one overcame plus one will stay. But once I get to negative one over K, that will cancel with one over K negative one over K minus one cancels with one over K minus one and so on in that direction. So we can see the positive numbers are just one a half and a third. And then we subtract one over K +11 over K plus two and one over K plus three. So let's go to the next page. We still have a limit. Okay, goes to infinity. One a half. These were the positive numbers. And then we subtracted one over K plus one one over K plus two and one over K plus three. When we take the limit, those last three fractures just go to zero because the denominators get really large, whereas the numerator is just one. Yeah, so we just add some fractions here so we could write that one in 6/6. We can write the one half 3/6 a third to over six, and then add that together we get 11/6. So let me get a little sloppy there, so it converges to 11/6. There's two answers here. The first question was whether it converges or diverges. We show that it can verges. And then the second part of the question was, If it does converse, what is it Converse to? We just found that out. It's 11/6. That's our final answer


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