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These are the user uploaded subtitles that are being translated: 1 00:00:00,510 --> 00:00:05,360 In this video we're talking about how to use the comparison test for Convergence to say whether a series 2 00:00:05,360 --> 00:00:07,040 converges or diverges. 3 00:00:07,250 --> 00:00:09,450 So first of all let's look at the comparison test. 4 00:00:09,560 --> 00:00:15,780 What it tells us is that if we have two series that are similar we'll call them a seven and B seven 5 00:00:16,160 --> 00:00:18,610 and if both of those series are always positive. 6 00:00:18,620 --> 00:00:21,280 So they're both always greater than zero. 7 00:00:21,380 --> 00:00:28,130 And if a 7 is always smaller than B 7 or B 7 is always larger than A. 8 00:00:28,490 --> 00:00:30,390 Then we can draw two conclusions. 9 00:00:30,530 --> 00:00:36,380 One we can say that if B 7 is convergent if we can show that piece of and converges then we can prove 10 00:00:36,380 --> 00:00:38,640 that a seben also converges. 11 00:00:38,810 --> 00:00:44,780 Or if we can prove that A7 diverges then we can also show that B-7 diverges. 12 00:00:44,810 --> 00:00:47,780 We can't draw conclusions if it's the other way around. 13 00:00:47,780 --> 00:00:54,730 In other words if we have this arrangement where a seben is smaller than B 7 if we can show that a 7 14 00:00:54,730 --> 00:00:59,750 converges it doesn't let us prove that B-7 also converges. 15 00:00:59,750 --> 00:01:02,810 And similarly if we can show that B-7 diverges. 16 00:01:02,810 --> 00:01:06,340 It doesn't let us conclude that a Subhan also diverges. 17 00:01:06,500 --> 00:01:08,680 In those cases the test is inconclusive. 18 00:01:08,690 --> 00:01:10,760 We can't do anything with that. 19 00:01:10,760 --> 00:01:16,550 So if we want to illustrate that graphically we can come down to these graphs here and what they show 20 00:01:16,670 --> 00:01:23,330 is that if we have a B sub n this here is the first conclusion we can draw here. 21 00:01:23,690 --> 00:01:28,970 And this is going to represent the second conclusion that we can come to use in the comparison series. 22 00:01:28,970 --> 00:01:35,750 So for this first conclusion if B-7 converges than A7 also converges Well we know A7 has to be smaller 23 00:01:35,750 --> 00:01:37,970 than B-7 but they're both always positive. 24 00:01:37,970 --> 00:01:40,120 So we have to stay above the x axis. 25 00:01:40,250 --> 00:01:45,860 So if B is convergent in other words if it looks something like this it's converging down toward the 26 00:01:45,860 --> 00:01:46,940 x axis. 27 00:01:47,000 --> 00:01:55,910 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 28 00:01:55,910 --> 00:01:57,750 also converge. 29 00:01:57,920 --> 00:02:04,760 If A7 was divergent it would come up this way and cross B seben And we'll that's not possible because 30 00:02:04,910 --> 00:02:07,680 a subset has to be less than B 7. 31 00:02:07,700 --> 00:02:11,930 So of B-7 Convergys it's going to force a 7 to also converge. 32 00:02:11,930 --> 00:02:18,380 And then for this second conclusion here we know that if a 7 diverges obviously B-7 has to be larger 33 00:02:18,380 --> 00:02:19,400 than A-7. 34 00:02:19,400 --> 00:02:28,490 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 35 00:02:28,630 --> 00:02:29,450 converge. 36 00:02:29,570 --> 00:02:36,170 It would come down this way cross the graph of A7 and we know that that's not possible because B-7 always 37 00:02:36,170 --> 00:02:39,710 has to be larger than a 7. 38 00:02:39,710 --> 00:02:44,600 So how does this help us if we want to use calculus to determine the convergence or divergence of A 39 00:02:44,600 --> 00:02:45,210 Series. 40 00:02:45,320 --> 00:02:49,460 Well in this particular problem we've been given this series from N equals 1 to infinity. 41 00:02:49,460 --> 00:02:53,080 The square root of N divided by quantity and squared plus. 42 00:02:53,090 --> 00:02:59,360 And so first of all what we want to recognize is that we have this series here squared of an over and 43 00:02:59,360 --> 00:03:06,650 squared plus and this series could be a seben or it could be B-7 it depends on the comparison series 44 00:03:06,650 --> 00:03:12,050 they were able to find because the comparison series we find might always be larger than the series 45 00:03:12,170 --> 00:03:15,090 or it might always be smaller than this series. 46 00:03:15,170 --> 00:03:22,930 We just want to find a comparison series that is consistently larger than or smaller than this series. 47 00:03:22,940 --> 00:03:25,000 So how do we go about finding compar. series. 