WEBVTT mathematics/calculus-ii/murray
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Hi, this is Will Murray and we are here today to talk about the comparison test.
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The way this works is that you will be given a series that we are going to call a < font size="-6" > n < /font > ,
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And you create your own series that we will call b < font size="-6" > n < /font > .
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Then you will try to compare these 2 series to each other.
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The way this works is, the one you create, if that one converges,
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And the one that you are given is smaller than it,
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Then you can say that the given series converges as well.
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On the other hand, if the one you create diverges,
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And the given series is bigger than the one you created,
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Then you can say the given series diverges.
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Now it is very important to get these inequalities going the right way.
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If these inequalities are going the wrong way, then the comparison test does not tell you anything.
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We will see some examples that illustrate the difference between those inequalities going the right way and going the wrong way.
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We will be using a second test called the limit comparison test.
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It starts out the same way,
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You will be given a series, then you create your own series,
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But then instead of checking the inequalities,
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What you do is you divide those 2 series together and then take the limit as n goes to infinity.
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If you get a finite and positive number, it cannot be 0, it has to be a positive number,
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Then you can say whatever the series you created does, converges or diverges,
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You can say that the given series does the same thing.
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let us check that out with some examples.
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The first one here is the series of 1/n+1.
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That is the given series, the a < font size="-6" > n < /font > .
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When you look at this series, the simplest series that this seems to resemble,
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Is the series 1/n.
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Because 1/n+1, is about the same as 1/n.
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So that is the series we create ourselves and we call the b < font size="-6" > n < /font > .
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Now let us try comparing those to each other.
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1/n+1 works as 1/n, well n+1 has a bigger denominator.
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Which means that 1/n is bigger.
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So, 1/n is bigger, the b < font size="-6" > n < /font > .
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So, the a < font size="-6" > n < /font > is less than the b < font size="-6" > n < /font > .
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The sum of b < font size="-6" > n < /font > , that is 1/n, diverges.
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We know that we saw that one before and that was the harmonic series.
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Or, you can think of it as the p series, with p = 1.
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We know that series diverges.
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What we have is a series that is less than it.
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If a series is less than a divergent series, that is not going the right way to use the comparison test.
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So, comparison fails to give us an answer here.
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Tell us nothing.
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So, the comparison test does not tell us what this series does.
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Instead, you have to try another test,
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For example, you could try the integral test and that actually will give you a good answer.
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You can check that out with the integral test.
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It turns out that this series diverges,
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But the important thing is that you cannot use the comparison test on this one.
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Let us try it out with another one.
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3^n/2^n - 1,
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That looks like the sum of 3^n/2^n.
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So a < font size="-6" > n < /font > is the given series, for our b < font size="-6" > n < /font > , we will use the sum of 3^n/2^n.
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Then we will compare those to each other.
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So, 3^n/2^n-1 vs. 3^n/2^n.
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Well 2^n-1 is a smaller denominator which means that 3^n/2^n-1 is actually a bigger number.
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What that is saying is that a < font size="-6" > n < /font > is bigger than b < font size="-6" > n < /font > .
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We know that the sum of the b < font size="-6" > n < /font > 's diverges.
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The reason we know that is because it is a geometric series.
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The common ratio is 3/2, because 3/n/2^n is just 3/2^n, and 3/2 is bigger than 1.
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So that is a geometric series that diverges,
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And here we have a bigger series, so this bigger series, we can say it diverges by the comparison test.
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In this case, the inequality did go the right way.
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So, we get a conclusion by the comparison test.
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Just a recap there, we looked at the series we were given, we tried to find the series that was similar to it,
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And simple enough that we could answer pretty quickly whether it converged or diverged.
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In this case it was a geometric series.
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Then we compared them to each other.
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The inequality does go the right way, so we can make this conclusion.
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Next example I wanted to try is the sum of sqrt(2n+17)/n.
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Now, this one most closely looks like,
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Well, the sqrt(2n+17) that is really more or less the sqrt(n).
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sqrt(n)/n, and so that in turn is sqrt(n)/n 1/sqrt(n).
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That is the series we are going to use as our b < font size="-6" > n < /font > .
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This series is our a < font size="-6" > n < /font > .
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What we are going to do is divide those together because we are going to try to use the limit comparison test this time.
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So we look at a < font size="-6" > N < /font > /b < font size="-6" > n < /font > ,
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And that is the sqrt(2n+17)/n.
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All of that divided by b < font size="-6" > n < /font > , which is 1/sqrt(n).
