WEBVTT 00:00:00.000 --> 00:00:02.420 align:middle line:90% [SQUEAKING] 00:00:02.420 --> 00:00:03.388 align:middle line:90% [RUSTLING] 00:00:03.388 --> 00:00:05.324 align:middle line:90% [CLICKING] 00:00:05.324 --> 00:00:13.833 align:middle line:90% 00:00:13.833 --> 00:00:15.250 align:middle line:84% CASEY RODRIGUEZ: So let's continue 00:00:15.250 --> 00:00:19.780 align:middle line:90% our discussion of series. 00:00:19.780 --> 00:00:25.750 align:middle line:84% So last time, we proved the comparison test 00:00:25.750 --> 00:00:27.530 align:middle line:90% at the end of last time. 00:00:27.530 --> 00:00:28.945 align:middle line:90% So this was the comparison test. 00:00:28.945 --> 00:00:35.270 align:middle line:90% 00:00:35.270 --> 00:00:41.680 align:middle line:84% So which is a statement about series with non-negative terms, 00:00:41.680 --> 00:00:43.420 align:middle line:84% with one being smaller than the other. 00:00:43.420 --> 00:00:46.180 align:middle line:90% 00:00:46.180 --> 00:00:51.580 align:middle line:84% And one of two things, or two things are true. 00:00:51.580 --> 00:00:55.900 align:middle line:84% If I have xn and yn non-negative with yn 00:00:55.900 --> 00:00:58.630 align:middle line:84% bigger than or equal to xn, then the conclusion 00:00:58.630 --> 00:01:06.220 align:middle line:84% is, if the bigger series converges, 00:01:06.220 --> 00:01:08.920 align:middle line:84% this implies that the smaller series converges. 00:01:08.920 --> 00:01:12.650 align:middle line:90% 00:01:12.650 --> 00:01:21.340 align:middle line:84% And the second statement is, if the smaller series diverges, 00:01:21.340 --> 00:01:29.690 align:middle line:90% then the larger series diverges. 00:01:29.690 --> 00:01:32.210 align:middle line:84% And we also proved not using the comparison test, 00:01:32.210 --> 00:01:39.650 align:middle line:84% but for p-series, that the series 1 sum from n 00:01:39.650 --> 00:01:46.505 align:middle line:84% equals 1 to infinity of 1 over n to the p, This converges. 00:01:46.505 --> 00:01:49.520 align:middle line:84% Well, I mean, we did use a comparison test 00:01:49.520 --> 00:01:50.870 align:middle line:90% for one direction, I guess. 00:01:50.870 --> 00:01:53.090 align:middle line:84% So 1 over n to the p converges if and only 00:01:53.090 --> 00:01:55.310 align:middle line:90% if p is bigger than 1. 00:01:55.310 --> 00:01:58.490 align:middle line:84% So this converging implied p has to be 00:01:58.490 --> 00:02:00.950 align:middle line:84% bigger than 1 by the comparison test 00:02:00.950 --> 00:02:03.530 align:middle line:84% and what we know about the harmonic series 00:02:03.530 --> 00:02:06.748 align:middle line:90% 1 over n, sum of 1 over n. 00:02:06.748 --> 00:02:09.500 align:middle line:90% 00:02:09.500 --> 00:02:12.560 align:middle line:84% So typically, in applications you use these two theorems 00:02:12.560 --> 00:02:14.570 align:middle line:90% together. 00:02:14.570 --> 00:02:18.670 align:middle line:84% To say something about series that don't look so simple. 00:02:18.670 --> 00:02:29.260 align:middle line:84% So for example, if I look at the series, 1 00:02:29.260 --> 00:02:39.480 align:middle line:84% over n squared plus 2020n, n equals 1 to infinity. 00:02:39.480 --> 00:02:43.340 align:middle line:84% So one could ask, does this converge, diverge? 00:02:43.340 --> 00:02:47.210 align:middle line:84% Well, this is a series with non-negative terms. 00:02:47.210 --> 00:02:51.020 align:middle line:84% And remember, I mean, the bigger thing 00:02:51.020 --> 00:02:55.830 align:middle line:84% has to converge to imply that the smaller thing converges. 00:02:55.830 --> 00:02:57.750 align:middle line:84% And if the smaller thing diverges, 00:02:57.750 --> 00:02:59.280 align:middle line:90% then the bigger thing diverges. 00:02:59.280 --> 00:03:03.030 align:middle line:84% So don't get the inequalities mixed up. 00:03:03.030 --> 00:03:11.110 align:middle line:84% So I have 1 over n squared plus 2020n. 00:03:11.110 --> 00:03:14.680 align:middle line:84% This 2020 times n is just making things bigger on the bottom 00:03:14.680 --> 00:03:16.340 align:middle line:90% and therefore smaller overall. 00:03:16.340 --> 00:03:20.090 align:middle line:84% So this is less than or equal to 1 over n squared. 