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These are the user uploaded subtitles that are being translated: 1 00:00:00,290 --> 00:00:04,620 In this video we're talking about how to use the ratio test to say whether or not a series converges 2 00:00:04,650 --> 00:00:05,930 or diverges. 3 00:00:05,930 --> 00:00:11,220 And in this particular problem we've been given the series the some from N equals 1 to infinity of three 4 00:00:11,220 --> 00:00:14,540 to the power divided by and squared. 5 00:00:14,700 --> 00:00:20,190 And as a reminder here we have the ratio test the ratio test is just one of our many convergence tests. 6 00:00:20,190 --> 00:00:25,830 And what this one says is that if we find a value L So this is our critical value here the important 7 00:00:25,830 --> 00:00:32,790 value when it comes to the ratio test L and L is equal to the limit is and goes to infinity of the absolute 8 00:00:32,790 --> 00:00:34,190 value of this quotient here. 9 00:00:34,230 --> 00:00:36,500 A seven plus one divided by a. 10 00:00:36,600 --> 00:00:43,890 Well a sub n is the original series we've been given so we can go ahead and say here that a 7 is this 11 00:00:43,950 --> 00:00:45,940 original series we've been given. 12 00:00:45,950 --> 00:00:48,840 So we'll plug this value here in for a second. 13 00:00:49,080 --> 00:00:54,990 A seven plus one is going to be whatever we get when we replace n with end plus 1 and this original 14 00:00:54,990 --> 00:00:57,660 series will take this original series everywhere. 15 00:00:57,690 --> 00:00:58,640 We have an N. 16 00:00:58,640 --> 00:01:00,570 We'll put an end plus one instead. 17 00:01:00,750 --> 00:01:04,050 And whatever value we get there we'll go in for a seven plus one. 18 00:01:04,140 --> 00:01:07,500 So then we'll have the absolute value of this quotient of series. 19 00:01:07,500 --> 00:01:12,420 We'll simplify it will take the limit is and goes to infinity and we'll find some value for L. 20 00:01:12,510 --> 00:01:18,480 Once we do we can use the value of L to say whether or not the series converges because if L is less 21 00:01:18,480 --> 00:01:21,490 than 1 by the ratio test the series converges. 22 00:01:21,690 --> 00:01:24,240 If Ellas greater than one then it diverges. 23 00:01:24,240 --> 00:01:30,030 And if L is equal to 1 then the ratio tests in particular is inconclusive and we'd have to try to use 24 00:01:30,270 --> 00:01:35,160 another test or just say that we can't draw a conclusion about the convergence of the series. 25 00:01:35,190 --> 00:01:37,290 So let's go ahead and work through this problem. 26 00:01:37,350 --> 00:01:39,230 Again the first thing we need to do is find L. 27 00:01:39,240 --> 00:01:46,620 So we'll go ahead and say L is equal to the limit as and goes to infinity at the absolute value of a 28 00:01:46,620 --> 00:01:47,790 sub and plus 1. 29 00:01:47,790 --> 00:01:51,890 So we're going to take this original series and we're going to replace and with and plus one. 30 00:01:51,900 --> 00:01:56,040 So instead of three to the end we're going to get three to the end plus one. 31 00:01:56,160 --> 00:02:01,140 And then in the denominator here instead of end squared we're going to get and plus 1 squared. 32 00:02:01,140 --> 00:02:07,830 We have to remember to put that in parentheses so we square the entire quantity and plus 1. 33 00:02:07,830 --> 00:02:09,890 So that's a seven plus one. 34 00:02:09,900 --> 00:02:14,670 Then we're going to divide that by a seven and a 7 is just the original series so we're going to divide 35 00:02:14,670 --> 00:02:20,150 this whole thing by a 7 which is three to the end divided by ends squared. 36 00:02:20,160 --> 00:02:22,160 We're taking the absolute value of this whole thing. 37 00:02:22,170 --> 00:02:26,430 Now what we realize is that in this case we have a fraction divided by a fraction. 38 00:02:26,460 --> 00:02:31,260 And I remember when we are dividing fractions what we can do is take this fraction in the numerator 39 00:02:31,260 --> 00:02:35,890 and leave it exactly as is who will say the limit as and goes to infinity. 