48 00:03:25,010 --> 00:03:30,470 Well when we have a rational function like this a fraction one of the easiest ways to do it is look 49 00:03:30,470 --> 00:03:36,020 at the term in the numerator that has the greatest effect on the numerator and the term in the denominator 50 00:03:36,030 --> 00:03:38,260 that has the greatest effect on the denominator. 51 00:03:38,390 --> 00:03:44,070 And just take those two terms put them into a new Fraction for the comparison series. 52 00:03:44,120 --> 00:03:50,020 So obviously in this case there's only one term in the numerator so we'll go ahead and take that term. 53 00:03:50,270 --> 00:03:54,890 And in the denominator at the end squared is going to have a greater effect on the denominator than 54 00:03:54,890 --> 00:03:59,690 just the end of the first because higher degree terms always have a greater effect than lower degree 55 00:03:59,690 --> 00:04:00,110 terms. 56 00:04:00,110 --> 00:04:02,500 So let's just take the and squared. 57 00:04:02,810 --> 00:04:07,110 We know that the square root of N is the same as any of the 1 1/2 power. 58 00:04:07,130 --> 00:04:11,200 So we can rewrite this as end of the one half divided by and squared. 59 00:04:11,570 --> 00:04:17,210 And then because we just have ended the one half in the numerator and we have end squared in the denominator 60 00:04:17,390 --> 00:04:21,350 we can say two minus one half is three halves. 61 00:04:21,350 --> 00:04:26,590 And so then we simplify what we end up with is 1 over and two the three halves. 62 00:04:26,780 --> 00:04:30,630 So this here is a potential comparison series. 63 00:04:30,650 --> 00:04:34,130 It's a similar but simplified version of our original series. 64 00:04:34,130 --> 00:04:39,320 So what we want to do now is test both of these series to see whether we can show that one is always 65 00:04:39,320 --> 00:04:40,930 going to be greater than the other. 66 00:04:40,940 --> 00:04:46,120 So let's take here squared of an over and squared plus an. 67 00:04:46,370 --> 00:04:55,250 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 68 00:04:55,250 --> 00:04:57,170 we'll see how these series compare. 69 00:04:57,260 --> 00:05:05,560 So four and equals one 4 9 and 16 which we'll use because they're easy to find the square root of. 70 00:05:05,800 --> 00:05:10,660 We'll plug these into the original series and the comparison series so plugging unequals one into the 71 00:05:10,660 --> 00:05:11,880 original series. 72 00:05:11,890 --> 00:05:17,000 We're going to get one squared is one divided by one plus one is two. 73 00:05:17,140 --> 00:05:21,910 So we end up with one half pleading for the original series we end up with the square to four which 74 00:05:21,910 --> 00:05:29,130 is two divided by 16 plus four is twenty two over 20 gives us 1 over 10. 75 00:05:29,140 --> 00:05:36,160 Plugging 9 into the original series we get 3 divided by 81 plus nine or ninety three over 90 is going 76 00:05:36,160 --> 00:05:47,800 to be one over 30 and then plugging in 16 we get 4 divided by 16 square 256 plus 16 is 272 4 divided 77 00:05:47,800 --> 00:05:52,280 by 272 is 1 over sixty eight. 78 00:05:52,360 --> 00:05:55,130 If we plug the same values into the comparison series. 79 00:05:55,180 --> 00:06:01,090 Keep in mind that the comparison series 1 over end of the three halves we could rewrite that as one 80 00:06:01,160 --> 00:06:07,930 over and to the one half which is the square root of N and then raised to the third power because we 81 00:06:07,930 --> 00:06:13,540 have ended the one half and then race to the third thats one half times three or three halves. 82 00:06:13,570 --> 00:06:19,670 So when we plug in these values one for nine and 16 we can just take the square root then cube the result 83 00:06:19,820 --> 00:06:21,250 and that will be the denominator. 84 00:06:21,340 --> 00:06:29,200 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 85 00:06:29,200 --> 00:06:29,840 in four. 86 00:06:29,860 --> 00:06:32,650 We're going to get square to 4 has 2 cubed is 8. 87 00:06:32,680 --> 00:06:36,130 So we end up with 1 over 8 plugging in 9. 88 00:06:36,160 --> 00:06:44,890 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 89 00:06:44,890 --> 00:06:47,140 with 1 over 64. 90 00:06:47,290 --> 00:06:52,060 So the first thing we notice when we look at these terms is that all of the terms of the series are 91 00:06:52,060 --> 00:06:54,120 always going to be positive. 92 00:06:54,130 --> 00:06:56,950 We plugged in N equals 1 4 9 16. 