1 00:00:00,000 --> 00:00:01,940 MALE SPEAKER: The following content is provided under a 2 00:00:01,940 --> 00:00:03,690 Creative Commons license. 3 00:00:03,690 --> 00:00:06,630 Your support will help MIT OpenCourseWare continue to 4 00:00:06,630 --> 00:00:09,980 offer high quality educational resources for free. 5 00:00:09,980 --> 00:00:12,830 To make a donation, or to view additional materials from 6 00:00:12,830 --> 00:00:16,760 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:16,760 --> 00:00:18,010 ocw.mit.edu. 8 00:00:31,480 --> 00:00:33,050 PROFESSOR: Hi. 9 00:00:33,050 --> 00:00:37,360 Last time, we had discussed series and sequences. 10 00:00:37,360 --> 00:00:41,010 Today, we're going to turn our attention to a rather special 11 00:00:41,010 --> 00:00:45,520 situation, a situation in which every term in our series 12 00:00:45,520 --> 00:00:46,720 is positive. 13 00:00:46,720 --> 00:00:49,630 For this reason, I have entitled today's lecture 14 00:00:49,630 --> 00:00:51,160 'Positive Series'. 15 00:00:51,160 --> 00:00:54,760 Before we can do full justice to positive series, however, 16 00:00:54,760 --> 00:00:58,770 there are a few topics that we must discuss as preliminaries. 17 00:00:58,770 --> 00:01:02,710 The first of these is simply the process of ordering. 18 00:01:02,710 --> 00:01:05,290 Now this is a rather strange situation. 19 00:01:05,290 --> 00:01:07,550 Because in the finite case, it turns out 20 00:01:07,550 --> 00:01:08,670 to be rather trivial. 21 00:01:08,670 --> 00:01:12,550 By way of illustration, let's suppose the set 'S' consists 22 00:01:12,550 --> 00:01:16,860 of the numbers 11, 8, 9, 7, and 10. 23 00:01:16,860 --> 00:01:18,510 Now there are many ways in which we could 24 00:01:18,510 --> 00:01:19,610 order the set 'S'. 25 00:01:19,610 --> 00:01:22,530 We can arrange them so that any one of these five elements 26 00:01:22,530 --> 00:01:24,340 comes first, any one of the remaining 27 00:01:24,340 --> 00:01:25,790 four second, et cetera. 28 00:01:25,790 --> 00:01:28,270 But suppose that we want to arrange these 29 00:01:28,270 --> 00:01:29,940 according to size. 30 00:01:29,940 --> 00:01:33,990 Observe that there is a rather straightforward, shall we say, 31 00:01:33,990 --> 00:01:37,240 binary technique, whereby we can order these elements. 32 00:01:37,240 --> 00:01:40,090 By binary, I mean, let's look at these two at the time. 33 00:01:40,090 --> 00:01:43,020 We look at 11 and 8, and we throw away the larger of the 34 00:01:43,020 --> 00:01:44,220 two, which is 11. 35 00:01:44,220 --> 00:01:47,020 Then we compare 8 with 9, throwaway 9 36 00:01:47,020 --> 00:01:48,180 because that's bigger. 37 00:01:48,180 --> 00:01:51,520 Compare 8 and 7, throw away 8. 38 00:01:51,520 --> 00:01:55,240 Compare 7 and 10, throw away 10. 39 00:01:55,240 --> 00:01:59,685 The survivor, being 7, is therefore the least member of 40 00:01:59,685 --> 00:02:00,790 our collection. 41 00:02:00,790 --> 00:02:04,810 In a similar way, we can delete 7 and start looking for 42 00:02:04,810 --> 00:02:06,770 the next number of our set. 43 00:02:06,770 --> 00:02:10,449 And in this way, eventually order the elements of 'S' 44 00:02:10,449 --> 00:02:11,390 according to size-- 45 00:02:11,390 --> 00:02:14,080 7, 8, 9, 10, 11. 46 00:02:14,080 --> 00:02:14,690 OK. 47 00:02:14,690 --> 00:02:20,690 Clearly, 7 is the smallest element of 'S' and 11 is the 48 00:02:20,690 --> 00:02:22,860 largest element of S. 49 00:02:22,860 --> 00:02:27,210 And in terms of nomenclature, we say that 7 is a lower bound 50 00:02:27,210 --> 00:02:30,840 for 'S', 11 is an upper bound for 'S'. 51 00:02:30,840 --> 00:02:34,070 You see, in terms of a picture, all we're saying is 52 00:02:34,070 --> 00:02:37,160 that when the elements of 'S' are ordered according to size, 53 00:02:37,160 --> 00:02:41,220 7 is the furthest to the left, 11 is the 54 00:02:41,220 --> 00:02:43,040 furthest to the right. 55 00:02:43,040 --> 00:02:46,490 We could even talk more about the nomenclature by saying-- 56 00:02:46,490 --> 00:02:48,210 and this sounds like a tongue twister. 57 00:02:48,210 --> 00:02:51,290 This is why I had you read this material first in the 58 00:02:51,290 --> 00:02:53,660 last assignment and the supplementary notes, so that 59 00:02:53,660 --> 00:02:55,990 part of this will at least seem like a review. 60 00:02:55,990 --> 00:02:59,740 Observe that 7 is called the greatest lower bound for 'S', 61 00:02:59,740 --> 00:03:04,570 simply because any number larger than 7 cannot be a 62 00:03:04,570 --> 00:03:08,350 lower bound for 'S', simply because 7 would be smaller 63 00:03:08,350 --> 00:03:09,880 than that particular number. 64 00:03:09,880 --> 00:03:13,690 And in a similar way, 11 is called the least upper bound 65 00:03:13,690 --> 00:03:17,650 for 'S', because any number smaller than 11 would be 66 00:03:17,650 --> 00:03:20,520 exceeded by 11, and hence could not be an 67 00:03:20,520 --> 00:03:22,260 upper bound for 'S'. 68 00:03:22,260 --> 00:03:24,110 Now the interesting point is this. 69 00:03:24,110 --> 00:03:27,000 Hopefully, at this stage of the game, you listened to what 70 00:03:27,000 --> 00:03:30,810 I've had to say, and you say, my golly, this is trivial. 71 00:03:30,810 --> 00:03:32,720 And the answer is, it is trivial. 72 00:03:32,720 --> 00:03:36,500 But remember what our main concern was in our last 73 00:03:36,500 --> 00:03:40,460 lecture when we introduced the concept of infinite series and 74 00:03:40,460 --> 00:03:41,490 sequences-- 75 00:03:41,490 --> 00:03:45,740 that many things that were trivial for the finite case 76 00:03:45,740 --> 00:03:49,800 became rather serious dilemmas for the infinite case. 77 00:03:49,800 --> 00:03:52,960 In other words, my claim is that these results are far 78 00:03:52,960 --> 00:03:55,260 more subtle for infinite sets. 79 00:03:55,260 --> 00:03:57,160 And I think the best way to do this is 80 00:03:57,160 --> 00:03:59,000 by means of an example. 81 00:03:59,000 --> 00:04:03,560 See now, let 'S' be the set of numbers where the n-th number 82 00:04:03,560 --> 00:04:05,860 is 'n' over 'n + 1'. 83 00:04:05,860 --> 00:04:08,480 In other words, the first member of 'S' will be 1/2, the 84 00:04:08,480 --> 00:04:13,300 second member, 2/3, the third member, 3/4, et cetera. 85 00:04:13,300 --> 00:04:16,050 Now, let's take look to see what the least upper 86 00:04:16,050 --> 00:04:17,350 bound for 'S' is. 87 00:04:17,350 --> 00:04:20,500 And sparing you the details, I think you can observe at this 88 00:04:20,500 --> 00:04:24,070 stage of the game, especially based on the homework of the 89 00:04:24,070 --> 00:04:30,390 last unit, that the limit of the sequence 'S' is 1. 