1 00:00:00,853 --> 00:00:01,454 Let's integrate. 2 00:00:01,454 --> 00:00:07,620 [MUSIC] 3 00:00:07,620 --> 00:00:11,880 Series and integrals are not entirely unrelated. 4 00:00:11,880 --> 00:00:16,408 Well, here I got a graph and I've drawn a bunch of boxes in red on my graph. 5 00:00:16,408 --> 00:00:20,220 Each of these boxes has width one and the heights 6 00:00:20,220 --> 00:00:23,570 of the boxes, given by the terms in a sequence. 7 00:00:23,570 --> 00:00:28,770 So the first box has height a sub 1, the next box has height a sub 2, the 8 00:00:28,770 --> 00:00:33,400 next box has height a sub 3, the next box has height a sub 4 and so on. 9 00:00:33,400 --> 00:00:38,420 Since the boxes each have width 1, it's easy to calculate their areas. 10 00:00:38,420 --> 00:00:42,720 The area of the first box is 1 times it's height, which is a sub 1. 11 00:00:42,720 --> 00:00:45,520 So the area of the first is a sub 1. 12 00:00:45,520 --> 00:00:49,510 The area of the second box is it's width 1 times it's height 13 00:00:49,510 --> 00:00:52,680 a sub 2, so the area of the next box is a sub 2. 14 00:00:52,680 --> 00:00:53,120 And so on. 15 00:00:53,120 --> 00:00:56,740 The area of the next box is a sub 3, the area of the next box is a sub 4. 16 00:00:56,740 --> 00:00:58,590 What if I want to calculate the area of all 17 00:00:58,590 --> 00:01:02,100 the red boxes, I imagine this red boxes go on forever. 18 00:01:02,100 --> 00:01:04,430 Well, that's really an infinite series, right? 19 00:01:04,430 --> 00:01:07,490 Its a sub 1 plus a sub 2 plus a sub 3 plus a sub 4 and 20 00:01:07,490 --> 00:01:10,530 so on, so I could write the total area 21 00:01:10,530 --> 00:01:13,657 in all the red boxes as these infinite series. 22 00:01:15,570 --> 00:01:17,700 There is more going on in this graph, I don't 23 00:01:17,700 --> 00:01:21,340 just have these boxes, I've also got this blue curve. 24 00:01:21,340 --> 00:01:24,410 That's the curve y equals f of x and 25 00:01:24,410 --> 00:01:28,040 one special feature about this function f is that, I've rigged it so 26 00:01:28,040 --> 00:01:32,760 that a stubborn is f of n and we can see that here. 27 00:01:32,760 --> 00:01:36,580 Because this blue curve passes through the corners of these boxes. 28 00:01:38,600 --> 00:01:41,680 Now what if I wanted to calculate the area under the 29 00:01:41,680 --> 00:01:46,550 blue curve and to the right of the line x equals 1? 30 00:01:46,550 --> 00:01:48,590 Well that would be this interval. 31 00:01:48,590 --> 00:01:50,910 The interval from one to infinity of f of x, dx. 32 00:01:50,910 --> 00:01:56,840 How does this blue region compare to this red region? 33 00:01:56,840 --> 00:02:01,499 Well this blue region sits inside all of the red boxes. 34 00:02:02,720 --> 00:02:03,400 What is that mean? 35 00:02:04,500 --> 00:02:09,160 That means that this is the red region, this is this blue 36 00:02:09,160 --> 00:02:14,550 region, but this blue region sits inside all of the red boxes. 37 00:02:15,630 --> 00:02:18,600 And that means that the area of the red region 38 00:02:18,600 --> 00:02:21,780 must be larger than the area of the blue region. 39 00:02:21,780 --> 00:02:23,770 Why does that matter, though? 40 00:02:23,770 --> 00:02:25,800 Well the takeaway message here is that 41 00:02:25,800 --> 00:02:29,640 this sum is even larger than this interval. 42 00:02:29,640 --> 00:02:35,170 And that means, if I know that this integral, say, is infinity, if 43 00:02:35,170 --> 00:02:41,050 the area under the blue curve and to the right of the line x equals 1 is infinite. 44 00:02:41,050 --> 00:02:44,250 Then what do I know about this series? 45 00:02:44,250 --> 00:02:48,120 Well that series then must diverge. 46 00:02:48,120 --> 00:02:52,590 The picture includes some as of yet unspecified assumptions. 47 00:02:52,590 --> 00:02:55,170 So more precisely, I need to assume 48 00:02:55,170 --> 00:03:01,110 that f is a decreasing and positive function. 49 00:03:01,110 --> 00:03:06,890 And I should explicitly mention that a sub n is equal to f of n and with those 50 00:03:06,890 --> 00:03:14,720 conditions, than if this integral diverges, then the sum also diverges. 