1 00:00:00,000 --> 00:00:02,591 The harmonic series again. 2 00:00:02,591 --> 00:00:08,677 [MUSIC] 3 00:00:08,677 --> 00:00:12,470 We've already seen that the harmonic series diverges. 4 00:00:12,470 --> 00:00:17,180 So we've shown that the harmonic series diverges by grouping together the terms 5 00:00:17,180 --> 00:00:22,650 and then showing that each of the groups is bigger than one half. 6 00:00:22,650 --> 00:00:27,930 So summing the harmonic series is even worse then summing 1 plus a half, 7 00:00:27,930 --> 00:00:31,140 plus a half, plus a half, plus a half, plus a half, and so on. 8 00:00:31,140 --> 00:00:34,060 We can also check this with the integral test, 9 00:00:34,060 --> 00:00:39,710 so let's say a sub n equal 1 over n and I am trying to analyze the convergence of 10 00:00:39,710 --> 00:00:45,070 the series n goes from one to infinity of a sub n, that's just the harmonic series. 11 00:00:45,070 --> 00:00:50,200 And to help me I'll look at the function f of x which is 1 over x. 12 00:00:50,200 --> 00:00:54,540 And note that a sub n is f of n. 13 00:00:54,540 --> 00:00:59,680 I want to make sure that f is positive and decreasing. 14 00:00:59,680 --> 00:01:04,420 Right, so I need the function f to be positive. 15 00:01:04,420 --> 00:01:09,300 And decreasing the interval that I care about, 16 00:01:09,300 --> 00:01:12,750 which is the interval from one to infinity. 17 00:01:12,750 --> 00:01:18,310 Well, I know that f is positive because if x is at least one, even if x is at 18 00:01:18,310 --> 00:01:22,130 least zero, alright, f of x, which 1 over x, 19 00:01:22,130 --> 00:01:24,730 is positive, I mean, the reciprocal of a positive number 20 00:01:24,730 --> 00:01:26,090 is positive. 21 00:01:26,090 --> 00:01:31,030 I need to check if f is decreasing, and what that means is that if a is bigger 22 00:01:31,030 --> 00:01:36,572 than b, then I need that f of a, is less than f of b. 23 00:01:36,572 --> 00:01:39,930 On the interval I'm thinking about here. 24 00:01:39,930 --> 00:01:41,860 Well, that's true because if I've got a 25 00:01:41,860 --> 00:01:44,740 and b, say they're both positive numbers even. 26 00:01:44,740 --> 00:01:50,160 Then how does 1 over a compared to 1 over b if a is bigger than b, or if 27 00:01:50,160 --> 00:01:56,140 I got a larger positive number but I take its reciprocal now its smaller. 28 00:01:56,140 --> 00:01:57,870 So this is telling me that F is a 29 00:01:57,870 --> 00:02:02,732 decreasing function, now I will check that the integral diverges. 30 00:02:02,732 --> 00:02:09,260 Well let's compute the integral from one to infinity of f of x dx. 31 00:02:09,260 --> 00:02:15,570 by definition this integral is the limit as big N approaches infinity 32 00:02:15,570 --> 00:02:23,110 of the integral from 1 to big N of f of x dx, now in this case f 33 00:02:23,110 --> 00:02:28,310 is just the one over function, right it's the function that sends x to one over x. 34 00:02:28,310 --> 00:02:32,000 So this is the limit as big N approaches infinity, of 35 00:02:32,000 --> 00:02:35,560 the integral from 1 to big N, just 1 over x dx. 36 00:02:36,660 --> 00:02:39,340 Now, to compute this definite integral, I'm 37 00:02:39,340 --> 00:02:41,550 going to use the fundamental theorem of calculus and 38 00:02:41,550 --> 00:02:48,740 recall that an anti-derivative for the 1 over x function is the natural logarithm. 39 00:02:48,740 --> 00:02:51,360 So, by the fundamental theorem of calculus, this is 40 00:02:51,360 --> 00:02:55,940 the limit as N approaches infinity of an anti-derivative, 41 00:02:55,940 --> 00:02:59,810 which is the natural log, at the one endpoint, 42 00:02:59,810 --> 00:03:05,210 minus the anti-derivative natural log at the other endpoint. 43 00:03:05,210 --> 00:03:06,680 And a log of 1 44 00:03:06,680 --> 00:03:09,770 is 0, so this is just asking. 45 00:03:09,770 --> 00:03:16,290 What's the limit as N approaches infinity, of the natural log of N, right? 46 00:03:16,290 --> 00:03:19,650 What happens when you take the natural log of an enormous number? 47 00:03:19,650 --> 00:03:24,230 Well, admittedly, it's less enormous, but it's still as large as you'd like, right. 48 00:03:24,230 --> 00:03:29,090 By choosing big N, big enough, I can make log N as large as you'd like. 49 00:03:29,090 --> 00:03:31,890 Which is to say that the limit of log N 50 00:03:31,890 --> 00:03:37,620 as N approaches infinity is infinity. So this integral diverges. 51 00:03:37,620 --> 00:03:40,490 So the same is true of the series. 52 00:03:40,490 --> 00:03:46,340 Right. So, by the integral 53 00:03:46,340 --> 00:03:51,840 test. The sum n goes 1 to infinity of' 1 over n, 54 00:03:51,840 --> 00:03:57,550 the harmonic series, diverges as well. 55 00:03:57,550 --> 00:04:01,440 So now you get two proofs that the harmonic series diverges. 56 00:04:02,670 --> 00:04:05,720 Any time in mathematics that you've got two different proofs of 57 00:04:05,720 --> 00:04:09,940 the same theorem, it's worth thinking about how those proofs relate. 58 00:04:09,940 --> 00:04:12,620 In this case which proof do you think is easier? 59 00:04:12,620 --> 00:04:15,390 I mean in a certain sense, the interval test is easier. 60 00:04:15,390 --> 00:04:17,810 Requires less creativity to just integrate one 61 00:04:17,810 --> 00:04:19,960 over x than to do that complicated grouping. 62 00:04:21,070 --> 00:04:22,590 And yet the grouping argument 63 00:04:22,590 --> 00:04:24,620 predates integration. 64 00:04:24,620 --> 00:04:27,770 Alright, the grouping argument is from the 1300s, 65 00:04:27,770 --> 00:04:31,850 and integration, right, wasn't really around until Newton, say. 66 00:04:32,890 --> 00:04:37,480 So yes, although the integral test seems like the easier way to show the harmonic 67 00:04:37,480 --> 00:04:43,770 series diverges, the integral test is dependent on the huge edifice of calculus. 68 00:04:43,770 --> 00:04:47,870 It is requiring more technology. And a grouping argument, 69 00:04:47,870 --> 00:04:50,468 how well somehow requiring more creativity. 70 00:04:50,468 --> 00:04:56,068 Doesn't 71 00:04:56,068 --> 00:05:03,730 require so much theory, to be built up.