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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 = 1}^{\infty} \frac {3}{n(n + 3)} $

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converges to $\frac{11}{6}$

Calculus 2 / BC

Chapter 11

Infinite Sequences and Series

Section 2

Series

Sequences

Campbell University

University of Michigan - Ann Arbor

Boston College

Lectures

01:59

In mathematics, a series is, informally speaking, the sum of the terms of an infinite sequence. The sum of a finite sequence of real numbers is called a finite series. The sum of an infinite sequence of real numbers may or may not have a well-defined sum, and may or may not be equal to the limit of the sequence, if it exists. The study of the sums of infinite sequences is a major area in mathematics known as analysis.

02:28

In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed. Like a set, it contains members (also called elements, or terms). The number of elements (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence. Formally, a sequence can be defined as a function whose domain is either the set of the natural numbers (for infinite sequences) or the set of the first "n" natural numbers (for a finite sequence). A sequence can be thought of as a list of elements with a particular order. Sequences are useful in a number of mathematical disciplines for studying functions, spaces, and other mathematical structures using the convergence properties of sequences. In particular, sequences are the basis for series, which are important in differential equations and analysis. Sequences are also of interest in their own right and can be studied as patterns or puzzles, such as in the study of prime numbers.

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Determine whether the seri…

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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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