51 00:03:14,720 --> 00:03:16,300 We're in a position to prove this. 52 00:03:16,300 --> 00:03:20,180 This is what I'd like to show. Suppose that little f is a decreasing but 53 00:03:20,180 --> 00:03:25,010 positive function and building a sequence a sub n out of the function f. 54 00:03:25,010 --> 00:03:28,280 And I'm supposing that the integral f of x dx goes from 1 55 00:03:28,280 --> 00:03:33,230 to infinity diverges and I want to show that the series diverges as well. 56 00:03:34,380 --> 00:03:36,610 Well to get a handle on a series, I'm going to look at this. 57 00:03:36,610 --> 00:03:39,140 The sum little n goes from 1 to big N the a 58 00:03:39,140 --> 00:03:43,400 sub n, and somehow, I want to relate this sum to an interval. 59 00:03:43,400 --> 00:03:45,280 I won't get there all at once but I'll do 60 00:03:45,280 --> 00:03:46,070 it in steps. 61 00:03:46,070 --> 00:03:49,410 First thing I can say is this, that the sum of 62 00:03:49,410 --> 00:03:51,740 the a sub n's is equal to the sum, little n 63 00:03:51,740 --> 00:03:57,770 goes from 1 to big N of the integral x goes from n to n plus 1 of f of n dx. 64 00:03:57,770 --> 00:04:03,110 And although this looks somewhat fearsome It's true for a not very deep reason. 65 00:04:03,110 --> 00:04:04,520 the upside here is what? 66 00:04:04,520 --> 00:04:08,170 F of n is equal to a sub n, 67 00:04:08,170 --> 00:04:11,630 and consequently, this is just the integral of a constant 68 00:04:11,630 --> 00:04:13,930 on an interval of length 1. 69 00:04:13,930 --> 00:04:17,240 So if you integrate a sub n, x goes from n 70 00:04:17,240 --> 00:04:20,480 to n plus 1, well, you're just getting a sub n. 71 00:04:20,480 --> 00:04:23,900 So this is the sum of a sub n which is exactly what I've got here. 72 00:04:24,960 --> 00:04:27,470 But when I write it like this, I'm 73 00:04:27,470 --> 00:04:29,300 giving myself some more freedom to mess around with 74 00:04:29,300 --> 00:04:34,930 this integral, instead of saying f of x I could say f of the floor of x. 75 00:04:35,940 --> 00:04:36,800 Well why's that 76 00:04:36,800 --> 00:04:39,540 helpful well what's floor of x. 77 00:04:39,540 --> 00:04:43,900 The floor of x is the greatest integer less than or equal to x. 78 00:04:43,900 --> 00:04:48,820 So if x is between n and n plus 1, n's an integer, then the floor of 79 00:04:48,820 --> 00:04:54,080 x is just equal to n and that means, except at the endpoints, f of the floor 80 00:04:54,080 --> 00:04:59,400 of x is equal to f of n on the interval between n and n plus 1. 81 00:04:59,400 --> 00:05:01,690 And that means this integral is just, 82 00:05:01,690 --> 00:05:03,710 again, a sub n. 83 00:05:03,710 --> 00:05:07,590 So the sum of the a sub n's is the sum of, well that's just a sub n. 84 00:05:09,560 --> 00:05:13,230 Now I'm writing a sum of a bunch of integrals, all the 85 00:05:13,230 --> 00:05:17,280 integrands look the same but the bounds of integration are changing, right. 86 00:05:17,280 --> 00:05:20,890 The first integral goes from 1 to 2, the next integral goes from 87 00:05:20,890 --> 00:05:24,780 2 to 3, the next integral goes from 3 to 4 and so on. 88 00:05:24,780 --> 00:05:26,760 But if you add up a bunch of integrals, 89 00:05:26,760 --> 00:05:31,320 all of which lie end-to-end like that, you can just write that as a single integral. 90 00:05:32,340 --> 00:05:37,140 So, this sum of integrals is equal to just the integral, x goes 91 00:05:37,140 --> 00:05:40,630 from 1 to big N plus 1, of f of the floor of x 92 00:05:43,520 --> 00:05:44,010 dx. All right. 93 00:05:44,010 --> 00:05:48,190 So I've replaced, this sum with this integral. 94 00:05:49,250 --> 00:05:49,960 Now what can I do? 95 00:05:51,100 --> 00:05:54,120 Oh, I know that f is a decreasing function, so if 96 00:05:54,120 --> 00:05:58,880 I plug in a bigger input, the functions output is smaller. 97 00:05:58,880 --> 00:06:04,650 That tells me that the integral of f of floor of x, x goes from 1 to big N plus 1. 98 00:06:04,650 --> 00:06:08,690 Is bigger than or equal to the integral cause f of x, x goes from 99 00:06:08,690 --> 00:06:14,110 1 to big N plus 1, because f of the floor of x is at least as big as f of x. 100 00:06:17,450 --> 00:06:22,640 How else do I know? I know that this integral diverges. 