90 00:04:30,390 --> 00:04:34,220 In fact, 1 is the smallest number which exceeds every 91 00:04:34,220 --> 00:04:35,690 member in this collection. 92 00:04:35,690 --> 00:04:38,740 In other words, 1 is the least upper bound for 'S'. 93 00:04:38,740 --> 00:04:41,945 But observe the interesting case that here-- 94 00:04:41,945 --> 00:04:43,820 and by the way, notice the abbreviation that we use in 95 00:04:43,820 --> 00:04:46,700 our notes. 'LUB', least upper bound. 96 00:04:46,700 --> 00:04:48,890 'GLB', greatest lower bound. 97 00:04:48,890 --> 00:04:51,730 But 1 is the least upper bound for 'S'. 98 00:04:51,730 --> 00:04:55,740 Yet the fact remains that the least upper bound is not a 99 00:04:55,740 --> 00:04:57,160 member of 'S' itself. 100 00:04:57,160 --> 00:05:00,960 In other words, there is no number 'n' such that 'n' over 101 00:05:00,960 --> 00:05:03,470 'n + 1' is equal to 1. 102 00:05:03,470 --> 00:05:06,320 You see, notice that as these numbers increase, they get 103 00:05:06,320 --> 00:05:09,620 bounded by 1, but 1 is not a member of 'S'. 104 00:05:09,620 --> 00:05:12,380 In other words, here's an example where the least upper 105 00:05:12,380 --> 00:05:16,110 bound of a set does not have to be a member of the set. 106 00:05:16,110 --> 00:05:18,740 And a companion to this would be an example where the 107 00:05:18,740 --> 00:05:21,950 greatest lower bound is not a member of the set. 108 00:05:21,950 --> 00:05:26,030 And to this end, simply let the n-th member of 'S' 109 00:05:26,030 --> 00:05:29,220 arranged by sequence be '1/n'. 110 00:05:29,220 --> 00:05:31,000 The n-th member is '1/n'. 111 00:05:31,000 --> 00:05:33,020 Therefore, 'S' is what? 112 00:05:33,020 --> 00:05:37,590 The set consisting of 1, 1/2, 1/3, et cetera. 113 00:05:37,590 --> 00:05:41,550 Observe that as the terms go further and further out, they 114 00:05:41,550 --> 00:05:44,650 get arbitrarily close to 0 in size-- 115 00:05:44,650 --> 00:05:47,310 every one of these terms, '1/n', no matter how big 'n' 116 00:05:47,310 --> 00:05:48,890 is is greater than 0. 117 00:05:48,890 --> 00:05:52,150 In other words, then, observe that 0 will be the greatest 118 00:05:52,150 --> 00:05:56,350 lower bound for 'S', but 0 is not a member of 'S'. 119 00:05:56,350 --> 00:05:59,690 In other words, it seems that things which are quite trivial 120 00:05:59,690 --> 00:06:02,550 for finite collections have certain degrees of 121 00:06:02,550 --> 00:06:04,870 sophistication for infinite collections. 122 00:06:04,870 --> 00:06:08,440 So what we're going to do now is to establish a few basic 123 00:06:08,440 --> 00:06:09,780 definitions. 124 00:06:09,780 --> 00:06:11,380 And we'll do it in a rather formal way. 125 00:06:11,380 --> 00:06:13,790 Our first definition is the following. 126 00:06:13,790 --> 00:06:17,000 Given the set of numbers, 'S', 'M' is called an upper bound 127 00:06:17,000 --> 00:06:21,020 for 'S', If 'M' is greater than or equal to 'x' for all 128 00:06:21,020 --> 00:06:22,410 'x' in 'S'. 129 00:06:22,410 --> 00:06:26,230 In other words, if 'M' exceeds every member of 'S', 'M' is 130 00:06:26,230 --> 00:06:29,920 called an upper bound for 'S'. 131 00:06:29,920 --> 00:06:32,590 As I say, these these definitions are very 132 00:06:32,590 --> 00:06:34,140 straightforward. 133 00:06:34,140 --> 00:06:35,430 The companion-- 134 00:06:35,430 --> 00:06:37,380 well, let's go one step further. 135 00:06:37,380 --> 00:06:39,900 By the way, observe that I have in the interest of 136 00:06:39,900 --> 00:06:43,100 brevity left out the corresponding definitions for 137 00:06:43,100 --> 00:06:43,740 lower bounds. 138 00:06:43,740 --> 00:06:45,680 But they are quite analogous. 139 00:06:45,680 --> 00:06:49,760 In other words, a lower bound for 'S' would be a number that 140 00:06:49,760 --> 00:06:53,080 was less than or equal to each member in 'S'. 141 00:06:53,080 --> 00:06:56,830 At any rate, then, 'M' is called a least upper bound for 142 00:06:56,830 --> 00:07:00,580 'S' if first of all, 'M' is an upper bound for 'S'. 143 00:07:00,580 --> 00:07:05,010 And secondly, if 'L' is less than 'M', 'L' is not an upper 144 00:07:05,010 --> 00:07:06,090 bound for 'S'. 145 00:07:06,090 --> 00:07:08,260 In other words, least upper bound means what? 146 00:07:08,260 --> 00:07:12,190 Anything smaller cannot be an upper bound. 147 00:07:12,190 --> 00:07:15,540 Notice in terms of our previous remark that the least 148 00:07:15,540 --> 00:07:19,630 upper bound need not belong to 'x'. 149 00:07:19,630 --> 00:07:21,860 The companion to this would be what? 150 00:07:21,860 --> 00:07:25,620 A greatest lower bound would be a number which is a lower 151 00:07:25,620 --> 00:07:28,790 bound such that anything greater could 152 00:07:28,790 --> 00:07:30,470 not be a lower bound. 153 00:07:30,470 --> 00:07:34,950 Again, all of this is easier to see in terms of a picture. 154 00:07:34,950 --> 00:07:39,220 Visualize 'S' as being this interval with little 'm' and 155 00:07:39,220 --> 00:07:41,910 capital 'M' being the endpoints of the interval. 156 00:07:41,910 --> 00:07:45,280 Observe that capital 'M' is the least upper bound. 157 00:07:45,280 --> 00:07:47,900 Little 'm' is the greatest lower bound. 158 00:07:47,900 --> 00:07:52,470 Anything smaller than little 'm' will be lower bound. 159 00:07:52,470 --> 00:07:56,690 Anything greater than capital 'M' will be an upper bound. 160 00:07:56,690 --> 00:08:00,770 And notice that nothing in the set 'S' itself can be either 161 00:08:00,770 --> 00:08:03,140 an upper bound or a lower bound. 162 00:08:03,140 --> 00:08:08,210 Because anything inside 'S' appears to the right of little 163 00:08:08,210 --> 00:08:11,230 'm' and to the left of capital 'M'. 164 00:08:11,230 --> 00:08:14,220 And one final definition as a preliminary. 165 00:08:14,220 --> 00:08:18,410 A set, 'S', is called bounded if it has both an upper and a 166 00:08:18,410 --> 00:08:19,790 lower bound. 167 00:08:19,790 --> 00:08:23,080 And the key property that we have to keep track of 168 00:08:23,080 --> 00:08:24,110 throughout this-- 169 00:08:24,110 --> 00:08:25,210 and we won't prove this. 170 00:08:25,210 --> 00:08:27,760 In other words, in more rigorous advanced math 171 00:08:27,760 --> 00:08:30,730 courses, this is proven as a theorem. 172 00:08:30,730 --> 00:08:33,739 For our purposes, it seems self-evident enough so that 173 00:08:33,739 --> 00:08:35,100 we're willing to accept it. 174 00:08:35,100 --> 00:08:37,830 And so rather than to belabor the point, let us just accept 175 00:08:37,830 --> 00:08:41,870 as a key property that every bounded set of numbers has a 176 00:08:41,870 --> 00:08:44,360 greatest lower bound and at least upper bound. 