93 00:06:57,040 --> 00:07:03,580 But if we continued plugging in N equals 25 and equals 36 or any value of n b is remember end starts 94 00:07:03,580 --> 00:07:08,170 at 1 and then we're just counting up so it's always going to be a positive value of n. 95 00:07:08,380 --> 00:07:13,660 So any positive value that we plug in we're always going to get a positive result for the term of the 96 00:07:13,660 --> 00:07:14,650 original series. 97 00:07:14,800 --> 00:07:19,630 And we're always going to get a positive result for the term of the comparison series which means both 98 00:07:19,630 --> 00:07:22,770 of these series are always going to be positive. 99 00:07:22,990 --> 00:07:27,910 Furthermore if we compare the terms of the two series together we can see that one half is less than 100 00:07:27,910 --> 00:07:28,480 one. 101 00:07:28,660 --> 00:07:36,700 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. 102 00:07:36,700 --> 00:07:42,940 In other words the terms of the original series are always going to be less than the terms of the comparison 103 00:07:42,940 --> 00:07:48,090 series since we want to meet this condition here for the comparison test. 104 00:07:48,100 --> 00:07:52,330 We've already shown that both of the series are positive so that meets this condition here that they 105 00:07:52,330 --> 00:07:54,550 both have to be greater than or equal to zero. 106 00:07:54,550 --> 00:07:58,350 We show that the original series was always less than the comparison series. 107 00:07:58,360 --> 00:08:03,050 So what that means is that we want this series to be the original series. 108 00:08:03,160 --> 00:08:09,910 We want this series to be the comparison series because our original series is less than the original 109 00:08:09,910 --> 00:08:11,170 series B 7. 110 00:08:11,500 --> 00:08:17,110 If we had found that the opposite was true if these terms were switched and the original series was 111 00:08:17,110 --> 00:08:23,630 always greater than the comparison series then we would want to call the Original Series B 7. 112 00:08:23,770 --> 00:08:26,100 And the comparison series a sedan. 113 00:08:26,110 --> 00:08:29,850 So just identify which terms are always smaller than the other. 114 00:08:29,920 --> 00:08:35,620 And then in your inequality here label which series should be which because a sideband is not always 115 00:08:35,620 --> 00:08:37,180 going to be the original series. 116 00:08:37,180 --> 00:08:39,560 It could be the original or comparison series. 117 00:08:39,790 --> 00:08:44,770 But in this case since the original series is always less than the comparison series we'll call the 118 00:08:44,770 --> 00:08:49,840 original series a Subban so will say the original series is a Sabahan. 119 00:08:49,840 --> 00:08:56,530 That means our comparison series then is going to be equal to when we can write this comparison series 120 00:08:56,560 --> 00:08:59,770 as some from unequals One to Infinity. 121 00:08:59,770 --> 00:09:04,780 The same as the original one since we started counting from inkwells 1 and we know that this is one 122 00:09:04,870 --> 00:09:11,580 over and two the three halves know what the comparison test we're always looking to draw a conclusion 123 00:09:11,610 --> 00:09:14,420 about the comparison series first. 124 00:09:14,430 --> 00:09:21,690 So since the comparison series is be Subhan that means we're hoping to draw conclusion one because conclusion 125 00:09:21,690 --> 00:09:24,320 one starts with the comparison series. 126 00:09:24,320 --> 00:09:30,750 So then so we're saying if B-7 converges then also converge is what we're hoping is that we can show 127 00:09:30,750 --> 00:09:32,470 that B-7 converges. 128 00:09:32,580 --> 00:09:36,290 So we look at B-7 and what we recognize is that it's a series. 129 00:09:36,300 --> 00:09:40,800 Remember a piece series is a series in the form one over and to the P.. 130 00:09:40,950 --> 00:09:51,150 If P is greater than one then this peace series converges and in our case P is equal to three halves 131 00:09:51,390 --> 00:09:54,140 and three halves is greater than 1. 132 00:09:54,150 --> 00:10:02,400 So what we can say is that B Subban Convergys by the pre-series test and then by the comparison test 133 00:10:02,760 --> 00:10:06,740 since we show that a seben and B Sabahan both greater than or equal to zero. 134 00:10:06,750 --> 00:10:13,090 Always that A7 was always going to be less than or equal to the comparison series piece had been. 135 00:10:13,290 --> 00:10:22,600 And since we showed that B-7 converges we can conclude by the comparison tests that sub also converges. 136 00:10:22,710 --> 00:10:27,880 And that's how you use the comparison test to say whether a series converges or diverges. 15285