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Use the partial fraction command on your CAS to find a convenient expression for the partial sum, and then use this expression to find the sum of the series. Check your answer by using CAS to sum the series directly.$ \displaystyle \sum_{n = 3}^{\infty} \frac {1}{n^5 - 5n^3 + 4n} $

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$\sum_{n=3}^{\infty} \frac{1}{n^{5}-5 n^{3}+4 n}=\frac{1}{96}$

Calculus 2 / BC

Chapter 11

Infinite Sequences and Series

Section 2

Series

Sequences

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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.

03:53

Use the partial fraction c…

04:16

Use partial fractions to r…

01:57

Express the following seri…

02:02

Use partial fractions to f…

03:23

$53-54$ Use the partial fr…

02:16

01:07

Use partial fractions to c…

06:34

01:20

let's use a computer algebra system to rewrite this home. So the first thing I'LL do is go to Wolfram and I'Ll find a different way to express this sum. So have all from Tab Open. And I see that one way. So let me maybe scroll up here to convince you that we have the right formula so less diffraction. And now one way to rewrite it is given by this first alternate forms. So let's write that out. So we have one over and minus two and minus one and and plus one and plus two now a trick here. But it's a common trick for this type of problem in which will end up telescoping. Let's go ahead and rewrite that one in the numerator using terms from the denominator. So what I mean by that is, for example, let's maybe rewrite this one. First of all, let me go ahead and multiply this by four and I'LL make up for it by dividing by four. Now I'LL rewrite this four as and plus two minus and minus two. The's are both terms from the denominator. We can see that and plus two and minus two so we'LL see in a second why this works out and then we still have that one over four. So let's just right over here the next step. After we write this, the nominator will be to just split up the two fractions. So this is just the sum. So one over for some three to infinity. Then we have N plus two. That's the first fraction we have one more to write. So we will have to write that denominator again this time and minus two up top and you can see in both fractions we will get some cancellation. This is why it's important to make sure you're choosing terms from the denominator up here and this step right here. Both of those terms came from the denominator, and this was what is allowing the cancellations happen. So if we don't use those terms, you most likely won't get cancellation. Here we can cancel the n plus twos. They're canceled and minus twos. Now let's has gone to the next page to rewrite this. We have one fourth, some three to infinity. Yeah, and then we just have the first fraction after we cancelled. And then the second one up top and then in the denominator and minus one and and plus one and then and plus two. So now wishes put parentheses around this to make sure we were adding The whole thing is inside the sun. Now we should rewrite this as a limit. This this's telescoping noticed that as you increase and tio en plus one that the denominator on the left turns into the denominator on the right. So here we will have lots of cancellation. This is the perfect time to do telescoping so full to telescoping. We should rewrite this infinite sum as a limit. A partial, some solace, right? As the Limited's cake goes to infinity of the partial, some from three decay. Now we'LL go ahead and just one more time right down these two fractions and then we're ready to telescope so and minus two and minus one and plus one second function, one over and minus one and and plus one and plus two. Now let's go ahead and value. So let's ignore the limit. For a moment, we'LL rewrite that limit. We're just simplifying everything inside filament. So everything in this large Prentice's including the sum. So let's go ahead. Let's evaluate that some. So plug in and equal three your starting point and try to be as careful as you can hear. So have one times two times three times for minus one over two times three times four times five. That's the first term. Then you increase and buy one. So now you're plugging for and you could see already will have some cancellation Lots of cancellation here if I want to write the next term. And once again I could see cancellation here so everything will be canceling out And then we'Ll keep adding in the other direction And then we'LL have looks like I'm getting a little low here on room But when we get to the very end so here you may want to keep track of this you could pause the screen, but I will have to include the last few part here of the sum on the next page. So going on to the next page, I would keep adding and then I have won over Okay, minus three K minus two. Okay, minus one K minus one over K minus two K minus one K K plus one and then that one more term out for that the very last term when you plug in n equals K down here and we could see once again cancellation and then relapsed left with the very last term. And so here, K minus one K K plus one K plus two, then everything that except the very first term in the last term, would have canceled. So remember this whole time we had pull out the limit and the one over four on the previous page. So we have one over four. Lim can goes to infinity after cancellation. All we're left with is one, two, three, four over here and then recent tracked. Okay, minus one k K plus one K plus two. Take that. Limited's cake goes to infinity. The second fraction will go to zero because the denominator goes to infinity. But the numerator is fixed. That one and then you just have one over four times one over one times, two times three times for so we would have to simplify this, and that is just one divided by ninety six. And that's our final answer

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