00:03:20.090 --> 00:03:31.670 align:middle line:84% And since this converges, this implies by the comparison test 00:03:31.670 --> 00:03:37.040 align:middle line:84% that sum from n equals 1 to infinity of 1 over n 00:03:37.040 --> 00:03:43.460 align:middle line:84% squared plus 2020n also converges. 00:03:43.460 --> 00:03:46.890 align:middle line:90% 00:03:46.890 --> 00:03:51.770 align:middle line:84% Now, a mistake I'm sure we've all made at some point 00:03:51.770 --> 00:03:56.000 align:middle line:84% is mixing up the inequality and not 00:03:56.000 --> 00:03:58.650 align:middle line:90% getting quite the right answer. 00:03:58.650 --> 00:04:03.440 align:middle line:84% So for example, it is also true that 1 over n 00:04:03.440 --> 00:04:09.740 align:middle line:84% squared plus 2020n is certainly less than or equal to 1 00:04:09.740 --> 00:04:18.329 align:middle line:84% over 2020 times n, which is also less than or equal to 1 over n. 00:04:18.329 --> 00:04:29.984 align:middle line:84% And so, you're tempted to say, since this diverges, 00:04:29.984 --> 00:04:39.950 align:middle line:84% it implies the other series diverges. 00:04:39.950 --> 00:04:48.630 align:middle line:90% But this is not right. 00:04:48.630 --> 00:04:50.670 align:middle line:90% Because the inequality is wrong. 00:04:50.670 --> 00:04:53.950 align:middle line:84% Remember, if you want to apply the comparison test, 00:04:53.950 --> 00:04:58.050 align:middle line:84% you either have to have a bigger series which converges 00:04:58.050 --> 00:05:00.660 align:middle line:84% or a smaller series which diverges. 00:05:00.660 --> 00:05:04.360 align:middle line:84% Here we came up with a bigger series which diverges, 00:05:04.360 --> 00:05:07.200 align:middle line:84% which gives us no information at all. 00:05:07.200 --> 00:05:10.050 align:middle line:84% If we have a bigger series, like 1 over n 00:05:10.050 --> 00:05:12.840 align:middle line:84% squared that we know converges, then 00:05:12.840 --> 00:05:14.865 align:middle line:84% do we get information about the original series. 00:05:14.865 --> 00:05:20.760 align:middle line:90% 00:05:20.760 --> 00:05:26.010 align:middle line:84% So let's do another well-known test, 00:05:26.010 --> 00:05:31.680 align:middle line:84% or at least test you should remember from calculus, 00:05:31.680 --> 00:05:32.790 align:middle line:90% so-called ratio test. 00:05:32.790 --> 00:05:39.420 align:middle line:90% 00:05:39.420 --> 00:05:41.070 align:middle line:90% So what is a statement of this? 00:05:41.070 --> 00:05:50.295 align:middle line:84% So suppose xn does not equal 0 for all n. 00:05:50.295 --> 00:05:53.840 align:middle line:90% 00:05:53.840 --> 00:05:59.030 align:middle line:84% And this limit l equals limit as n 00:05:59.030 --> 00:06:04.370 align:middle line:84% goes to infinity of x to the n plus 1 over x to the n, 00:06:04.370 --> 00:06:06.530 align:middle line:90% an absolute value exists. 00:06:06.530 --> 00:06:13.460 align:middle line:90% 00:06:13.460 --> 00:06:28.030 align:middle line:84% Then, if l is less than 1 this implies that the series xn 00:06:28.030 --> 00:06:29.350 align:middle line:90% converges absolutely. 00:06:29.350 --> 00:06:34.280 align:middle line:90% 00:06:34.280 --> 00:06:40.115 align:middle line:84% And if l is bigger than 1, then the series diverges. 00:06:40.115 --> 00:06:44.460 align:middle line:90% 00:06:44.460 --> 00:06:46.130 align:middle line:90% Now, what about l equals 1? 00:06:46.130 --> 00:06:48.050 align:middle line:84% There's no information for l equals 1. 00:06:48.050 --> 00:06:58.080 align:middle line:90% 00:06:58.080 --> 00:07:04.530 align:middle line:84% Meaning you could have a series so that this l equals 1. 00:07:04.530 --> 00:07:08.043 align:middle line:84% And you could have a series where it diverges. 00:07:08.043 --> 00:07:10.710 align:middle line:84% And you could also have a series where l equals 1 and the series 00:07:10.710 --> 00:07:11.710 align:middle line:90% converges. 00:07:11.710 --> 00:07:23.510 align:middle line:84% So for example, if xn equals 1 over n, for example. 00:07:23.510 --> 00:07:30.640 align:middle line:84% Or I mean, even worse, let's say 1 for all n. 00:07:30.640 --> 00:07:35.810 align:middle line:90% 00:07:35.810 --> 00:07:44.720 align:middle line:84% But we know that the series 1 diverges. 00:07:44.720 --> 00:07:47.420 align:middle line:84% Because the individual terms do not converge to 0. 