177 00:08:44,360 --> 00:08:48,140 In other words, if a set is bounded, we can find a 178 00:08:48,140 --> 00:08:52,680 smallest upper bound and a largest lower bound. 179 00:08:52,680 --> 00:08:55,390 And this completes the first portion of 180 00:08:55,390 --> 00:08:57,550 our preliminary material. 181 00:08:57,550 --> 00:09:00,050 The next portion of our preliminary material before 182 00:09:00,050 --> 00:09:04,690 studying positive series involves what we mean by a 183 00:09:04,690 --> 00:09:09,010 monotonic non-decreasing sequence. 184 00:09:09,010 --> 00:09:12,580 A sequence is called monotonic non-decreasing-- 185 00:09:12,580 --> 00:09:16,160 and if you don't frighten at these words, it's almost 186 00:09:16,160 --> 00:09:18,100 self-evident what this thing means. 187 00:09:18,100 --> 00:09:21,120 It means that no term can be smaller than the 188 00:09:21,120 --> 00:09:22,580 one that came before. 189 00:09:22,580 --> 00:09:26,370 In other words, the n-th term is less than or equal to the 190 00:09:26,370 --> 00:09:29,070 'n plus first term' for each 'n'. 191 00:09:29,070 --> 00:09:32,070 And wording that more explicitly, it says what? 192 00:09:32,070 --> 00:09:35,010 'a sub 1' is less than or equal to 'a sub 2' is less 193 00:09:35,010 --> 00:09:37,380 than or equal to 'a sub 3', et cetera. 194 00:09:37,380 --> 00:09:40,030 And I hope that it's clear by this time that it's not 195 00:09:40,030 --> 00:09:42,560 self-evident that a sequence has to have this. 196 00:09:42,560 --> 00:09:45,510 Remember the subscripts simply tell you the order in which 197 00:09:45,510 --> 00:09:46,610 the terms appear. 198 00:09:46,610 --> 00:09:49,850 It has no bearing on the size of the term. 199 00:09:49,850 --> 00:09:52,860 For example, in an arbitrary sequence, recall that the 200 00:09:52,860 --> 00:09:56,100 second term can be smaller in magnitude than the first term. 201 00:09:56,100 --> 00:10:00,540 However, if the terms are non-decreasing sequentially 202 00:10:00,540 --> 00:10:03,350 this way, the sequence is called monotonic 203 00:10:03,350 --> 00:10:04,560 non-decreasing. 204 00:10:04,560 --> 00:10:07,260 And the problem that comes up is, or the important question 205 00:10:07,260 --> 00:10:11,130 that comes up is, what's so important about monotonic 206 00:10:11,130 --> 00:10:13,000 non-decreasing sequences? 207 00:10:13,000 --> 00:10:15,730 And the answer is that for such sequences, two 208 00:10:15,730 --> 00:10:17,530 possibilities exist. 209 00:10:17,530 --> 00:10:21,600 In other words, either the terms can keep getting larger 210 00:10:21,600 --> 00:10:23,695 and larger without bound-- 211 00:10:23,695 --> 00:10:24,850 see? 212 00:10:24,850 --> 00:10:26,940 In other words, that the sequences 'a sub n' has no 213 00:10:26,940 --> 00:10:28,160 upper bound. 214 00:10:28,160 --> 00:10:31,140 In which case, we say that the limit of 'a sub n' as 'n' 215 00:10:31,140 --> 00:10:34,070 approaches infinity is infinity. 216 00:10:34,070 --> 00:10:36,320 And for example, the ordinary counting 217 00:10:36,320 --> 00:10:37,950 sequence has this property. 218 00:10:37,950 --> 00:10:42,400 See, 1, 2, 3, 4, 5, et cetera, is a monotonic 219 00:10:42,400 --> 00:10:43,565 non-decreasing sequence. 220 00:10:43,565 --> 00:10:45,930 In fact, it's monotonic increasing. 221 00:10:45,930 --> 00:10:49,000 Every member of the sequence is greater than the one that 222 00:10:49,000 --> 00:10:49,980 came before. 223 00:10:49,980 --> 00:10:52,440 But as you go further and further out, the terms 224 00:10:52,440 --> 00:10:54,750 increase without upper bound. 225 00:10:54,750 --> 00:10:56,000 OK? 226 00:10:56,000 --> 00:11:00,170 Now what is the other possibility for a monotonic 227 00:11:00,170 --> 00:11:01,980 non-decreasing sequence? 228 00:11:01,980 --> 00:11:06,450 After all, the opposite, or the only other possibility, is 229 00:11:06,450 --> 00:11:09,520 that if the 'a sub n'-- if it's false, that the sequence 230 00:11:09,520 --> 00:11:12,460 doesn't have an upper bound, then of course it must have an 231 00:11:12,460 --> 00:11:12,960 upper bound. 232 00:11:12,960 --> 00:11:14,350 That's the second case. 233 00:11:14,350 --> 00:11:16,950 In other words, suppose the sequence has an upper bound. 234 00:11:16,950 --> 00:11:19,710 Then the interesting point is that the limit of this 235 00:11:19,710 --> 00:11:21,790 sequence exists. 236 00:11:21,790 --> 00:11:25,980 And not only does it exist, but the limit itself is the 237 00:11:25,980 --> 00:11:28,500 least upper bound of the sequence. 238 00:11:28,500 --> 00:11:32,060 In other words, in the case where the sequence is 239 00:11:32,060 --> 00:11:35,680 non-decreasing, if it's bounded-- 240 00:11:35,680 --> 00:11:36,900 if it's bounded-- 241 00:11:36,900 --> 00:11:39,180 the least upper bound will be the limit. 242 00:11:39,180 --> 00:11:41,150 Now you see, here's where we use that key property. 243 00:11:41,150 --> 00:11:45,020 Namely, if a sequence is bounded, it must have a least 244 00:11:45,020 --> 00:11:45,990 upper bound. 245 00:11:45,990 --> 00:11:48,650 Let's call that least upper bound 'L'. 246 00:11:48,650 --> 00:11:50,780 And by the way, you'll notice I wrote the word proof in 247 00:11:50,780 --> 00:11:51,910 quotation marks. 248 00:11:51,910 --> 00:11:54,920 It's simply to indicate that I prefer to give you a geometric 249 00:11:54,920 --> 00:11:57,630 proof here rather than an analytic one. 250 00:11:57,630 --> 00:12:00,060 But the analytic proof follows word for 251 00:12:00,060 --> 00:12:02,780 word from this picture. 252 00:12:02,780 --> 00:12:05,530 In other words, it just translates in the usual way. 253 00:12:05,530 --> 00:12:08,750 And we'll drill on this with the exercises and the notes 254 00:12:08,750 --> 00:12:09,770 and the textbook. 255 00:12:09,770 --> 00:12:11,690 But at any rate, the idea is this. 256 00:12:11,690 --> 00:12:14,960 To prove that 'L' is a limit, what must I do? 257 00:12:14,960 --> 00:12:18,700 I must show that if I choose any epsilon greater than 0, 258 00:12:18,700 --> 00:12:22,860 that all the terms beyond a certain one lie between 'L 259 00:12:22,860 --> 00:12:25,920 minus epsilon' and 'L plus epsilon'. 260 00:12:25,920 --> 00:12:27,320 Now here's the way this works. 261 00:12:27,320 --> 00:12:29,360 Let's see if we can just read the diagram and get this thing 262 00:12:29,360 --> 00:12:30,390 very quickly. 263 00:12:30,390 --> 00:12:35,060 First of all, at least one term in my sequence has to be 264 00:12:35,060 --> 00:12:37,590 between 'L minus epsilon' and 'L'. 265 00:12:37,590 --> 00:12:39,730 And the reason for that is simply this. 266 00:12:39,730 --> 00:12:41,800 Remember 'L' is a least upper bound. 267 00:12:41,800 --> 00:12:44,890 Because 'L' is a least upper bound, 'L minus epsilon' 268 00:12:44,890 --> 00:12:46,580 cannot be an upper bound. 