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These are the user uploaded subtitles that are being translated: 1 00:00:00,510 --> 00:00:04,140 Today we're going to be talking about how to determine whether or not the sequence converges. 2 00:00:04,140 --> 00:00:06,750 And if it does converge how to find its limit. 3 00:00:07,020 --> 00:00:12,750 And in this particular problem what we've been given the sequence A7 is equal to cosign squared of N 4 00:00:13,080 --> 00:00:16,020 divided by 2 to the end power. 5 00:00:16,020 --> 00:00:21,450 Now before we start in on this problem I want to talk a little bit about the instructions of the problem. 6 00:00:21,450 --> 00:00:25,970 These are the instructions that we've been given for this particular problem but I feel like they're 7 00:00:25,980 --> 00:00:31,650 a little bit misleading and often we're asked to find whether or not a sequence converges and then if 8 00:00:31,650 --> 00:00:36,690 it does to find its limit but that can be a little misleading because when it comes to sequences and 9 00:00:36,690 --> 00:00:41,270 keep in mind that this has nothing to do at all with series series are completely separate. 10 00:00:41,280 --> 00:00:45,300 This is for sequences only for sequences only. 11 00:00:45,450 --> 00:00:52,650 We can almost reverse these directions these instructions because if we can find the limit of a sequence 12 00:00:53,120 --> 00:00:57,210 and that limit exists then by definition the sequence converges. 13 00:00:57,210 --> 00:01:02,730 So these instructions could just as easily say find the limit of the sequence if it exists and then 14 00:01:02,790 --> 00:01:07,830 use your answer to determine whether or not the sequence converges So that's what we're going to do. 15 00:01:07,830 --> 00:01:13,230 I'm also going to show you how to use squeeze theorem quickly to determine first whether or not the 16 00:01:13,230 --> 00:01:14,580 sequence converges. 17 00:01:14,580 --> 00:01:16,320 And then to draw a conclusion about the limit. 18 00:01:16,320 --> 00:01:18,490 If you feel more comfortable doing it that way. 19 00:01:18,900 --> 00:01:22,060 But the first thing we're going to do here is try to find the limit. 20 00:01:22,060 --> 00:01:29,340 So we're talking about the limit as and approaches infinity of our sequence of the limit as it approaches 21 00:01:29,340 --> 00:01:36,890 infinity of a sob in that case of course is always just equal to the limit as and goes to infinity of 22 00:01:36,920 --> 00:01:37,930 our sequence here. 23 00:01:37,950 --> 00:01:43,120 Co-sign squared of n over 2 to the end power. 24 00:01:43,230 --> 00:01:48,120 Now the easiest way to try to find the limit when you have a rational function like this is to deal 25 00:01:48,120 --> 00:01:51,140 with the numerator and denominator separately. 26 00:01:51,330 --> 00:01:57,360 So if we look at the numerator here first cosigners squared of n and we're trying to figure out what 27 00:01:57,360 --> 00:02:00,780 the value is of this as and goes to infinity. 28 00:02:00,780 --> 00:02:03,660 We need to think about the cosigned function. 29 00:02:03,660 --> 00:02:10,170 Remember that the coastline function oscillates back and forth between the values negative 1 and positive 30 00:02:10,170 --> 00:02:12,510 one we know that from the graph of cosign. 31 00:02:12,540 --> 00:02:18,600 We also know it from the unit circle that the smallest value cosign ever attained is negative 1 the 32 00:02:18,600 --> 00:02:24,480 largest value it attains is positive ones so we can almost say cosign of N is going to be greater than 33 00:02:24,480 --> 00:02:28,820 or equal to negative 1 and less than or equal to positive one. 34 00:02:28,830 --> 00:02:30,700 So we already know that. 35 00:02:30,780 --> 00:02:36,270 Now if we square the cosign function right because we have cosigned squared of and here if we square 36 00:02:36,270 --> 00:02:41,240 the cosigned function all the values of cosigned between negative 1 and 0. 37 00:02:41,250 --> 00:02:44,980 All those negative values when we square them they become positive. 38 00:02:45,120 --> 00:02:48,940 But they're all going to be positive values between zero and positive one. 39 00:02:48,960 --> 00:02:49,170 Right. 40 00:02:49,170 --> 00:02:55,800 If we square the left limit here negative 1 we get positive 1 if we square zero we just get zero. 41 00:02:56,010 --> 00:03:00,260 And everything in between all the way from negative 1 to zero. 42 00:03:00,360 --> 00:03:05,880 All those you know negative points by values we square those they become positive values between 0 and 43 00:03:05,880 --> 00:03:06,570 1. 44 00:03:06,570 --> 00:03:12,570 So what we can say is that cosigners squared of N on the other hand will always be less than or equal 45 00:03:12,570 --> 00:03:15,620 to 1 but greater than or equal to zero. 46 00:03:15,630 --> 00:03:20,460 All the values now are going to fall in between 0 and 1. 47 00:03:20,460 --> 00:03:25,680 So that's our conclusion about cosigned squared of N that this numerator here is just going to oscillate 48 00:03:25,680 --> 00:03:29,850 back and forth between 0 and 1 and I'll never be less than zero. 49 00:03:29,880 --> 00:03:31,820 It'll never be greater than 1. 50 00:03:31,890 --> 00:03:38,130 So that's going to be some constant value maybe we can call it k we'll say the limit as an goes to infinity 51 00:03:38,880 --> 00:03:42,750 of K because it's always going to be between 0 and 1. 