00:07:47.420 --> 00:07:50.270 align:middle line:90% They're just 1 the whole time. 00:07:50.270 --> 00:07:55.890 align:middle line:84% But for xn equals, let's say, 1 over n 00:07:55.890 --> 00:08:00.690 align:middle line:84% squared, so the one we just saw a minute ago, 00:08:00.690 --> 00:08:03.030 align:middle line:84% implies l, which is the limit as n 00:08:03.030 --> 00:08:09.180 align:middle line:84% goes to infinity of n plus 1 squared over n squared 00:08:09.180 --> 00:08:12.990 align:middle line:84% equals the limit as n goes to infinity. 00:08:12.990 --> 00:08:19.490 align:middle line:84% Dividing through, one plus 1 over n squared, 00:08:19.490 --> 00:08:25.460 align:middle line:84% this equals 1 plus 0 squared equals 1. 00:08:25.460 --> 00:08:35.374 align:middle line:84% So for this series, the root or the ratio gives this l is 1. 00:08:35.374 --> 00:08:39.730 align:middle line:90% 00:08:39.730 --> 00:08:41.679 align:middle line:90% And this series converges. 00:08:41.679 --> 00:08:45.530 align:middle line:90% 00:08:45.530 --> 00:08:48.380 align:middle line:84% So again, for the case that l equals 1, 00:08:48.380 --> 00:08:49.340 align:middle line:90% we have no information. 00:08:49.340 --> 00:08:52.760 align:middle line:90% We can't say anything based on-- 00:08:52.760 --> 00:08:54.884 align:middle line:84% this theorem gives us no information. 00:08:54.884 --> 00:09:03.430 align:middle line:90% 00:09:03.430 --> 00:09:10.130 align:middle line:84% So we're going to prove this theorem basically by-- 00:09:10.130 --> 00:09:12.860 align:middle line:90% 00:09:12.860 --> 00:09:15.140 align:middle line:84% so we're not exactly going to use a comparison test. 00:09:15.140 --> 00:09:20.300 align:middle line:84% But we are still going to compare this series satisfying 00:09:20.300 --> 00:09:23.150 align:middle line:84% one of these two assumptions to a series we know, 00:09:23.150 --> 00:09:26.240 align:middle line:84% which is namely the ratio test for number one. 00:09:26.240 --> 00:09:31.405 align:middle line:84% So first off, let's get number two out of the way. 00:09:31.405 --> 00:09:37.610 align:middle line:90% 00:09:37.610 --> 00:09:43.740 align:middle line:84% So let's suppose l is bigger than 1. 00:09:43.740 --> 00:09:47.740 align:middle line:90% 00:09:47.740 --> 00:09:55.390 align:middle line:84% Let alpha be a number between 1 and l. 00:09:55.390 --> 00:09:59.730 align:middle line:90% 00:09:59.730 --> 00:10:01.410 align:middle line:84% Actually, we don't even need to do that. 00:10:01.410 --> 00:10:06.570 align:middle line:90% 00:10:06.570 --> 00:10:15.650 align:middle line:84% So here's-- here is 1, here is l. 00:10:15.650 --> 00:10:24.500 align:middle line:84% So since xn plus 1 over xn absolute value converges to l, 00:10:24.500 --> 00:10:29.510 align:middle line:90% there exists an integer M0. 00:10:29.510 --> 00:10:32.180 align:middle line:90% So that's a natural number. 00:10:32.180 --> 00:10:38.090 align:middle line:84% So that for all n bigger than or equal to M0, 00:10:38.090 --> 00:10:42.260 align:middle line:84% xn plus 1 over xn in absolute value 00:10:42.260 --> 00:10:44.040 align:middle line:90% is bigger than or equal to-- 00:10:44.040 --> 00:10:48.020 align:middle line:90% so here, you could write 1 as-- 00:10:48.020 --> 00:10:53.340 align:middle line:90% this is equal to l plus-- 00:10:53.340 --> 00:10:55.300 align:middle line:90% I should say, minus-- 00:10:55.300 --> 00:10:57.400 align:middle line:90% l minus 1. 00:10:57.400 --> 00:11:00.190 align:middle line:84% So think of this as being epsilon in the definition 00:11:00.190 --> 00:11:01.240 align:middle line:90% of convergence. 00:11:01.240 --> 00:11:04.810 align:middle line:90% 00:11:04.810 --> 00:11:12.950 align:middle line:84% And so, this is over here. l plus l minus 1. 00:11:12.950 --> 00:11:15.290 align:middle line:84% So by the definition of convergence, 00:11:15.290 --> 00:11:18.560 align:middle line:84% for all n sufficiently large, xn plus 1 00:11:18.560 --> 00:11:21.180 align:middle line:84% over xn in absolute value has to be in this interval. 00:11:21.180 --> 00:11:22.970 align:middle line:84% So it has to be bigger than or equal to 1. 