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We can flip that fraction in the denominator up the numerator,
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So we get sqrt(2n+17).
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The sqrt(2n+17) × sqrt(n)/n,
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Which is sqrt(2n²) + 17n/n.
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If we look at the top and bottom there, the biggest terms we have in the top, we have an n² with a square root over it.
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That is essentially n and in the bottom we have n.
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So we would divide top and bottom by n, and we get,
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If we bring that top end under the square root, we get 2+17/n.
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Because when the n comes under the square root it turns into an n².
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Then just 1 in the denominator.
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17/n goes to 0.
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So, the whole thing goes to sqrt(2).
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The important thing about the sqrt(2) is that it is a finite number, it is not infinity.
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It is not 0, so the limit comparison test applies.
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Since the sum of b < font size="-6" > n < /font > diverges, well how do we know that,
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That is because we can think of 1/sqrt(n) as 1/n^1/2.
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That is a p series, with p equal to 1/2.
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The key thing there is that 1/2 < or = 1.
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That means that b < font size="-6" > n < /font > is a divergent series.
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The limit comparison test says that the given series a < font size="-6" > n < /font > does the same thing that your series b < font size="-6" > n < /font > does.
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So, we can conclude that the sum of a < font size="-6" > n < /font > also diverges by the limit comparison test.
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Again, the key point there is that we look at the series that we are given, and we try to find,
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So that is the given series,
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We try to find the given series that behaves like the series that we are given but is simpler to deal with.
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What I did here was essentially strip away the 2 and the 17,
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Because those were not so important, and then I called the new series b < font size="-6" > n < /font > .
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Then we try to compare those series to each other.
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This time by dividing.
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If we get this finite non-zero number at the end of it,
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That says that both series behave the same way.
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Since we know the b < font size="-6" > n < /font > diverges, because it is a p series,
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Then we can say the a < font size="-6" > n < /font > diverges as well and justify that conclusion using the limit comparison test.
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So, another example I would like to look at is the sum of sin(1/n).
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The way we might think about that is,
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let us think about the graph of sin(x).
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As n goes to infinity, 1/n goes to 0.
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So let us look at the graph of sin(x) when x is very near 0.
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What we see is that the graph of sin(x),
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This is sin(x), is very close to the graph of x as x approaches 0.
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Sin(x) is almost the same as x.
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Sin(1/n) the a < font size="-6" > n < /font > might behave like the series b < font size="-6" > n < /font > = 1/n.
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Because sin(x) behaves like x.
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Let us try that out.
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We will take the series b < font size="-6" > n < /font > = 1/n, and then we will look at a < font size="-6" > n < /font > /b < font size="-6" > n < /font > .
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Is sin(1/n)/1/n.
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Remember as n goes to infinity, 1/n goes to 0.
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So this is sin(0) which goes to 0.
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1/n also goes to 0, so we have a 0/0 situation.
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That is a situation where you can use l'Hopital's rule.
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So l'Hopital's rule says you can take the derivatives of the top and bottom,
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I will take the derivative of the bottom first because it is easier.
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The derivative of 1/n is just 1/-n².
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The derivative of sin(1/n) is cos(1/n).
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× derivative of 1/n by the chain rule which is -1/n².
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Those -1/n²'s cancel.
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We get cos(1/n) and if we take the limit of that,
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As n goes to infinity, that is cos(0) which is 1.
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The key thing about 1 here is only that it is a finite non-zero number.
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If it is a finite non-zero number, that says whatever one series does, the other series does.
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So, let us ask ourselves,
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The sum of b < font size="-6" > n < /font > does what?
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Well b < font size="-6" > n < /font > is just 1/n, and we know that is the harmonic series.
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And it is also a p series.
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So, either one of those is just vacationed since we already showed that it diverges.
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It is a p series with p = 1.
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So that is a divergent series.
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Since that series diverges,
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We can say that the given series a < font size="-6" > n < /font > also diverges by the limit comparison test.
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So that one was a little bit trickier, probably was not so obvious what series we should compare it too.
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The key thing there was realizing that sin(1/n) is a lot like 1/n when n goes to infinity.
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Because 1/n was going to 0.
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sin(x) was very similar to x.
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Once we figure out which series we want to compare it to, we divide them together, take the limit,
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Which uses l'Hopital's rule, we get a finite non-zero number,
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And that says the two series do the same thing.
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Since one of them diverges, we can say that the given series diverges as well.