269 00:12:46,580 --> 00:12:50,080 Now if no term can get beyond 'L minus epsilon', then 270 00:12:50,080 --> 00:12:53,320 certainly 'L minus epsilon' would be an upper bound. 271 00:12:53,320 --> 00:12:55,920 The fact that 'L minus epsilon' isn't an upper bound, 272 00:12:55,920 --> 00:13:00,770 therefore, means that at least one 'a sub n', say 'a sub 273 00:13:00,770 --> 00:13:04,220 capital N', is in this interval here. 274 00:13:04,220 --> 00:13:09,340 Notice also that because 'L' is an upper bound, no 'a sub 275 00:13:09,340 --> 00:13:11,320 n' can get beyond here. 276 00:13:11,320 --> 00:13:13,840 In other words, no 'a sub n' lies between 277 00:13:13,840 --> 00:13:15,700 'L' and 'L plus epsilon'. 278 00:13:15,700 --> 00:13:18,560 Because in that particular case, if that were to happen, 279 00:13:18,560 --> 00:13:20,960 'L' couldn't be an upper bound. 280 00:13:20,960 --> 00:13:22,720 OK, so far, so good. 281 00:13:22,720 --> 00:13:24,980 That follows just from the definition of 282 00:13:24,980 --> 00:13:26,370 the least upper bound. 283 00:13:26,370 --> 00:13:30,320 Now, we use the fact that the sequence is non-decreasing. 284 00:13:30,320 --> 00:13:31,660 And that says what? 285 00:13:31,660 --> 00:13:36,165 That if little 'n' is greater than capital 'N', 'a sub 286 00:13:36,165 --> 00:13:39,430 little n' is greater than or equal to 'a sub capital N'. 287 00:13:39,430 --> 00:13:42,360 In other words, what this means is if you list the 'a 288 00:13:42,360 --> 00:13:46,740 sub n's along this line, as 'n' increases, the terms move 289 00:13:46,740 --> 00:13:49,500 progressively from left to right. 290 00:13:49,500 --> 00:13:56,680 In other words, notice then that if 'n' is greater than or 291 00:13:56,680 --> 00:14:04,190 equal to capital 'N', 'L minus epsilon' is less than 'a sub 292 00:14:04,190 --> 00:14:07,220 capital N', which in turn is less than or equal to 'a sub 293 00:14:07,220 --> 00:14:10,220 little n', which in turn is less than or equal to 'L'. 294 00:14:10,220 --> 00:14:12,900 And that's just another geometric way of saying that 295 00:14:12,900 --> 00:14:16,050 'a sub n' is within epsilon of 'L'. 296 00:14:16,050 --> 00:14:19,630 And that's exactly the definition that the 'a sub n's 297 00:14:19,630 --> 00:14:21,250 converge to the limit 'L'. 298 00:14:21,250 --> 00:14:24,520 Now I went over this rather quickly simply because this is 299 00:14:24,520 --> 00:14:25,840 done in the textbook. 300 00:14:25,840 --> 00:14:28,500 And all I'm trying to do with this lecture is to give you a 301 00:14:28,500 --> 00:14:31,040 quick overview of what's going on. 302 00:14:31,040 --> 00:14:33,970 Well, you see, we're now roughly 15 minutes into our 303 00:14:33,970 --> 00:14:37,460 lecture and we haven't come to the main topic yet. 304 00:14:37,460 --> 00:14:40,410 My claim is that the main topic works very, very 305 00:14:40,410 --> 00:14:44,140 smoothly once we understand these preliminaries. 306 00:14:44,140 --> 00:14:47,600 The main topic, you see, is called positive series. 307 00:14:47,600 --> 00:14:49,850 And the definition of a positive series is just as the 308 00:14:49,850 --> 00:14:51,230 name implies. 309 00:14:51,230 --> 00:14:55,890 If each term in the series is at least as great as 0-- in 310 00:14:55,890 --> 00:14:58,470 other words, if each term is non-negative-- 311 00:14:58,470 --> 00:15:01,080 then the series is called positive. 312 00:15:01,080 --> 00:15:04,010 Now why are positive series important? 313 00:15:04,010 --> 00:15:07,320 And why do they tie in with our previous discussion? 314 00:15:07,320 --> 00:15:10,290 Well, let's answer the last question. 315 00:15:10,290 --> 00:15:13,140 The reason they tie in with our previous discussion is 316 00:15:13,140 --> 00:15:17,990 that we have already seen that by the sum of a series, we 317 00:15:17,990 --> 00:15:20,820 mean the limit of the sequence of partial sums. 318 00:15:20,820 --> 00:15:24,150 To go from one partial sum to the next, you add on the next 319 00:15:24,150 --> 00:15:25,290 term in the series. 320 00:15:25,290 --> 00:15:28,430 If each of the 'a sub n's is positive, or at least 321 00:15:28,430 --> 00:15:32,590 non-negative, notice then that the sequence of partial sums 322 00:15:32,590 --> 00:15:34,860 is monotonic non-decreasing. 323 00:15:34,860 --> 00:15:35,670 Why? 324 00:15:35,670 --> 00:15:38,790 Because to go from the n-th partial sums to the 'n plus 325 00:15:38,790 --> 00:15:43,460 first' partial sum, you add on 'a sub 'n plus 1''. 326 00:15:43,460 --> 00:15:47,560 And since 'a sub 'n plus 1'' is at least as big as 0, it 327 00:15:47,560 --> 00:15:51,040 means that the 'n plus first' partial sum can be no smaller 328 00:15:51,040 --> 00:15:52,650 than the n-th partial sum. 329 00:15:52,650 --> 00:15:57,860 In other words, therefore, if the series summation 'n' goes 330 00:15:57,860 --> 00:16:01,780 from 1 to infinity, 'a sub n' is a positive series, it must 331 00:16:01,780 --> 00:16:05,900 either diverge to infinity, or else it converges to the limit 332 00:16:05,900 --> 00:16:09,400 'L', where 'L' is the least upper bound for the sequence 333 00:16:09,400 --> 00:16:11,140 of partial sums. 334 00:16:11,140 --> 00:16:11,660 OK? 335 00:16:11,660 --> 00:16:15,470 Now qualitatively, that's the end of story. 336 00:16:15,470 --> 00:16:16,820 In other words, we now know what? 337 00:16:16,820 --> 00:16:20,380 For a positive series, it either diverges to infinity, 338 00:16:20,380 --> 00:16:24,570 or else it converges to a sum, a limit. 339 00:16:24,570 --> 00:16:25,820 And that limit is what? 340 00:16:25,820 --> 00:16:29,460 The least upper bound for the sequence of partial sums. 341 00:16:29,460 --> 00:16:32,610 The problem is that quantitatively, we would like 342 00:16:32,610 --> 00:16:36,590 to have some criteria for determining whether a positive 343 00:16:36,590 --> 00:16:38,610 series falls into the first category 344 00:16:38,610 --> 00:16:40,190 or the second category. 345 00:16:40,190 --> 00:16:42,950 Notice, how can we tell whether the series converges 346 00:16:42,950 --> 00:16:44,900 or whether it diverges? 347 00:16:44,900 --> 00:16:47,670 And you see, the reading material in the text for this 348 00:16:47,670 --> 00:16:52,360 assignment focuses attention on three major tests. 349 00:16:52,360 --> 00:16:55,060 And these are the ones I'd like to go over with you once 350 00:16:55,060 --> 00:16:57,080 lightly, so to speak. 351 00:16:57,080 --> 00:17:01,300 The first test is called the comparison test. 352 00:17:01,300 --> 00:17:03,140 And it almost sounds self-evident. 