101 00:06:22,640 --> 00:06:24,240 Which means what? 102 00:06:24,240 --> 00:06:27,380 Well, it means that as long as n is big 103 00:06:27,380 --> 00:06:31,990 enough, I can make this as big as you'd like. 104 00:06:33,392 --> 00:06:35,740 All right, all you need to do is tell me how big you want this to 105 00:06:35,740 --> 00:06:39,280 be and I'll tell you some big enough ends, so that this thing is really that large. 106 00:06:41,260 --> 00:06:42,790 But, this thing is 107 00:06:42,790 --> 00:06:45,360 smaller than this sum. 108 00:06:45,360 --> 00:06:49,710 So what that means is that, the sum can be made as big 109 00:06:49,710 --> 00:06:52,370 as you like, as long as you choose big N and big enough. 110 00:06:53,860 --> 00:06:58,010 But if you can make the sum really, really large, that means 111 00:06:58,010 --> 00:07:03,080 the sequence of partial sums isn't getting close to any finite value. 112 00:07:03,080 --> 00:07:07,030 And what that means, is that this series diverges. 113 00:07:07,030 --> 00:07:08,700 We can go the other way too. 114 00:07:08,700 --> 00:07:11,720 We can integrate to prove convergence. 115 00:07:11,720 --> 00:07:17,470 So again, I've got this integral and this series and I somehow want to relate them. 116 00:07:17,470 --> 00:07:21,080 Well let's suppose that this integral is finite. 117 00:07:21,080 --> 00:07:25,950 Then I'd like to be able to conclude that this series converges. 118 00:07:25,950 --> 00:07:27,510 With all the same conditions as before, right, I 119 00:07:27,510 --> 00:07:30,950 want f to be a positive decreasing function and such. 120 00:07:30,950 --> 00:07:32,190 Let's prove it. 121 00:07:32,190 --> 00:07:33,870 So here's what I've got. I've got that 122 00:07:33,870 --> 00:07:36,430 this integral is finite, converges, that I 123 00:07:36,430 --> 00:07:39,290 want to know that the series converges as well. 124 00:07:39,290 --> 00:07:40,020 So here's the integral. 125 00:07:40,020 --> 00:07:42,790 The integral from one to infinity of f of x dx. 126 00:07:42,790 --> 00:07:46,590 How does that integral compare just the integral from one to big N of f of x dx. 127 00:07:47,590 --> 00:07:50,050 Well f's a positive function. 128 00:07:50,050 --> 00:07:52,430 So these things differ by the integral from 129 00:07:52,430 --> 00:07:55,160 n to infinity, which is some positive number. 130 00:07:55,160 --> 00:07:58,250 So this thing here is definitely an over estimate of this thing. 131 00:07:58,250 --> 00:07:59,090 So the integral from 132 00:07:59,090 --> 00:08:05,180 1 to infinity is bigger than the integral from 1 to big N of f of x, dx. 133 00:08:05,180 --> 00:08:08,660 I'm going to relate that integral to this integral, the integral from 134 00:08:08,660 --> 00:08:11,330 1 to big N of f of the ceiling of x. 135 00:08:11,330 --> 00:08:13,530 Remember what the ceiling is, the ceiling is 136 00:08:13,530 --> 00:08:16,630 the smallest integer bigger than equal to x. 137 00:08:16,630 --> 00:08:20,210 So how does f of x relate to f the ceiling of x? 138 00:08:20,210 --> 00:08:23,440 Well the ceiling of x is always at least just as big as x. 139 00:08:23,440 --> 00:08:24,210 But because f is 140 00:08:24,210 --> 00:08:28,890 decreasing, f sends bigger inputs to smaller outputs. 141 00:08:28,890 --> 00:08:32,660 So f for the ceiling of x is less than or equal to f of x. 142 00:08:32,660 --> 00:08:35,820 And that means this intergral is an overestimate of this intergral. 143 00:08:35,820 --> 00:08:41,170 Now I can split this integral up and instead of integrating from 1 144 00:08:41,170 --> 00:08:46,250 to big N all in one fell swoop, I can instead integrate like this. 145 00:08:46,250 --> 00:08:49,360 I can integrate from little n to little n plus 1 and 146 00:08:49,360 --> 00:08:54,470 add up all those integrals as n goes from 1 to big n minus 1. 147 00:08:54,470 --> 00:08:59,438 But on each of these intervals, I know exactly what the ceiling of x is. 148 00:08:59,438 --> 00:09:08,052 If x is between little n and little n plus 1, then the ceiling of x is just n plus 1. 149 00:09:08,052 --> 00:09:12,309 So now I've got the sum of the integral of f of n plus 1. 150 00:09:12,309 --> 00:09:14,692 But what is f of n plus 1? 