40 00:02:36,090 --> 00:02:43,300 Of this fraction in the numerator so three to the end plus 1 divided by and plus 1 quantities squared. 41 00:02:43,530 --> 00:02:49,290 And then instead of dividing by this fraction in the denominator we can multiply by a reciprocal those 42 00:02:49,290 --> 00:02:51,120 two things are the same. 43 00:02:51,120 --> 00:02:55,360 So we just flip this upside down instead of 3 to the end over end squared. 44 00:02:55,470 --> 00:03:01,300 We get and squared over three to the end and we just change the division to multiplication. 45 00:03:01,500 --> 00:03:05,280 Now when you're doing the ratio test because this is always going to be the case when you have a fraction 46 00:03:05,280 --> 00:03:06,480 over a fraction. 47 00:03:06,510 --> 00:03:12,720 A lot of people just go straight to the second step and take the a seven plus one and multiply by the 48 00:03:12,720 --> 00:03:16,730 reciprocal of a 7 and skip this first step. 49 00:03:16,740 --> 00:03:18,390 Here you can do that if you want. 50 00:03:18,390 --> 00:03:24,240 If you feel comfortable or you can write out the original quotient of fractions here and then flip this 51 00:03:24,240 --> 00:03:29,220 denominator upside down find it's reciprocal and change the division to multiplication whichever one 52 00:03:29,610 --> 00:03:30,300 you prefer. 53 00:03:30,360 --> 00:03:34,800 But ultimately we're getting here to this second step and once we're to this point what we want to do 54 00:03:34,800 --> 00:03:39,480 is look for terms that are similar and usually that means terms that have a like base. 55 00:03:39,510 --> 00:03:43,910 So for example here we have three race to the end plus one the base is three. 56 00:03:43,920 --> 00:03:45,330 Here we have three to the end. 57 00:03:45,330 --> 00:03:46,920 The base is still three. 58 00:03:46,920 --> 00:03:53,250 So what we can do is say that these two terms here are similar to one another. 59 00:03:53,310 --> 00:03:59,220 Therefore we want to put them together in the scene fraction and you'll get better at recognizing terms 60 00:03:59,220 --> 00:04:04,900 that are similar in these types of problems so what we're going to do is we're going to pair like terms. 61 00:04:04,920 --> 00:04:10,310 So because these fractions are multiplied together we can swap numerators and denominators easily. 62 00:04:10,320 --> 00:04:14,580 So we're just going to swap these two denominators to get like terms together so we're going to say 63 00:04:15,020 --> 00:04:16,670 three is the end plus one. 64 00:04:16,680 --> 00:04:22,140 We're going to put this denominator with this numerator and say three to the N and then multiply by 65 00:04:22,590 --> 00:04:27,080 and squared over and plus one quantity squared. 66 00:04:27,120 --> 00:04:32,310 And now you can see how these two terms are similar and these two terms are similar more so than the 67 00:04:32,310 --> 00:04:34,560 arrangement that we had in this second step. 68 00:04:34,560 --> 00:04:39,900 The reason that this is helpful is because when terms have like bases you can subtract the exponent 69 00:04:39,900 --> 00:04:42,480 in the denominator from the exponent in the numerator. 70 00:04:42,480 --> 00:04:48,540 So for example here this three to the end plus one divided by three to the N is the same as 3 to the 71 00:04:48,540 --> 00:04:54,570 end plus 1 the exponent from the numerator minus the exponent in the denominator which is just N and 72 00:04:54,570 --> 00:04:55,020 then we have. 73 00:04:55,050 --> 00:04:58,200 And plus one minus Anwyl and minus N is zero. 74 00:04:58,200 --> 00:05:04,290 Those two things cancel and we just have three to the first power three to the one which is just three. 75 00:05:04,450 --> 00:05:08,800 So this whole fraction here this whole first fraction becomes three. 76 00:05:08,810 --> 00:05:15,130 And because now we've eliminated ends we can pull that 3 out in front of the limit so we'll go and say 77 00:05:15,130 --> 00:05:21,940 that Al is going to be equal to three times the limit as and goes to infinity of the absolute value 78 00:05:22,030 --> 00:05:26,590 of square divided by and plus one quantity squared. 