353 00:17:03,140 --> 00:17:05,869 I just want to outline how these proofs go. 354 00:17:05,869 --> 00:17:08,869 Because I think that once you hear these things spoken, as 355 00:17:08,869 --> 00:17:12,780 you read the material, the formal proofs will fit into a 356 00:17:12,780 --> 00:17:16,010 pattern much more nicely than if you haven't heard the stuff 357 00:17:16,010 --> 00:17:18,060 said out loud, you see. 358 00:17:18,060 --> 00:17:19,079 The idea is this. 359 00:17:19,079 --> 00:17:22,270 Let's suppose that we know that summation 'n' goes from 1 360 00:17:22,270 --> 00:17:27,150 to infinity, 'C sub n', is a convergent positive series. 361 00:17:27,150 --> 00:17:29,070 In other words, all of these are positive, 362 00:17:29,070 --> 00:17:30,920 and the series converges. 363 00:17:30,920 --> 00:17:33,700 Suppose I now have another sequence of numbers, 'u sub 364 00:17:33,700 --> 00:17:37,210 n', where each 'u sub n' is positive-- 365 00:17:37,210 --> 00:17:39,800 see, it's at least as big as 0-- but can be no bigger than 366 00:17:39,800 --> 00:17:42,170 'C sub n' for each 'n'. 367 00:17:42,170 --> 00:17:45,660 Then the statement is that the series formed by adding up the 368 00:17:45,660 --> 00:17:48,080 'u sub n's must also converge. 369 00:17:48,080 --> 00:17:50,810 Notice what you're saying is, here's a bunch of positive 370 00:17:50,810 --> 00:17:53,920 terms that can't get too large. 371 00:17:53,920 --> 00:17:56,360 And these terms in magnitude are less 372 00:17:56,360 --> 00:17:57,560 than or equal to these. 373 00:17:57,560 --> 00:17:59,760 Therefore, what you're saying is that this sum can't get too 374 00:17:59,760 --> 00:18:01,030 large either. 375 00:18:01,030 --> 00:18:03,770 And if you want to verbalize that so that it becomes more 376 00:18:03,770 --> 00:18:06,080 formal, the idea is this. 377 00:18:06,080 --> 00:18:11,700 Let 'T sub n' denote the n-th partial sums of the series 378 00:18:11,700 --> 00:18:12,980 involving the 'C sub n's. 379 00:18:12,980 --> 00:18:17,220 In other words, let 'T sub n' be 'C sub 1' plus et cetera, 380 00:18:17,220 --> 00:18:19,020 up to 'C sub n'. 381 00:18:19,020 --> 00:18:22,780 And let 'S sub n' be the n-th term in the sequence of 382 00:18:22,780 --> 00:18:25,470 partial sums of the series involving the 'u's. 383 00:18:25,470 --> 00:18:29,010 In other words, let 'S sub n' be 'u1' plus et cetera 384 00:18:29,010 --> 00:18:31,080 up to 'u sub n'. 385 00:18:31,080 --> 00:18:32,630 Now the idea is-- lookit. 386 00:18:32,630 --> 00:18:35,870 We know that each of the 'u's in magnitude is smaller than 387 00:18:35,870 --> 00:18:37,120 each of the 'C's. 388 00:18:37,120 --> 00:18:40,350 Consequently, the sum of all the 'u's must be no greater 389 00:18:40,350 --> 00:18:42,520 than the sum of all the 'C's. 390 00:18:42,520 --> 00:18:46,460 In other words, n-th partial sum, 'S sub n', is less than 391 00:18:46,460 --> 00:18:50,800 or equal to the partial sum, 'T sub n' for each 'n'. 392 00:18:50,800 --> 00:18:56,230 Now, by our previous result, knowing that these series 393 00:18:56,230 --> 00:18:59,990 summation 'C sub n' converges, it converges to its least 394 00:18:59,990 --> 00:19:02,140 upper bound of partial sums. 395 00:19:02,140 --> 00:19:05,450 In other words, the limit of 'T sub n' as 'n' approaches 396 00:19:05,450 --> 00:19:08,970 infinity is some number 'T', where 'T' is the least upper 397 00:19:08,970 --> 00:19:12,680 bound of the sequence of partial sums of the series. 398 00:19:12,680 --> 00:19:16,520 Well in particular, then, since each 'S sub n' is less 399 00:19:16,520 --> 00:19:20,570 than or equal to 'T sub n', 'S sub n' itself certainly cannot 400 00:19:20,570 --> 00:19:23,030 exceed the least upper bound, namely 'T'. 401 00:19:23,030 --> 00:19:25,130 In other words, 'S sub n' can be no bigger than 402 00:19:25,130 --> 00:19:26,760 'T' for each 'n'. 403 00:19:26,760 --> 00:19:28,700 Well, what does this mean? 404 00:19:28,700 --> 00:19:31,350 It means, then, that the sequence 'S sub 405 00:19:31,350 --> 00:19:32,900 n' itself is bounded. 406 00:19:32,900 --> 00:19:34,990 Well, it's bounded. 407 00:19:34,990 --> 00:19:38,130 It's a monotonic non-decreasing sequence. 408 00:19:38,130 --> 00:19:40,820 Therefore, its limit exists. 409 00:19:40,820 --> 00:19:43,840 Not only does it exist, but it's the least upper bound of 410 00:19:43,840 --> 00:19:45,430 the sequence of partial sums. 411 00:19:45,430 --> 00:19:47,660 In other words, this proof is given in the text. 412 00:19:47,660 --> 00:19:51,700 All I want you to see is that step by step, what this proof 413 00:19:51,700 --> 00:19:55,780 really does is it compares magnitudes of terms. 414 00:19:55,780 --> 00:19:59,430 In other words, if one batch of terms can't get lodged, if 415 00:19:59,430 --> 00:20:01,220 term by term, everything-- 416 00:20:01,220 --> 00:20:05,310 another sequence is less than these terms, then that second 417 00:20:05,310 --> 00:20:07,830 sum can't get too large either. 418 00:20:07,830 --> 00:20:10,080 And this is just a formalization of that. 419 00:20:10,080 --> 00:20:13,590 There are a few notes that we should make first of all. 420 00:20:13,590 --> 00:20:18,930 Namely, the condition that 'u sub n' be between 0 and 'C sub 421 00:20:18,930 --> 00:20:23,980 n' for all 'n' can be weakened to cover the case where this 422 00:20:23,980 --> 00:20:27,290 is true only beyond a certain point. 423 00:20:27,290 --> 00:20:29,380 Look, let me show you what I mean by this. 424 00:20:29,380 --> 00:20:36,030 Let's suppose I look at the sequence 1 plus 1/2 plus 1/3-- 425 00:20:36,030 --> 00:20:37,640 let's do at a different one. 426 00:20:37,640 --> 00:20:43,410 1 plus 1/2 plus 1/4 plus 1/8 plus 1/16, et cetera. 427 00:20:43,410 --> 00:20:45,880 In other words, the geometric series each of 428 00:20:45,880 --> 00:20:47,810 whose terms is 1/2. 429 00:20:47,810 --> 00:20:50,950 This series we know converges OK. 430 00:20:50,950 --> 00:20:53,700 Now what the comparison test says is, suppose you have 431 00:20:53,700 --> 00:20:54,950 something like this-- 432 00:20:54,950 --> 00:21:05,400 1 plus 1/3 plus 1/5 plus 1/9 plus 1/17, et cetera. 433 00:21:05,400 --> 00:21:07,980 See, notice that each of these terms is less than the 434 00:21:07,980 --> 00:21:10,050 corresponding term here. 435 00:21:10,050 --> 00:21:13,380 Consequently, since these terms add up to a finite 436 00:21:13,380 --> 00:21:16,210 amount, these terms here must also add 437 00:21:16,210 --> 00:21:17,600 up to a finite amount. 438 00:21:17,600 --> 00:21:20,950 But suppose for the sake of argument I decide to replace 439 00:21:20,950 --> 00:21:25,120 the first term here by 10 to the sixth. 440 00:21:25,120 --> 00:21:26,530 I'll make this a million. 441 00:21:26,530 --> 00:21:29,720 Now notice that this sum is going to become much larger. 