52 00:03:42,750 --> 00:03:44,850 Now we have to deal with two and here. 53 00:03:44,850 --> 00:03:49,310 So as we take the limit is and goes to infinity of 2 to the n. 54 00:03:49,410 --> 00:03:53,980 You can almost think about plugging in infinity here for and we have to do the Infinity well. 55 00:03:54,150 --> 00:04:00,760 Either way the larger you make this exponent the larger the denominator gets right 2 squared is four. 56 00:04:00,810 --> 00:04:03,120 But two to five is 32. 57 00:04:03,190 --> 00:04:07,770 And that are you just going to keep getting larger and larger as we take and larger and larger so the 58 00:04:07,770 --> 00:04:14,370 limit as and approaches infinity of this to the end here is going to be an infinitely large number so 59 00:04:14,370 --> 00:04:19,560 we can almost just think about it as k over infinity some huge number. 60 00:04:19,740 --> 00:04:24,210 Well whenever that's the case this is just going to be some value between 0 and 1. 61 00:04:24,210 --> 00:04:29,540 Either way the smallest value we would ever have here is zero over infinity. 62 00:04:29,550 --> 00:04:30,090 Right. 63 00:04:30,090 --> 00:04:35,650 And that's obviously just going to be 0 or the largest value would have for k would be 1. 64 00:04:35,700 --> 00:04:40,230 And when we divide that by infinity that's also going to give us zero because when we take any constant 65 00:04:40,230 --> 00:04:46,140 number and we divide it by an infinitely large number as this becomes larger and larger and larger This 66 00:04:46,140 --> 00:04:47,250 is always going to go to zero. 67 00:04:47,250 --> 00:04:54,480 So what this tells us is that the limit as end goes to infinity of our sequence here that this is going 68 00:04:54,480 --> 00:05:03,990 to be 0 that the limit as and goes to in the are sequence a Sabun is going to be equal to zero. 69 00:05:03,990 --> 00:05:07,680 This tells us that the limit exists that the limit is zero. 70 00:05:07,680 --> 00:05:16,320 And because the limit exists and it's zero we automatically know therefore that the sequence converges. 71 00:05:16,500 --> 00:05:22,800 And so notice that we instead of following the instructions in the order that they gave them to us we 72 00:05:22,800 --> 00:05:24,110 first found the limit. 73 00:05:24,120 --> 00:05:26,650 We proved the limit was equal to zero. 74 00:05:26,670 --> 00:05:32,760 We found that it existed and because it existed we know by the definition of convergence for a sequence 75 00:05:32,970 --> 00:05:35,160 that the sequence does in fact converge. 76 00:05:35,160 --> 00:05:41,520 So you can just find it that way if you want to try to prove without the limit necessarily that the 77 00:05:41,520 --> 00:05:42,450 sequence converges. 78 00:05:42,480 --> 00:05:47,250 You can always try to use something like squeeze theorem and it would look something like this that 79 00:05:47,250 --> 00:05:52,620 we already started right where we said that cosign of and was always going to be greater than equal 80 00:05:52,620 --> 00:05:57,030 to negative 1 and less than or equal to 1 we squared cosign again. 81 00:05:57,030 --> 00:06:04,440 Now if we multiply this through by one over two to the end because we're trying to get this middle function 82 00:06:04,440 --> 00:06:07,330 here back to our original sequence. 83 00:06:07,350 --> 00:06:12,120 So when you multiply through one or two to the end times zero is still zero. 84 00:06:12,540 --> 00:06:19,860 The middle function here becomes cosign squared of an over 2 to the n r sequence and then less than 85 00:06:19,860 --> 00:06:23,520 or equal to we just get 1 over 2 to the end. 86 00:06:23,610 --> 00:06:30,360 Now if we take the limit of this whole inequality as and goes to infinity we take the limit of the whole 87 00:06:30,360 --> 00:06:31,020 thing. 88 00:06:31,080 --> 00:06:38,460 What we can do is distribute this limit across at each one so the limit as end goes to infinity of zero 89 00:06:38,790 --> 00:06:40,270 is still just 0. 90 00:06:40,290 --> 00:06:48,470 OK so less than or equal to than the limit as an goes to infinity of our function here cosign squared 91 00:06:48,540 --> 00:06:59,050 of an over to the N or sequence less than or equal to the limit as and NGOs to infinity of 1 over 2 92 00:06:59,060 --> 00:06:59,930 to the end. 93 00:07:00,010 --> 00:07:06,180 Well we already know from looking at this equation here that when we have some constant divided by some 94 00:07:06,270 --> 00:07:11,700 infinitely large number when we plug in infinity here for n this denominator is going to get infinitely 95 00:07:11,700 --> 00:07:12,510 large. 96 00:07:12,540 --> 00:07:18,310 We know that this value right here is going to go to zero that's going to become zero. 97 00:07:18,480 --> 00:07:23,730 So what this tells us is that the limit as and goes to infinity of our sequence because it's always 98 00:07:23,730 --> 00:07:28,380 going to be greater than or equal to zero and less than or equal to zero by the squeeze theorem. 99 00:07:28,380 --> 00:07:31,860 We've proven that this converges to zero. 100 00:07:31,890 --> 00:07:33,640 That the limit is equal to zero. 101 00:07:33,690 --> 00:07:35,160 So that's another way. 102 00:07:35,370 --> 00:07:37,250 Without first finding the limit. 103 00:07:37,270 --> 00:07:41,490 Indirectly that's another way to prove that the sequence converges. 104 00:07:41,640 --> 00:07:46,200 And then to find that the limit is also equal to zero by the squeeze them. 11787