151 00:09:14,692 --> 00:09:18,529 Remember f of n plus 1 is just a sub n plus 1. 152 00:09:18,529 --> 00:09:23,569 And I'm integrating this constant of an interval of length 1, so this 153 00:09:23,569 --> 00:09:28,710 integral just ends up being a sub n plus 1 and that means these are equal. 154 00:09:28,710 --> 00:09:31,826 Now I could re-index this, you know, the first 155 00:09:31,826 --> 00:09:35,018 time that I'm adding here, the n equals 1 term, 156 00:09:35,018 --> 00:09:39,958 is a sub 2, the next term is a sub 3, and eventually I plug in a little n equals 157 00:09:39,958 --> 00:09:40,946 big N minus 1. 158 00:09:40,946 --> 00:09:44,996 Which gives me a sub big N here, so I could instead write that 159 00:09:44,996 --> 00:09:48,420 as just the sum little n equals 2 to big N of a sub n. 160 00:09:49,480 --> 00:09:52,976 So all together, right, what I have shown is that this 161 00:09:52,976 --> 00:09:57,365 integral, which has some finite value is an overestimate of this. 162 00:09:57,365 --> 00:10:01,421 Which is almost the partial sum, it's just missing a sub 1, 163 00:10:01,421 --> 00:10:05,243 but what that means is of a sequence of partial sums for this 164 00:10:05,243 --> 00:10:08,909 series Is bounded and I already know that it's monotone 165 00:10:08,909 --> 00:10:13,400 because all the things that I'm adding up are positive. 166 00:10:13,400 --> 00:10:15,450 So I've got a sequence which is bounded 167 00:10:15,450 --> 00:10:19,890 and monotone and therefore, by the monotone conversions theoreom, 168 00:10:19,890 --> 00:10:23,030 this sequence of partial sums converges, which is exactly 169 00:10:23,030 --> 00:10:25,850 what it means to say that this series converges. 170 00:10:25,850 --> 00:10:29,220 We're doing more than just proving convergence. 171 00:10:29,220 --> 00:10:30,480 What I mean is that I've got 172 00:10:30,480 --> 00:10:32,500 this inequality here. 173 00:10:32,500 --> 00:10:37,530 The big deal here is that I could add a sub 1 to both sides 174 00:10:37,530 --> 00:10:41,450 and if I add a sub 1 to both sides, now I've got this inequality. 175 00:10:41,450 --> 00:10:44,200 But this is just a sub 1 plus a sub 2 plus a sub 176 00:10:44,200 --> 00:10:48,440 3 plus dot dot dot plus a sub big N, which I already know. 177 00:10:49,800 --> 00:10:55,570 Is at least as large as the integral from 1 to big N plus 1 of f of x dx. 178 00:10:55,570 --> 00:10:59,870 Let's now take a look at this. I could now take 179 00:10:59,870 --> 00:11:03,810 a limit as big N approaches infinity and 180 00:11:03,810 --> 00:11:08,930 here I would be getting what. Well this would just become the series, 181 00:11:08,930 --> 00:11:14,950 the sum of all the a sub n's. And this is now the integral from 1 to 182 00:11:14,950 --> 00:11:20,920 infinity of f of x dx. So I really know is that the value of this 183 00:11:20,920 --> 00:11:27,580 series, is trapped between this integral and the same integral plus the first term. 184 00:11:27,580 --> 00:11:30,040 Let me summarize what we have. 185 00:11:30,040 --> 00:11:33,300 Same setup as always, I'm supposing I've got some function little f which 186 00:11:33,300 --> 00:11:37,940 is decreasing but positive and I built my sequence out of that function. 187 00:11:38,980 --> 00:11:42,920 Then if I've got that, this integral 188 00:11:42,920 --> 00:11:46,300 converges, if that integral has a finite value, 189 00:11:46,300 --> 00:11:52,150 that happens if and only if this series converges, right? 190 00:11:52,150 --> 00:11:54,300 Because you've shown both directions, I've shown that 191 00:11:54,300 --> 00:11:59,010 if this integral diverges, then the series diverges. 192 00:11:59,010 --> 00:12:02,700 But I've also shown that if this interval has a finite value, then 193 00:12:02,700 --> 00:12:04,290 the sequence of partial sums is 194 00:12:04,290 --> 00:12:08,340 monitored and bounded therefore the series converges. 195 00:12:08,340 --> 00:12:10,610 And more than that, I've also got these bounds. 196 00:12:10,610 --> 00:12:11,705 I also 197 00:12:11,705 --> 00:12:18,713 know that the value of this sequence is 198 00:12:18,713 --> 00:12:25,302 trapped between these two values. 199 00:12:25,302 --> 00:12:26,771 [SOUND].