Original subtitles

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In this video we're talking about how to use the comparison test for Convergence to say whether a series

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converges or diverges.

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So first of all let's look at the comparison test.

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What it tells us is that if we have two series that are similar we'll call them a seven and B seven

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and if both of those series are always positive.

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So they're both always greater than zero.

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And if a 7 is always smaller than B 7 or B 7 is always larger than A.

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Then we can draw two conclusions.

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One we can say that if B 7 is convergent if we can show that piece of and converges then we can prove

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that a seben also converges.

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Or if we can prove that A7 diverges then we can also show that B-7 diverges.

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We can't draw conclusions if it's the other way around.

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In other words if we have this arrangement where a seben is smaller than B 7 if we can show that a 7

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converges it doesn't let us prove that B-7 also converges.

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And similarly if we can show that B-7 diverges.

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It doesn't let us conclude that a Subhan also diverges.

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In those cases the test is inconclusive.

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We can't do anything with that.

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So if we want to illustrate that graphically we can come down to these graphs here and what they show

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is that if we have a B sub n this here is the first conclusion we can draw here.

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And this is going to represent the second conclusion that we can come to use in the comparison series.

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So for this first conclusion if B-7 converges than A7 also converges Well we know A7 has to be smaller

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than B-7 but they're both always positive.

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So we have to stay above the x axis.

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So if B is convergent in other words if it looks something like this it's converging down toward the

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x axis.

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If a 7 has to be underneath B 7 but above the x axis then of course b 7 is going to force a seben to

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also converge.

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If A7 was divergent it would come up this way and cross B seben And we'll that's not possible because

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a subset has to be less than B 7.

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So of B-7 Convergys it's going to force a 7 to also converge.

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And then for this second conclusion here we know that if a 7 diverges obviously B-7 has to be larger

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than A-7.

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In other words it has to stay above A-7 which means A7 is going to force be 7 to also diverge if B-7

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converge.

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It would come down this way cross the graph of A7 and we know that that's not possible because B-7 always

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has to be larger than a 7.

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So how does this help us if we want to use calculus to determine the convergence or divergence of A

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Series.

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Well in this particular problem we've been given this series from N equals 1 to infinity.

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The square root of N divided by quantity and squared plus.

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And so first of all what we want to recognize is that we have this series here squared of an over and

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squared plus and this series could be a seben or it could be B-7 it depends on the comparison series

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they were able to find because the comparison series we find might always be larger than the series

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or it might always be smaller than this series.

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We just want to find a comparison series that is consistently larger than or smaller than this series.

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So how do we go about finding compar. series.

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Well when we have a rational function like this a fraction one of the easiest ways to do it is look

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at the term in the numerator that has the greatest effect on the numerator and the term in the denominator

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that has the greatest effect on the denominator.

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And just take those two terms put them into a new Fraction for the comparison series.

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So obviously in this case there's only one term in the numerator so we'll go ahead and take that term.

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And in the denominator at the end squared is going to have a greater effect on the denominator than

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just the end of the first because higher degree terms always have a greater effect than lower degree

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terms.

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So let's just take the and squared.

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We know that the square root of N is the same as any of the 1 1/2 power.

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So we can rewrite this as end of the one half divided by and squared.

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And then because we just have ended the one half in the numerator and we have end squared in the denominator

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we can say two minus one half is three halves.

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And so then we simplify what we end up with is 1 over and two the three halves.

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So this here is a potential comparison series.

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It's a similar but simplified version of our original series.

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So what we want to do now is test both of these series to see whether we can show that one is always

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going to be greater than the other.

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So let's take here squared of an over and squared plus an.

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And we'll take one over and to the three halves and what we'll do is we'll plug in a few values and

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we'll see how these series compare.

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So four and equals one 4 9 and 16 which we'll use because they're easy to find the square root of.

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We'll plug these into the original series and the comparison series so plugging unequals one into the

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original series.

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We're going to get one squared is one divided by one plus one is two.

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So we end up with one half pleading for the original series we end up with the square to four which

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is two divided by 16 plus four is twenty two over 20 gives us 1 over 10.

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Plugging 9 into the original series we get 3 divided by 81 plus nine or ninety three over 90 is going

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to be one over 30 and then plugging in 16 we get 4 divided by 16 square 256 plus 16 is 272 4 divided

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by 272 is 1 over sixty eight.

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If we plug the same values into the comparison series.

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Keep in mind that the comparison series 1 over end of the three halves we could rewrite that as one

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over and to the one half which is the square root of N and then raised to the third power because we

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have ended the one half and then race to the third thats one half times three or three halves.

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So when we plug in these values one for nine and 16 we can just take the square root then cube the result

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and that will be the denominator.

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So plugging in any was one we're gonna get square to 1 is 1 1 cubed is 1 1 divided by 1 is 1 plugging

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in four.

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We're going to get square to 4 has 2 cubed is 8.

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So we end up with 1 over 8 plugging in 9.

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We get three and then cubed is 27 so 1 over 27 and plugging in 16 we get 4 cubed is 64 so we end up

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with 1 over 64.

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So the first thing we notice when we look at these terms is that all of the terms of the series are

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always going to be positive.

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We plugged in N equals 1 4 9 16.

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But if we continued plugging in N equals 25 and equals 36 or any value of n b is remember end starts

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at 1 and then we're just counting up so it's always going to be a positive value of n.

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So any positive value that we plug in we're always going to get a positive result for the term of the

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original series.

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And we're always going to get a positive result for the term of the comparison series which means both

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of these series are always going to be positive.

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Furthermore if we compare the terms of the two series together we can see that one half is less than

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one.

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One tenth is less than 1 8 1 30th is less than 1 over 27 and 1 over 68 is less than 1 over 64.

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In other words the terms of the original series are always going to be less than the terms of the comparison

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series since we want to meet this condition here for the comparison test.

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We've already shown that both of the series are positive so that meets this condition here that they

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both have to be greater than or equal to zero.

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We show that the original series was always less than the comparison series.

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So what that means is that we want this series to be the original series.

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We want this series to be the comparison series because our original series is less than the original

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series B 7.

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If we had found that the opposite was true if these terms were switched and the original series was

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always greater than the comparison series then we would want to call the Original Series B 7.

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And the comparison series a sedan.

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So just identify which terms are always smaller than the other.

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And then in your inequality here label which series should be which because a sideband is not always

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going to be the original series.

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It could be the original or comparison series.

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But in this case since the original series is always less than the comparison series we'll call the

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original series a Subban so will say the original series is a Sabahan.

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That means our comparison series then is going to be equal to when we can write this comparison series

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as some from unequals One to Infinity.

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The same as the original one since we started counting from inkwells 1 and we know that this is one

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over and two the three halves know what the comparison test we're always looking to draw a conclusion

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about the comparison series first.

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So since the comparison series is be Subhan that means we're hoping to draw conclusion one because conclusion

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one starts with the comparison series.

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So then so we're saying if B-7 converges then also converge is what we're hoping is that we can show

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that B-7 converges.

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So we look at B-7 and what we recognize is that it's a series.

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Remember a piece series is a series in the form one over and to the P..

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If P is greater than one then this peace series converges and in our case P is equal to three halves

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and three halves is greater than 1.

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So what we can say is that B Subban Convergys by the pre-series test and then by the comparison test

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since we show that a seben and B Sabahan both greater than or equal to zero.

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Always that A7 was always going to be less than or equal to the comparison series piece had been.

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And since we showed that B-7 converges we can conclude by the comparison tests that sub also converges.

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And that's how you use the comparison test to say whether a series converges or diverges.