442 00:21:29,720 --> 00:21:33,680 But the point is when I change this from a 1 to 10 to the 443 00:21:33,680 --> 00:21:37,230 sixth, even though I made the sum larger, I didn't change 444 00:21:37,230 --> 00:21:38,620 the finiteness of it. 445 00:21:38,620 --> 00:21:41,800 In other words, if I replace the first four terms here by 446 00:21:41,800 --> 00:21:45,150 fantastically large numbers and then keep the rest of the 447 00:21:45,150 --> 00:21:49,470 series intact, sure, I've made to sum very, very large. 448 00:21:49,470 --> 00:21:52,600 But I've only change it by a finite amount, which will in 449 00:21:52,600 --> 00:21:54,840 effect, not change the convergence. 450 00:21:54,840 --> 00:21:58,140 That's all I was saying over here, that the comparison test 451 00:21:58,140 --> 00:21:59,550 really hinges on what? 452 00:21:59,550 --> 00:22:02,360 Beyond a certain term, you could stop making the 453 00:22:02,360 --> 00:22:03,800 comparison. 454 00:22:03,800 --> 00:22:06,640 And the second observation is the converse to what we're 455 00:22:06,640 --> 00:22:07,730 talking about. 456 00:22:07,730 --> 00:22:12,590 Namely, if we know that 'u sub n' is at least as big as 'd 457 00:22:12,590 --> 00:22:16,730 sub n' for each 'n', where the series summation 'n' goes from 458 00:22:16,730 --> 00:22:21,990 1 to infinity, 'd sub n' is a positive divergent series, 459 00:22:21,990 --> 00:22:27,360 then this series, summation 'u sub n', must also diverge 460 00:22:27,360 --> 00:22:31,510 since its convergence would imply the convergence of this. 461 00:22:31,510 --> 00:22:35,980 In other words, notice that by the comparison test, if 462 00:22:35,980 --> 00:22:41,290 summation 'u sub n' converged, the 'd n's, being less than 463 00:22:41,290 --> 00:22:44,230 the 'u n's would mean that summation 'dn' end would have 464 00:22:44,230 --> 00:22:45,220 to converge also. 465 00:22:45,220 --> 00:22:47,990 At any rate, these are the two portions of 466 00:22:47,990 --> 00:22:49,760 the comparison test. 467 00:22:49,760 --> 00:22:52,620 And this is what goes into the comparison test. 468 00:22:52,620 --> 00:22:54,540 Now the interesting thing, or the draw back to the 469 00:22:54,540 --> 00:22:56,620 comparison test is simply this-- 470 00:22:56,620 --> 00:23:00,760 that 99 times out of 100, if you can find a series to 471 00:23:00,760 --> 00:23:04,500 compare a given series with, you probably would have known 472 00:23:04,500 --> 00:23:06,730 whether the given series converged or diverged in the 473 00:23:06,730 --> 00:23:07,870 first place. 474 00:23:07,870 --> 00:23:10,050 In other words, somehow or other, to find the right 475 00:23:10,050 --> 00:23:13,570 series to compare something with is a rather subtle thing 476 00:23:13,570 --> 00:23:16,440 if you didn't already know the right answer to the problem. 477 00:23:16,440 --> 00:23:19,370 Well, at any rate, what I'm trying to drive at is that the 478 00:23:19,370 --> 00:23:25,170 comparison test has as one of its features a proof for a 479 00:23:25,170 --> 00:23:28,525 more interesting test-- a test that's far less intuitive-- 480 00:23:28,525 --> 00:23:30,980 called the 'ratio test'. 481 00:23:30,980 --> 00:23:32,970 The ratio test says the following. 482 00:23:32,970 --> 00:23:35,600 Let's suppose again you're given a positive series. 483 00:23:35,600 --> 00:23:39,610 What you do now is form a sequence whereby each term in 484 00:23:39,610 --> 00:23:42,200 the sequence is the ratio between two 485 00:23:42,200 --> 00:23:44,190 terms in the series. 486 00:23:44,190 --> 00:23:48,150 In other words, what I do is I form the sequence by taking 487 00:23:48,150 --> 00:23:51,050 the second term divided by the first term, the third term 488 00:23:51,050 --> 00:23:53,910 divided by the second term, the fourth term divided by the 489 00:23:53,910 --> 00:23:54,590 third term. 490 00:23:54,590 --> 00:23:58,100 Now let's call that general term 'u sub n'. 491 00:23:58,100 --> 00:24:00,360 This seems a little bit abstract for you. 492 00:24:00,360 --> 00:24:03,230 Let's look at a more tangible example. 493 00:24:03,230 --> 00:24:06,380 Suppose I take the series summation 'n' goes from 1 to 494 00:24:06,380 --> 00:24:10,210 infinity, '10 to the n' over 'n factorial'. 495 00:24:10,210 --> 00:24:14,620 Notice that in this case, the n-th term is '10 to the n' 496 00:24:14,620 --> 00:24:16,340 over 'n factorial'. 497 00:24:16,340 --> 00:24:20,910 The 'n plus first' term is '10 to the 'n plus 1'' over ''n 498 00:24:20,910 --> 00:24:22,770 plus 1' factorial'. 499 00:24:22,770 --> 00:24:26,830 So 'u sub n' is the ratio of the 'n plus first' 500 00:24:26,830 --> 00:24:28,850 term to the n-th term. 501 00:24:28,850 --> 00:24:32,120 In other words, '10 to the 'n plus 1'' over ''n plus 1' 502 00:24:32,120 --> 00:24:36,700 factorial' divided by '10 to the n' over 'n factorial'. 503 00:24:36,700 --> 00:24:40,100 '10 to the 'n plus 1'' divided by '10 to the n' is simply 10. 504 00:24:40,100 --> 00:24:43,740 And ''n plus 1' factorial' divided by 'n factorial' is 505 00:24:43,740 --> 00:24:45,560 simply 'n plus 1'. 506 00:24:45,560 --> 00:24:48,220 In other words, observe the structure of the factorials 507 00:24:48,220 --> 00:24:51,900 that you get from the n-th to the 'n plus first' simply by 508 00:24:51,900 --> 00:24:54,200 multiplying by 'n plus 1'. 509 00:24:54,200 --> 00:24:56,350 Again, the computational details will be 510 00:24:56,350 --> 00:24:58,870 left for the exercises. 511 00:24:58,870 --> 00:25:01,570 At any rate, then, in this particular case, 'u sub n' 512 00:25:01,570 --> 00:25:03,950 would be '10 over 'n plus 1''. 513 00:25:03,950 --> 00:25:06,220 Now here's what the ratio test says. 514 00:25:06,220 --> 00:25:09,480 Assuming that the limit 'u sub n' as 'n' approaches infinity 515 00:25:09,480 --> 00:25:12,220 exists, call it 'rho'. 516 00:25:12,220 --> 00:25:17,660 Then, the series converges if rho less than 1 and diverges 517 00:25:17,660 --> 00:25:19,490 if rho is greater than 1. 518 00:25:19,490 --> 00:25:22,750 And the test fails if rho equals 1. 519 00:25:22,750 --> 00:25:26,830 In other words, if the terms get progressively smaller, so 520 00:25:26,830 --> 00:25:30,040 that the ratio and the limit stays less than 1, then the 521 00:25:30,040 --> 00:25:31,450 series converges. 522 00:25:31,450 --> 00:25:34,590 If on the other hand, the ratio in the limit exceeds 1, 523 00:25:34,590 --> 00:25:37,170 that means the terms are getting big fast enough so 524 00:25:37,170 --> 00:25:39,020 that the series diverges. 525 00:25:39,020 --> 00:25:42,430 Let me point out an important observation here. 526 00:25:42,430 --> 00:25:46,310 Notice the difference between the limit equaling 1 and each 527 00:25:46,310 --> 00:25:48,490 term of the sequence being less than 1. 