00:11:22.970 --> 00:11:31.930 align:middle line:90% 00:11:31.930 --> 00:11:37.990 align:middle line:84% Which implies that for all n bigger than or equal to M0, 00:11:37.990 --> 00:11:41.410 align:middle line:84% xn plus 1 is bigger than or equal to xn. 00:11:41.410 --> 00:11:49.370 align:middle line:90% 00:11:49.370 --> 00:11:55.910 align:middle line:84% And so, I could write this as saying that x M0 is less than 00:11:55.910 --> 00:12:03.320 align:middle line:84% or equal to x M0 plus 1, is less than or equal to x M0 plus 2, 00:12:03.320 --> 00:12:04.040 align:middle line:90% and so on. 00:12:04.040 --> 00:12:06.790 align:middle line:90% 00:12:06.790 --> 00:12:20.170 align:middle line:84% But this implies that these xn's cannot converge to 0 as n goes 00:12:20.170 --> 00:12:20.680 align:middle line:90% to infinity. 00:12:20.680 --> 00:12:23.950 align:middle line:84% Because for all n bigger than or equal to M0, 00:12:23.950 --> 00:12:27.070 align:middle line:90% they are increasing. 00:12:27.070 --> 00:12:29.410 align:middle line:84% And the only way for an increasing sequence 00:12:29.410 --> 00:12:31.750 align:middle line:84% which is non-negative to converge to zeros for them 00:12:31.750 --> 00:12:34.630 align:middle line:90% all to be 0. 00:12:34.630 --> 00:12:36.055 align:middle line:84% And we're assuming they're not 0. 00:12:36.055 --> 00:12:45.640 align:middle line:90% 00:12:45.640 --> 00:12:48.025 align:middle line:90% So that proves two. 00:12:48.025 --> 00:12:49.000 align:middle line:90% Let's prove one. 00:12:49.000 --> 00:12:53.860 align:middle line:90% 00:12:53.860 --> 00:13:00.700 align:middle line:90% So suppose l is now less than 1. 00:13:00.700 --> 00:13:08.320 align:middle line:84% And now, let alpha be a number between l and 1. 00:13:08.320 --> 00:13:12.060 align:middle line:84% So I just want to give myself a little bit of room, 00:13:12.060 --> 00:13:14.770 align:middle line:84% to work as you'll see in just a second. 00:13:14.770 --> 00:13:20.830 align:middle line:84% So by the same reasoning as before, 00:13:20.830 --> 00:13:25.620 align:middle line:90% so here's l, alpha 1 alpha. 00:13:25.620 --> 00:13:32.130 align:middle line:84% I can write as l plus alpha minus l. 00:13:32.130 --> 00:13:34.995 align:middle line:84% So I should think of this as kind of an epsilon. 00:13:34.995 --> 00:13:38.350 align:middle line:90% 00:13:38.350 --> 00:13:47.590 align:middle line:84% Then since xn plus 1 over xn in absolute value converges 00:13:47.590 --> 00:13:53.060 align:middle line:84% to l, which is less than alpha, there 00:13:53.060 --> 00:13:56.315 align:middle line:90% exists in M0 natural number. 00:13:56.315 --> 00:13:59.240 align:middle line:90% 00:13:59.240 --> 00:14:05.146 align:middle line:84% So that, for all M bigger than or equal to M0, 00:14:05.146 --> 00:14:10.240 align:middle line:84% xn plus 1 over xn in absolute value 00:14:10.240 --> 00:14:12.235 align:middle line:90% is less than or equal to alpha. 00:14:12.235 --> 00:14:15.370 align:middle line:90% 00:14:15.370 --> 00:14:19.030 align:middle line:84% So here I have alpha just a little bit bigger than l. 00:14:19.030 --> 00:14:20.890 align:middle line:90% And if I draw-- 00:14:20.890 --> 00:14:25.660 align:middle line:84% so this is l minus alpha minus l. 00:14:25.660 --> 00:14:29.440 align:middle line:84% So since I have this sequence here converging 00:14:29.440 --> 00:14:33.040 align:middle line:84% to l for all insufficiently large, 00:14:33.040 --> 00:14:35.500 align:middle line:90% this should be in this interval. 00:14:35.500 --> 00:14:39.220 align:middle line:84% Meaning this should be less than or equal to alpha. 00:14:39.220 --> 00:14:49.490 align:middle line:90% 00:14:49.490 --> 00:14:52.250 align:middle line:84% Then for all n bigger than or equal to 0, 00:14:52.250 --> 00:14:55.760 align:middle line:84% so let me just write this slightly differently, 00:14:55.760 --> 00:15:03.290 align:middle line:84% x sub n plus 1 is less than or equal to alpha times 00:15:03.290 --> 00:15:06.710 align:middle line:90% the absolute value of x of n. 00:15:06.710 --> 00:15:14.330 align:middle line:90% Now, let's see what this means. 00:15:14.330 --> 00:15:18.750 align:middle line:90% 00:15:18.750 --> 00:15:22.080 align:middle line:84% For all n bigger than or equal M sub 0. 