79 00:05:26,620 --> 00:05:32,890 At this point now we want to go ahead and expand the denominator of this fraction. 80 00:05:33,040 --> 00:05:34,780 So we have and plus one quantity squared. 81 00:05:34,780 --> 00:05:38,380 That's the same thing as Plus one times and plus one right. 82 00:05:38,560 --> 00:05:44,590 So we want to foil this out so end times and it's going to give us and squared and then we're going 83 00:05:44,590 --> 00:05:50,420 to have plus and plus and it's going to be plus two and then one time's one plus one. 84 00:05:50,530 --> 00:05:52,540 So we end up with this value here. 85 00:05:52,660 --> 00:05:58,540 Now at this point I want to take the highest degree variable in the denominator and divide through both 86 00:05:58,540 --> 00:06:01,180 the numerator and denominator by that value. 87 00:06:01,180 --> 00:06:05,650 So the highest degree variable in the denominator is and squared. 88 00:06:05,710 --> 00:06:10,210 That means we want to divide every term in the numerator and denominator by end squared so what we can 89 00:06:10,210 --> 00:06:17,950 really say here is we can multiply the numerator here by one over and squared and we're going to multiply 90 00:06:17,950 --> 00:06:21,670 the denominator by 1 over and squared like this. 91 00:06:21,670 --> 00:06:25,090 This is the same thing because the numerator and denominator are the same. 92 00:06:25,090 --> 00:06:27,420 This is really the same thing as multiplying by 1. 93 00:06:27,430 --> 00:06:29,650 So we're not actually changing the value. 94 00:06:29,710 --> 00:06:36,040 But the reason that this helps us out is because we get limit and to infinity is because here we take 95 00:06:36,100 --> 00:06:43,180 and squared times one over and squared the squares here cancel and we're just left with one in the numerator 96 00:06:43,600 --> 00:06:51,670 and denominator we have to take this one over and squared and multiply it by all three of these terms 97 00:06:51,670 --> 00:06:52,050 here. 98 00:06:52,060 --> 00:06:58,300 The end squared the two ends and the one so one over and squared times and squared the square it's cancell 99 00:06:58,300 --> 00:07:06,580 and we get 1 2 and times 1 over and squared we would have to an over and squared this and here would 100 00:07:06,580 --> 00:07:09,890 cancel and we'd be left with 1 and we'd have 2 over. 101 00:07:09,930 --> 00:07:17,380 And so we'll get plus two over an and then 1 times 1 over and square it is of course just won over and 102 00:07:17,380 --> 00:07:18,900 squared like this. 103 00:07:18,910 --> 00:07:25,660 Now when we evaluated the limit as and goes to infinity we have just and in these denominators here 104 00:07:25,750 --> 00:07:31,540 and then we have ended going to infinity in the denominator and just a constant in the numerator a constant 105 00:07:31,540 --> 00:07:34,400 divided by infinity is going to become zero. 106 00:07:34,420 --> 00:07:44,170 So what we're left with is 1 over 1 plus 0 plus 0 because both of these become zero when we evaluate 107 00:07:44,170 --> 00:07:45,540 them at infinity. 108 00:07:45,550 --> 00:07:52,100 So obviously here we just have one divided by one or one the absolute value of 1 is still just 1. 109 00:07:52,240 --> 00:07:56,580 So we end up with 1 times 3 which is going to be equal to 3. 110 00:07:56,590 --> 00:07:59,970 So now we can say that L is equal to 3. 111 00:07:59,980 --> 00:08:05,290 This value of L here that we are trying to find from the beginning is going to be 3 so now we go over 112 00:08:05,290 --> 00:08:09,010 here to our ratio test and in our case we can say L is equal to 3. 113 00:08:09,010 --> 00:08:10,780 Well how does 3 relate to 1. 114 00:08:10,780 --> 00:08:12,270 3 is greater than 1. 115 00:08:12,310 --> 00:08:16,590 So we can say our is greater than 1 for this particular problem. 116 00:08:16,720 --> 00:08:20,540 Since I was greater than 1 and the ratio test tells us that the series diverges. 117 00:08:20,650 --> 00:08:26,380 We can say that this original series here three to the end over end squared is a divergent series by 118 00:08:26,380 --> 00:08:27,340 the ratio test. 13377