528 00:25:48,490 --> 00:25:51,030 In other words, notice that even if 'u sub n' is less than 529 00:25:51,030 --> 00:25:55,970 1 for every 'n', rho may still equal 1. 530 00:25:55,970 --> 00:25:59,350 For example, look at the terms 'n' over 'n plus 1''. 531 00:25:59,350 --> 00:26:02,980 For each 'n', 'n' over 'n plus 1' is less than 1. 532 00:26:02,980 --> 00:26:05,810 Yet the limit as 'n' approaches infinity is exactly 533 00:26:05,810 --> 00:26:07,360 equal to 1. 534 00:26:07,360 --> 00:26:10,810 Now again, the formal proof of this is given in the book. 535 00:26:10,810 --> 00:26:13,160 But I thought if I just take a few minutes to show you 536 00:26:13,160 --> 00:26:16,150 geometrically what's happening here, you'll get a better 537 00:26:16,150 --> 00:26:19,060 picture to understand what's happening in the text. 538 00:26:19,060 --> 00:26:22,170 Let's prove this in the case that rho is less than 1. 539 00:26:22,170 --> 00:26:25,310 Pictorially, if rho is less than 1, that means there's a 540 00:26:25,310 --> 00:26:27,310 space between rho and 1. 541 00:26:27,310 --> 00:26:30,600 That means I can choose an epsilon such that rho plus 542 00:26:30,600 --> 00:26:34,660 epsilon, which I'll call it 'r', is a positive number, but 543 00:26:34,660 --> 00:26:36,500 still less than 1. 544 00:26:36,500 --> 00:26:39,370 Now, by definition of rho being the limit of the 545 00:26:39,370 --> 00:26:43,470 sequence 'a sub 'n plus 1'' over 'a sub n', that means I 546 00:26:43,470 --> 00:26:47,350 can find the capital 'N' for this given epsilon, such that 547 00:26:47,350 --> 00:26:51,820 whenever 'n' is greater than capital 'N', that 'a sub 'n 548 00:26:51,820 --> 00:26:56,040 plus 1' over 'a sub n' is less than rho plus epsilon. 549 00:26:56,040 --> 00:26:59,180 In other words, is less than 'r', where 'r' is some fixed 550 00:26:59,180 --> 00:27:01,380 number now less than 1. 551 00:27:01,380 --> 00:27:05,450 Now let me apply this to successive values of 'n'. 552 00:27:05,450 --> 00:27:09,150 In other words, taking 'n' to be capital-- 553 00:27:09,150 --> 00:27:13,010 see, looking at this thing here, taking 'n' to be capital 554 00:27:13,010 --> 00:27:18,030 'N', I have 'a' over capital 'N plus 1', 'a sub capital 'N 555 00:27:18,030 --> 00:27:20,990 plus 1'' over 'a sub N' is less than 'r'. 556 00:27:20,990 --> 00:27:24,960 In other words, 'a sub capital 'N plus 1'' is less than 'r' 557 00:27:24,960 --> 00:27:26,500 times 'a sub N'. 558 00:27:26,500 --> 00:27:31,720 Similarly, 'a sub capital 'N plus 2'' over 'a sub capital 559 00:27:31,720 --> 00:27:34,240 'N plus 1'' is also less than 'r'. 560 00:27:34,240 --> 00:27:38,190 In other words, 'a sub capital 'N plus 2'' is less than 'r' 561 00:27:38,190 --> 00:27:41,920 times 'a sub capital 'N plus 1''. 562 00:27:41,920 --> 00:27:45,890 But 'a sub 'N plus 1'' in turn is less than 'r' 563 00:27:45,890 --> 00:27:47,300 times 'a sub N'. 564 00:27:47,300 --> 00:27:51,290 Putting this in here, 'a sub 'N plus 2' is less than 'r 565 00:27:51,290 --> 00:27:53,240 squared' times 'a sub N'.. 566 00:27:53,240 --> 00:27:57,170 At any rate, if I now sum these inequalities, you see 567 00:27:57,170 --> 00:27:58,450 what I get is what? 568 00:27:58,450 --> 00:28:01,430 The limit as we go from 'n plus 1' to infinity-- in other 569 00:28:01,430 --> 00:28:03,880 words, the tail end of this particular sum 570 00:28:03,880 --> 00:28:05,340 is less than what? 571 00:28:05,340 --> 00:28:08,310 Well, I can factor out the 'a sub N' over here. 572 00:28:08,310 --> 00:28:10,680 And what's left inside is what? 573 00:28:10,680 --> 00:28:14,780 'r' plus 'r squared' plus 'r cubed', et cetera. 574 00:28:14,780 --> 00:28:19,550 But this particular series is a convergent geometric series. 575 00:28:19,550 --> 00:28:22,720 In other words, this must converge because this 576 00:28:22,720 --> 00:28:25,870 converges, and this is less than, term by term, 577 00:28:25,870 --> 00:28:27,280 the terms over here. 578 00:28:27,280 --> 00:28:30,910 In other words, notice that the proof of the ratio test 579 00:28:30,910 --> 00:28:34,690 hinges on knowing two things, the comparison test and the 580 00:28:34,690 --> 00:28:37,110 convergence of a geometric series. 581 00:28:37,110 --> 00:28:40,490 Again, the reason I go through this as rapidly as I do is 582 00:28:40,490 --> 00:28:44,360 that every detail is done magnificently in the textbook. 583 00:28:44,360 --> 00:28:47,640 All I want you to see here is the overview of how these 584 00:28:47,640 --> 00:28:49,730 tests come about. 585 00:28:49,730 --> 00:28:54,060 Finally, we have another powerful test called the 586 00:28:54,060 --> 00:28:55,360 'integral test'. 587 00:28:55,360 --> 00:28:58,740 And the integral test essentially equates positive 588 00:28:58,740 --> 00:29:01,480 series with improper integrals. 589 00:29:01,480 --> 00:29:05,500 By the way, I have presented the material, so to speak, in 590 00:29:05,500 --> 00:29:08,640 the order of appearance in the textbook. 591 00:29:08,640 --> 00:29:12,940 You see, the comparison test, ratio test, and integral test 592 00:29:12,940 --> 00:29:15,690 are given in that order in the text. 593 00:29:15,690 --> 00:29:17,760 And so, I kept the same order here. 594 00:29:17,760 --> 00:29:20,470 However, it's interesting to point out that the integral 595 00:29:20,470 --> 00:29:24,450 test, in a way, is a companion of the comparison test. 596 00:29:24,450 --> 00:29:26,600 And let me show you first of all what the 597 00:29:26,600 --> 00:29:28,130 integral test says. 598 00:29:28,130 --> 00:29:31,570 It says, and I've written out the formal statements here. 599 00:29:31,570 --> 00:29:34,250 I'll show you pictorially what this means in a moment. 600 00:29:34,250 --> 00:29:37,040 But it says, suppose there is a decreasing continuous 601 00:29:37,040 --> 00:29:39,410 function, 'f of x'. 602 00:29:39,410 --> 00:29:40,760 Notice the word continuous in here. 603 00:29:40,760 --> 00:29:43,610 That guarantees that we can integrate 'f of x'. 604 00:29:43,610 --> 00:29:48,360 Such that 'f' evaluated at each integral value of 'x', 605 00:29:48,360 --> 00:29:52,240 say 'x' equals 'n', is 'u sub n', where 'u sub n' is the 606 00:29:52,240 --> 00:29:54,860 n-th term of the positive series 'u1' 607 00:29:54,860 --> 00:29:56,650 plus 'u2', et cetera. 608 00:29:56,650 --> 00:29:59,830 Then, what the integral test says is that the series 609 00:29:59,830 --> 00:30:03,760 summation 'n' goes from 1 to infinity 'u sub n', and the 610 00:30:03,760 --> 00:30:08,220 integral 1 to infinity, ''f of x' dx', either converge 611 00:30:08,220 --> 00:30:11,140 together or diverge together. 