00:15:22.080 --> 00:15:24.795 align:middle line:84% Let's say I look at the absolute value of x sub n. 00:15:24.795 --> 00:15:27.880 align:middle line:90% 00:15:27.880 --> 00:15:30.100 align:middle line:84% So now, this n has nothing to do with these n. 00:15:30.100 --> 00:15:41.410 align:middle line:84% It's just-- so then the absolute value of x of n is less than 00:15:41.410 --> 00:15:45.025 align:middle line:90% or equal to x-- 00:15:45.025 --> 00:15:48.782 align:middle line:90% 00:15:48.782 --> 00:15:50.225 align:middle line:90% so maybe I'm off by-- 00:15:50.225 --> 00:15:58.800 align:middle line:90% 00:15:58.800 --> 00:16:02.520 align:middle line:90% OK, so let's increase that by 1. 00:16:02.520 --> 00:16:06.780 align:middle line:90% 00:16:06.780 --> 00:16:10.440 align:middle line:84% Well, so let's not be confusing here. 00:16:10.440 --> 00:16:14.630 align:middle line:90% Let's make this l. 00:16:14.630 --> 00:16:15.890 align:middle line:90% Let's make that l. 00:16:15.890 --> 00:16:21.140 align:middle line:84% So for all l bigger than or equal to M0 plus 1. 00:16:21.140 --> 00:16:25.390 align:middle line:84% So think of l as playing the role of being n plus 1. 00:16:25.390 --> 00:16:31.330 align:middle line:84% This is less than or equal to x to the l minus 1, which 00:16:31.330 --> 00:16:35.650 align:middle line:84% is less than or equal to x to the l minus 2, 00:16:35.650 --> 00:16:37.030 align:middle line:90% now with a square. 00:16:37.030 --> 00:16:44.540 align:middle line:84% So what I have here is now alpha times alpha x to the l minus 2. 00:16:44.540 --> 00:16:49.450 align:middle line:90% 00:16:49.450 --> 00:16:57.660 align:middle line:90% And now, let's drop again by 1. 00:16:57.660 --> 00:17:04.170 align:middle line:84% And I can do this as long as this quantity here 00:17:04.170 --> 00:17:07.700 align:middle line:84% is bigger than or equal to M. So I 00:17:07.700 --> 00:17:11.540 align:middle line:84% get this is, if I keep doing that, this is less than 00:17:11.540 --> 00:17:23.029 align:middle line:84% or equal to alpha to the l minus M0, x to the M0. 00:17:23.029 --> 00:17:46.670 align:middle line:90% 00:17:46.670 --> 00:17:49.655 align:middle line:84% And so, actually let's, we shifted things-- 00:17:49.655 --> 00:17:57.890 align:middle line:90% 00:17:57.890 --> 00:17:58.390 align:middle line:90% yeah. 00:17:58.390 --> 00:18:03.610 align:middle line:90% 00:18:03.610 --> 00:18:13.280 align:middle line:84% So now we're going to use this to bound 00:18:13.280 --> 00:18:16.280 align:middle line:90% the partial sums of the x of n. 00:18:16.280 --> 00:18:25.130 align:middle line:90% So let M be a natural number. 00:18:25.130 --> 00:18:31.790 align:middle line:84% And if we look at partial sum, sum from n equals 1 to M 00:18:31.790 --> 00:18:36.230 align:middle line:84% of x sub n, this is equal to sum n equals 1. 00:18:36.230 --> 00:18:39.410 align:middle line:90% 00:18:39.410 --> 00:18:53.790 align:middle line:84% to M0 xn plus sum from n equals M0 plus 1 to M. x sub-- 00:18:53.790 --> 00:18:58.740 align:middle line:84% now let me just change dummy variables for signing, 00:18:58.740 --> 00:19:02.180 align:middle line:90% for indexing this. 00:19:02.180 --> 00:19:07.490 align:middle line:90% 00:19:07.490 --> 00:19:12.350 align:middle line:84% So this is equal to the sum from n equals 1 to M0 of xn. 00:19:12.350 --> 00:19:18.600 align:middle line:84% And now, this, I'm going to use this inequality here. 00:19:18.600 --> 00:19:21.515 align:middle line:90% So this is kind of a mess. 00:19:21.515 --> 00:19:40.660 align:middle line:90% 00:19:40.660 --> 00:19:49.890 align:middle line:84% So this is the inequality that I get from this. 00:19:49.890 --> 00:19:51.480 align:middle line:90% Remember, alpha's less than 1. 00:19:51.480 --> 00:19:55.950 align:middle line:90% 00:19:55.950 --> 00:20:13.770 align:middle line:90% Plus-- OK. 00:20:13.770 --> 00:20:23.460 align:middle line:84% So time's l equals M0 plus 1 to M. OK, everything's fine. 00:20:23.460 --> 00:20:30.390 align:middle line:84% Times alpha to the l minus M0 plus 1. 