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These are the user uploaded subtitles that are being translated: 1 00:00:00,400 --> 00:00:04,830 In this video we're talking about how to use the limit comparison test to say whether a series converges 2 00:00:05,010 --> 00:00:06,240 or diverges. 3 00:00:06,240 --> 00:00:10,950 And in this particular problem we've been given the series from equals 1 to infinity of end squared 4 00:00:10,950 --> 00:00:13,890 divided by quantity and cute plus two. 5 00:00:13,890 --> 00:00:18,000 Now as a reminder the limit comparison test relies on a couple of conditions. 6 00:00:18,030 --> 00:00:23,790 First of all both the original series A7 and the comparison series be seben have to be greater than 7 00:00:24,030 --> 00:00:24,460 zero. 8 00:00:24,480 --> 00:00:29,760 So all the terms of the series have to be positive and in addition to that the limit as and goes to 9 00:00:29,760 --> 00:00:33,160 infinity of a sideband divided by B-7. 10 00:00:33,160 --> 00:00:37,590 So we're literally going to take the original series and divide it by the comparison series. 11 00:00:37,590 --> 00:00:43,020 If that limit the value of that limit is greater than zero then we can draw conclusions. 12 00:00:43,110 --> 00:00:49,510 We can say if the comparison series B-7 converges then the original series a seben also converges. 13 00:00:49,710 --> 00:00:55,960 Or if the comparison series B-7 diverges than the original series A7 also diverges. 14 00:00:56,100 --> 00:01:02,040 So we're going to first do is make sure that our original series and our comparison series meet these 15 00:01:02,040 --> 00:01:05,840 conditions before we go on to drawing any conclusion. 16 00:01:05,850 --> 00:01:08,780 So what will be our comparison series. 17 00:01:08,880 --> 00:01:14,370 Well what we usually want to do if we have a rational function or a fraction is we want to take the 18 00:01:14,370 --> 00:01:19,710 term from the numerator and denominator that's going to have the greatest effect on the numerator and 19 00:01:19,710 --> 00:01:22,510 denominator or the term with the greatest magnitude. 20 00:01:22,650 --> 00:01:26,820 So in this case there's only one term in the numerator it's end squared. 21 00:01:26,820 --> 00:01:32,520 So we're going to go ahead and take and squared for the new numerator and then for the denominator and 22 00:01:32,520 --> 00:01:37,390 cubed has a greater magnitude or greater degree than just the constant 2. 23 00:01:37,410 --> 00:01:41,010 So we're going to take and cubed only for the denominator. 24 00:01:41,010 --> 00:01:45,870 This is going to simplify we're going to be able to cancel and squared from both the numerator and denominator 25 00:01:46,230 --> 00:01:52,410 and this will simplify to 1 over n that means we're going to call our comparison series 1 over and so 26 00:01:52,410 --> 00:01:59,310 let's go ahead and say that this series from N equals 1 to infinity of 1 over n and we'll go ahead and 27 00:01:59,310 --> 00:02:05,680 call the original series a 7 and we'll call the comparison Series B 7. 28 00:02:06,030 --> 00:02:10,510 So now we want to do is show that both of these series are always going to be positive. 29 00:02:10,530 --> 00:02:15,570 So if we look at the original series we know we're starting at and equals 1 and we're counting up and 30 00:02:15,570 --> 00:02:17,730 equals 2 equals 3 and equals four. 31 00:02:17,850 --> 00:02:19,580 All the way up to infinity. 32 00:02:19,700 --> 00:02:24,440 Well if we plug in N equals 1 in the numerator here we can get 1 squared which is 1. 33 00:02:24,480 --> 00:02:29,430 Here we're going to get one plus two is three so we're going to end up with one third. 34 00:02:29,610 --> 00:02:35,490 If we plug in and equals 2 we're going to get 4 over 8 plus 2 is 10 for over 10. 35 00:02:35,490 --> 00:02:37,920 Or two over five. 36 00:02:38,250 --> 00:02:43,540 If we plug in N equals three we're going to get eight over twenty seven plus two is 29. 37 00:02:43,590 --> 00:02:47,640 So we're going to get 8 over 29 and we could keep going. 