612 00:30:11,140 --> 00:30:13,420 In other words, we can test a particular series for 613 00:30:13,420 --> 00:30:16,110 convergence by knowing whether a particular 614 00:30:16,110 --> 00:30:18,200 improper integral converges. 615 00:30:18,200 --> 00:30:20,390 And to show you what this thing means pictorially, 616 00:30:20,390 --> 00:30:21,860 simply observe this. 617 00:30:21,860 --> 00:30:26,220 See, what we're saying is, suppose that when you plot the 618 00:30:26,220 --> 00:30:27,800 terms of the series-- 619 00:30:27,800 --> 00:30:31,300 so, 'u1', 'u2', 'u3', 'u4'-- 620 00:30:31,300 --> 00:30:34,800 that these happen to be the integral values of a 621 00:30:34,800 --> 00:30:37,770 continuous curve, 'y' equals 'f of x', which not only 622 00:30:37,770 --> 00:30:41,060 passes through these points, but is a continuous decreasing 623 00:30:41,060 --> 00:30:43,100 function as this happens. 624 00:30:43,100 --> 00:30:47,300 Then what the statement is is that the sum of these lengths 625 00:30:47,300 --> 00:30:50,200 converges if and only if the area under 626 00:30:50,200 --> 00:30:52,610 the curve is finite. 627 00:30:52,610 --> 00:30:54,200 And the proof go something like this. 628 00:30:54,200 --> 00:30:56,530 It's a rather ingenious type of thing. 629 00:30:56,530 --> 00:31:00,470 You see, notice that if this height is 'u sub 1', and the 630 00:31:00,470 --> 00:31:04,070 base of the rectangle is 1, notice that numerically, the 631 00:31:04,070 --> 00:31:06,600 area of the rectangle-- it's quite in general. 632 00:31:06,600 --> 00:31:11,000 If the base of a rectangle is 1, numerically the area of the 633 00:31:11,000 --> 00:31:14,080 rectangle equals the height, because the area is the base 634 00:31:14,080 --> 00:31:14,930 times the height. 635 00:31:14,930 --> 00:31:17,620 If the base is 1, the area equals the height. 636 00:31:17,620 --> 00:31:19,510 So the idea is simply this. 637 00:31:19,510 --> 00:31:20,370 Lookit. 638 00:31:20,370 --> 00:31:23,060 Suppose, for example, that we look at this 639 00:31:23,060 --> 00:31:24,630 diagram over here. 640 00:31:24,630 --> 00:31:28,490 Notice that in this diagram, if we stop at n, the area 641 00:31:28,490 --> 00:31:32,500 under this curve is the integral from 1 to 'n', or 1 642 00:31:32,500 --> 00:31:35,350 to 'n plus 1', because of how these lines are drawn. 643 00:31:35,350 --> 00:31:40,320 See, notice that the first height stops at the number 2. 644 00:31:40,320 --> 00:31:43,850 The second base stops at number 3, et cetera. 645 00:31:43,850 --> 00:31:46,950 The idea is this, though, that the area under the curve in 646 00:31:46,950 --> 00:31:51,130 general, from 1 to 'n plus 1', is integral from 1 to 'n plus 647 00:31:51,130 --> 00:31:53,060 1', ''f of x' dx'. 648 00:31:53,060 --> 00:31:58,320 On the other hand, the area of the rectangles are what? 649 00:31:58,320 --> 00:32:02,970 'u1' plus 'u2' plus 'u3', up to 'u n'. 650 00:32:02,970 --> 00:32:07,190 In other words, for any given 'n', 'u1' up to u n' is 651 00:32:07,190 --> 00:32:10,110 greater than or equal to this particular integral. 652 00:32:10,110 --> 00:32:14,290 Consequently, taking the limit as 'n' goes to 653 00:32:14,290 --> 00:32:16,220 infinity, we get what? 654 00:32:16,220 --> 00:32:20,990 The summation 'u n' is greater than or equal to integral from 655 00:32:20,990 --> 00:32:23,380 one to infinity ''f of x' dx'. 656 00:32:23,380 --> 00:32:26,530 Consequently, if this integral diverges, meaning that this 657 00:32:26,530 --> 00:32:30,030 gets very large, the fact that this can be no less than this 658 00:32:30,030 --> 00:32:33,150 means that this too must also diverge. 659 00:32:33,150 --> 00:32:37,340 Correspondingly, if we now do the same problem, but draw the 660 00:32:37,340 --> 00:32:41,980 thing slightly differently, notice that now in this 661 00:32:41,980 --> 00:32:47,770 particular picture, the area under the curve is integral 662 00:32:47,770 --> 00:32:50,780 from 1 to 'n', ''f of x' dx'. 663 00:32:50,780 --> 00:32:53,400 On the other hand, the area of the rectangles are what? 664 00:32:53,400 --> 00:32:56,840 It's 'u2' plus 'u3' up to 'u n'. 665 00:32:56,840 --> 00:33:01,220 But now you see the rectangles are inscribed under the curve. 666 00:33:01,220 --> 00:33:05,220 Consequently, 'u2' plus et cetera, up to 'u n', is less 667 00:33:05,220 --> 00:33:06,860 than this integral. 668 00:33:06,860 --> 00:33:10,767 Therefore, if I add 'u1' onto both sides, the sum 'u1', et 669 00:33:10,767 --> 00:33:14,920 cetera, up to 'u n', is less than 'u1' plus integral 1 to 670 00:33:14,920 --> 00:33:17,090 'n', ''f of x' dx'. 671 00:33:17,090 --> 00:33:20,680 If I now take the limit as n goes to infinity, you see this 672 00:33:20,680 --> 00:33:23,920 becomes summation un. 673 00:33:23,920 --> 00:33:26,330 n goes from 1 to infinity. 674 00:33:26,330 --> 00:33:32,880 This becomes integral from 1 to infinity, ''f of x' dx'. 675 00:33:32,880 --> 00:33:37,120 And therefore, if this now converges, the sum on the 676 00:33:37,120 --> 00:33:38,630 right is finite. 677 00:33:38,630 --> 00:33:41,780 Since the sum on the left cannot exceed the sum on the 678 00:33:41,780 --> 00:33:45,240 right, the sum on the left must also be finite. 679 00:33:45,240 --> 00:33:48,960 Consequently, the convergence of the integral implies the 680 00:33:48,960 --> 00:33:50,930 convergence of the series. 681 00:33:50,930 --> 00:33:54,320 Again, I apologize for doing this this rapidly. 682 00:33:54,320 --> 00:33:57,780 All I wanted you to do was to get an idea of what's 683 00:33:57,780 --> 00:33:58,510 happening here. 684 00:33:58,510 --> 00:34:01,760 Because as I say, the book is magnificent in this section. 685 00:34:01,760 --> 00:34:03,960 The proofs are very well self-contained. 686 00:34:03,960 --> 00:34:07,790 At any rate, this gives us three rather powerful tests, 687 00:34:07,790 --> 00:34:11,190 which I will drill you on in the exercises for testing 688 00:34:11,190 --> 00:34:13,350 convergence of positive series. 689 00:34:13,350 --> 00:34:16,590 What we're going to do next time is to come to grips with 690 00:34:16,590 --> 00:34:17,989 a much more serious problem. 691 00:34:17,989 --> 00:34:22,969 And that is, what do you do if the series isn't positive? 692 00:34:22,969 --> 00:34:24,380 But that's another story. 693 00:34:24,380 --> 00:34:26,360 And so until next time, good bye. 694 00:34:29,170 --> 00:34:32,370 Funding for the publication of this video was provided by the 695 00:34:32,370 --> 00:34:36,420 Gabriella and Paul Rosenbaum Foundation. 696 00:34:36,420 --> 00:34:40,590 Help OCW continue to provide free and open access to MIT 697 00:34:40,590 --> 00:34:44,800 courses by making a donation at ocw.mit.edu/donate.