00:20:30.390 --> 00:20:36.030 align:middle line:84% So all I did here was replace this absolute value of x to 00:20:36.030 --> 00:20:37.740 align:middle line:90% of x sub l. 00:20:37.740 --> 00:20:40.590 align:middle line:84% l is just a dummy variable for indexing these guys, 00:20:40.590 --> 00:20:44.820 align:middle line:90% but using this inequality here. 00:20:44.820 --> 00:20:48.910 align:middle line:84% So, sorry, I kind of messed that up a little bit. 00:20:48.910 --> 00:20:52.020 align:middle line:84% But the important thing is that we have this inequality, which 00:20:52.020 --> 00:20:55.980 align:middle line:84% somehow tells you this series is not 00:20:55.980 --> 00:20:58.050 align:middle line:84% that far from being a geometric series, 00:20:58.050 --> 00:20:59.460 align:middle line:90% at least when l is big enough. 00:20:59.460 --> 00:21:02.280 align:middle line:90% 00:21:02.280 --> 00:21:04.680 align:middle line:84% So then, when we take an arbitrary partial sum, 00:21:04.680 --> 00:21:07.080 align:middle line:90% we split it up into two parts. 00:21:07.080 --> 00:21:09.840 align:middle line:84% The stuff that comes up to this integer 00:21:09.840 --> 00:21:11.460 align:middle line:84% sub 0, which we don't care about, 00:21:11.460 --> 00:21:15.640 align:middle line:84% that's just a fixed number, plus this part that we care about, 00:21:15.640 --> 00:21:19.380 align:middle line:84% which is after this fixed number M0. 00:21:19.380 --> 00:21:21.600 align:middle line:84% And we use this inequality to replace this 00:21:21.600 --> 00:21:26.850 align:middle line:84% by x sub M0 plus 1, alpha to the l minus M0 plus 1. 00:21:26.850 --> 00:21:29.370 align:middle line:84% The thing to remember is that little m 00:21:29.370 --> 00:21:31.500 align:middle line:90% is the thing that's changing. 00:21:31.500 --> 00:21:36.960 align:middle line:84% So we're trying to bound this independent of little m. 00:21:36.960 --> 00:21:39.240 align:middle line:84% Capital M sub 0, that's just something fixed. 00:21:39.240 --> 00:21:43.510 align:middle line:90% That could be 1,000. 00:21:43.510 --> 00:21:49.780 align:middle line:84% So this is equal to sum from n equals 1 to M0 00:21:49.780 --> 00:21:54.660 align:middle line:90% xn, plus x sub M0 plus 1. 00:21:54.660 --> 00:21:57.360 align:middle line:90% 00:21:57.360 --> 00:22:00.790 align:middle line:84% And now, if I change the variable again, 00:22:00.790 --> 00:22:06.420 align:middle line:84% so l starts at M0 plus 1 and ends at M. 00:22:06.420 --> 00:22:10.720 align:middle line:90% So this is now a sum if I go-- 00:22:10.720 --> 00:22:18.540 align:middle line:84% so now, for the second sum, n is equal to l minus M0 plus 1. 00:22:18.540 --> 00:22:28.540 align:middle line:84% So now, n starts at 0 and ends at M minus M0 plus 1, 00:22:28.540 --> 00:22:31.890 align:middle line:90% alpha to the n. 00:22:31.890 --> 00:22:35.370 align:middle line:84% Again, Emma, is the little-- is the thing that's changing here. 00:22:35.370 --> 00:22:38.610 align:middle line:84% And we're trying to bound this independently of little m. 00:22:38.610 --> 00:22:39.810 align:middle line:90% But now we're in good shape. 00:22:39.810 --> 00:22:42.480 align:middle line:84% Because this looks like a geometric series. 00:22:42.480 --> 00:22:44.880 align:middle line:90% Alpha, remember, is less than 1. 00:22:44.880 --> 00:22:49.260 align:middle line:84% So this is less than or equal to sum from n 00:22:49.260 --> 00:22:59.100 align:middle line:84% equals 1 to M0 x sub n plus x sub n 0 plus 1 times-- 00:22:59.100 --> 00:23:06.870 align:middle line:84% now, instead of just being a sum from 0 up to M minus M0 plus 1, 00:23:06.870 --> 00:23:11.640 align:middle line:84% why not throw all of it in there. 00:23:11.640 --> 00:23:25.410 align:middle line:84% And this is equal to n equals 0, xn, plus x to M0 plus 1 times 1 00:23:25.410 --> 00:23:26.490 align:middle line:90% over 1 minus alpha. 00:23:26.490 --> 00:23:31.730 align:middle line:84% Remember, alpha is a number that we fixed to be less than 1. 00:23:31.730 --> 00:23:34.190 align:middle line:84% If you try to do what we did before 00:23:34.190 --> 00:23:38.000 align:middle line:84% and not fix the alpha just a little bit to the left of 1, 00:23:38.000 --> 00:23:40.190 align:middle line:84% and try to do everything with 1, you 00:23:40.190 --> 00:23:43.340 align:middle line:84% would have wound up with 1 to the n here. 00:23:43.340 --> 00:23:45.890 align:middle line:84% And that wouldn't have finished the proof. 00:23:45.890 --> 00:23:47.390 align:middle line:84% That wouldn't have closed the proof. 