38 00:02:47,640 --> 00:02:51,040 Let's go ahead and look at though the comparison Series B 7. 39 00:02:51,150 --> 00:02:59,610 We're also starting at any one equals one we get 1 over 1 or 1 at N equals 2 we get 1 1/2 and equals 40 00:02:59,610 --> 00:03:02,690 3 we get one third and we could see that we could keep going. 41 00:03:02,700 --> 00:03:04,910 But in either case there's no value of. 42 00:03:04,920 --> 00:03:10,530 And then we could plug in if we're counting up from N equals 1 2 3 4 et cetera that we could plug into 43 00:03:10,530 --> 00:03:14,190 either series that would make any of these terms negative. 44 00:03:14,250 --> 00:03:20,220 And so we can say then that a 7 is always going to be greater than zero and we can say that B-7 is always 45 00:03:20,220 --> 00:03:21,670 going to be greater than zero. 46 00:03:21,930 --> 00:03:26,160 So those two conditions have been met and the next thing we want to show is that the limit is and goes 47 00:03:26,160 --> 00:03:30,670 to infinity of the ratio of these series is going to be greater than zero. 48 00:03:30,690 --> 00:03:38,160 So if we take the limit as and approaches infinity of a sob in our original series and squared over 49 00:03:38,580 --> 00:03:44,310 and cubed plus two and we divide it by our comparison series we're going to divide this whole thing 50 00:03:44,580 --> 00:03:45,800 by one over and. 51 00:03:45,810 --> 00:03:51,300 But keep in mind that that's exactly the same thing as multiplying by an over 1. 52 00:03:51,300 --> 00:03:57,870 The result then here is going to be the limit as and goes to infinity will multiply and squared by and 53 00:03:57,920 --> 00:04:02,180 we'll get and cubed over and cubed plus two. 54 00:04:02,190 --> 00:04:07,140 Now in order to take the limit what we want to do is we want to multiply both the numerator and denominator 55 00:04:07,470 --> 00:04:14,390 by the highest degree variables so we have one over and cubed and one over and cubed like this. 56 00:04:14,390 --> 00:04:20,580 What that's going to do when we say one over and cubed times and cubed we're going to get one in the 57 00:04:20,680 --> 00:04:26,760 numerator so we're going to have limit as and approaches infinity we're going to have one in the numerator 58 00:04:27,210 --> 00:04:31,730 and denominator one over and cubed times and cubed is going to give us 1. 59 00:04:32,010 --> 00:04:38,610 Then we're going to have here plus two over and cubed that way when we take the limit as and goes to 60 00:04:38,610 --> 00:04:44,640 infinity the denominator right here is going to get extremely large to divided by an infinitely large 61 00:04:44,640 --> 00:04:46,490 number is going to be zero. 62 00:04:46,500 --> 00:04:50,670 So this term is going to go to zero which means it's going to become zero there. 63 00:04:50,880 --> 00:04:57,240 And we're just going to end up with one over 1 or that this limit here if we call this limit L that 64 00:04:57,330 --> 00:05:02,440 L is going to be equal to 1 and one is greater than zero. 65 00:05:02,460 --> 00:05:07,990 So we've shown that this condition here has also been met because Al is greater than zero. 66 00:05:08,100 --> 00:05:11,990 So now that those conditions have been met we can go on to our conclusions. 67 00:05:12,060 --> 00:05:17,290 What we want to do is evaluate be stubborn and say whether B-7 converges or diverges. 68 00:05:17,370 --> 00:05:23,130 If we look here at B-7 we notice that it's the harmonic series one divided by and is a harmonic series 69 00:05:23,490 --> 00:05:26,640 and we know that the harmonic series diverges. 70 00:05:26,880 --> 00:05:31,790 So since B Subban diverges we're looking at this conclusion right here. 71 00:05:31,890 --> 00:05:35,570 If B-7 diverges then A7 also diverges. 72 00:05:35,670 --> 00:05:44,790 So since we've shown that B Subban diverges we can conclude that a Sabun also diverges and that's how 73 00:05:44,790 --> 00:05:49,800 you use the limit comparison test to say whether a series converges or diverges. 8657