00:23:47.390 --> 00:23:49.130 align:middle line:84% But given yourself a little bit of room, 00:23:49.130 --> 00:23:50.855 align:middle line:90% which is why we fix this alpha. 00:23:50.855 --> 00:23:54.910 align:middle line:90% 00:23:54.910 --> 00:23:57.980 align:middle line:84% Now, this number here is independent of little m. 00:23:57.980 --> 00:24:04.060 align:middle line:90% 00:24:04.060 --> 00:24:06.340 align:middle line:90% That's the whole point. 00:24:06.340 --> 00:24:20.010 align:middle line:90% So what have we proven? 00:24:20.010 --> 00:24:23.040 align:middle line:84% That for all natural numbers little m, 00:24:23.040 --> 00:24:30.290 align:middle line:84% the nth partial sum is less than or equal to a fixed number 00:24:30.290 --> 00:24:33.230 align:middle line:90% given by sum from-- 00:24:33.230 --> 00:24:42.280 align:middle line:90% 00:24:42.280 --> 00:24:42.780 align:middle line:90% OK. 00:24:42.780 --> 00:24:46.180 align:middle line:90% 00:24:46.180 --> 00:24:48.880 align:middle line:84% And therefore, this sequence of partial sums 00:24:48.880 --> 00:25:00.620 align:middle line:84% is bounded and therefore converges. 00:25:00.620 --> 00:25:22.460 align:middle line:90% 00:25:22.460 --> 00:25:28.320 align:middle line:84% So kind of the simplest application of this 00:25:28.320 --> 00:25:32.920 align:middle line:84% is maybe a series that looks familiar. 00:25:32.920 --> 00:25:37.380 align:middle line:90% So for all x-- 00:25:37.380 --> 00:25:40.140 align:middle line:84% so this is I guess a theorem slash example. 00:25:40.140 --> 00:25:46.840 align:middle line:84% But for all x in R, the series x to the n over n factorial, 00:25:46.840 --> 00:25:57.275 align:middle line:84% n equals 0 to infinity converges absolutely here. 00:25:57.275 --> 00:25:59.990 align:middle line:90% 00:25:59.990 --> 00:26:01.880 align:middle line:90% 0 factorial is, of course, 1. 00:26:01.880 --> 00:26:05.580 align:middle line:90% 00:26:05.580 --> 00:26:08.400 align:middle line:90% And you just use the ratio test. 00:26:08.400 --> 00:26:19.740 align:middle line:84% So of dx to the n plus 1, or the n plus first term 00:26:19.740 --> 00:26:20.970 align:middle line:90% over x to the n. 00:26:20.970 --> 00:26:33.260 align:middle line:84% So absolute value, and this equals the limit as n 00:26:33.260 --> 00:26:37.430 align:middle line:84% goes to infinity of n plus 1 factorial 00:26:37.430 --> 00:26:40.320 align:middle line:84% is equal to n plus 1 times n factorial. 00:26:40.320 --> 00:26:42.230 align:middle line:84% So this cancels with that n factorial. 00:26:42.230 --> 00:26:47.960 align:middle line:84% And I get x over n plus 1, limit as n 00:26:47.960 --> 00:26:50.870 align:middle line:84% goes to infinity of just this fixed number over n 00:26:50.870 --> 00:26:52.460 align:middle line:90% plus 1 equals 0. 00:26:52.460 --> 00:26:56.790 align:middle line:84% And this is certainly less than 1. 00:26:56.790 --> 00:26:59.710 align:middle line:84% And therefore, by the ratio test, 00:26:59.710 --> 00:27:01.380 align:middle line:84% this series converges absolutely. 00:27:01.380 --> 00:27:07.360 align:middle line:90% 00:27:07.360 --> 00:27:15.400 align:middle line:84% So I got a little hung up on exactly the indexes 00:27:15.400 --> 00:27:17.930 align:middle line:90% and matching them up precisely. 00:27:17.930 --> 00:27:23.200 align:middle line:84% But the important thing to take home from this proof 00:27:23.200 --> 00:27:28.920 align:middle line:84% was that when this ratio is less than 1, then 00:27:28.920 --> 00:27:35.400 align:middle line:84% this series behaves very much like a geometric series 00:27:35.400 --> 00:27:38.340 align:middle line:90% for as the terms get very-- 00:27:38.340 --> 00:27:43.510 align:middle line:84% as you go far enough out in the terms. 00:27:43.510 --> 00:27:48.620 align:middle line:84% And this idea of just trying to relate the series 00:27:48.620 --> 00:27:52.970 align:middle line:84% to a simple series that you know a lot about, I mean, 00:27:52.970 --> 00:27:56.960 align:middle line:84% basically the only series that you know everything about, 00:27:56.960 --> 00:28:03.320 align:middle line:84% namely even how to sum it, this is how you get this test. 00:28:03.320 --> 00:28:05.060 align:middle line:84% And this is also even simpler how 00:28:05.060 --> 00:28:07.280 align:middle line:84% you get the next test, which is the root test. 00:28:07.280 --> 00:28:23.000 align:middle line:90% 00:28:23.000 --> 00:28:29.600 align:middle line:84% So the root test, let's take a series. 00:28:29.600 --> 00:28:35.480 align:middle line:90% 00:28:35.480 --> 00:28:47.170 align:middle line:84% And suppose this l, this limit l equals the limit as n 00:28:47.170 --> 00:28:53.565 align:middle line:84% goes to infinity of x sub entity 1 over n exists. 00:28:53.565 --> 00:28:59.550 align:middle line:90% 00:28:59.550 --> 00:29:05.100 align:middle line:84% Then, two conclusions, just kind of just like in ratio test. 00:29:05.100 --> 00:29:09.950 align:middle line:90% 00:29:09.950 --> 00:29:18.230 align:middle line:84% If l is less than 1, and this implies that the series 00:29:18.230 --> 00:29:22.220 align:middle line:90% converges absolutely. 00:29:22.220 --> 00:29:30.870 align:middle line:90% 00:29:30.870 --> 00:29:42.070 align:middle line:84% And if l is bigger than 1, then this series converges-- 00:29:42.070 --> 00:29:43.345 align:middle line:90% no, the series diverges. 00:29:43.345 --> 00:29:49.240 align:middle line:90% 00:29:49.240 --> 00:29:52.000 align:middle line:84% And again, just like in the ratio test, 00:29:52.000 --> 00:29:55.510 align:middle line:90% no information for l equals 1. 00:29:55.510 --> 00:30:04.310 align:middle line:90% 00:30:04.310 --> 00:30:06.470 align:middle line:84% You take the same series that we looked at before. 00:30:06.470 --> 00:30:10.140 align:middle line:90% 00:30:10.140 --> 00:30:11.140 align:middle line:90% Did I leave it up there? 00:30:11.140 --> 00:30:11.980 align:middle line:90% Yes. 00:30:11.980 --> 00:30:15.280 align:middle line:84% So xn-- x sub n equals 1 for all n. 00:30:15.280 --> 00:30:18.730 align:middle line:90% This limit here, l, exists. 00:30:18.730 --> 00:30:19.600 align:middle line:90% It's equal to 1. 00:30:19.600 --> 00:30:21.320 align:middle line:90% And that series diverges. 00:30:21.320 --> 00:30:27.790 align:middle line:84% If you look at 1 over n squared, and I take this limit, 00:30:27.790 --> 00:30:28.960 align:middle line:90% I again get 1. 00:30:28.960 --> 00:30:30.640 align:middle line:90% But that series converges. 00:30:30.640 --> 00:30:35.747 align:middle line:84% So for capital l equals 1, we get no information. 00:30:35.747 --> 00:30:47.440 align:middle line:90% 00:30:47.440 --> 00:30:53.340 align:middle line:84% So let's-- and in this case, it's even clearer how 00:30:53.340 --> 00:30:57.880 align:middle line:84% we're relating the series to a geometric series. 00:30:57.880 --> 00:31:05.670 align:middle line:84% So let's-- in my notes, I proved two first both times. 00:31:05.670 --> 00:31:11.340 align:middle line:84% So perhaps I should have written it one, two. 00:31:11.340 --> 00:31:12.820 align:middle line:90% I'll know for the future. 00:31:12.820 --> 00:31:19.650 align:middle line:90% So suppose l is bigger than 1. 00:31:19.650 --> 00:31:22.050 align:middle line:84% And now, we're going to show that this series diverges, 00:31:22.050 --> 00:31:25.080 align:middle line:84% again, by showing that the terms do not converge to 0. 00:31:25.080 --> 00:31:29.800 align:middle line:90% So it's the same idea as before. 00:31:29.800 --> 00:31:32.430 align:middle line:90% Here's l, here's 1. 00:31:32.430 --> 00:31:39.390 align:middle line:84% So for all n sufficiently large, x to the n to the 1 over n 00:31:39.390 --> 00:31:41.220 align:middle line:90% has to be inside this interval. 00:31:41.220 --> 00:31:46.180 align:middle line:90% 00:31:46.180 --> 00:31:52.200 align:middle line:84% And since x to the n, x sub n to the 1 over n converges 00:31:52.200 --> 00:31:54.530 align:middle line:84% to l, which is bigger than 1, this 00:31:54.530 --> 00:31:58.260 align:middle line:84% implies there exists an integer in 0 00:31:58.260 --> 00:32:05.600 align:middle line:84% such that for all n bigger than or equal to M0, x 00:32:05.600 --> 00:32:12.430 align:middle line:84% sub n to the 1 over n is bigger than 1, which implies 00:32:12.430 --> 00:32:17.160 align:middle line:84% all n bigger than or equal to M0, x sub n in absolute value 00:32:17.160 --> 00:32:20.880 align:middle line:84% is bigger than by just taking powers of both sides 00:32:20.880 --> 00:32:23.700 align:middle line:90% is bigger than 1. 00:32:23.700 --> 00:32:27.390 align:middle line:84% Now, since all of the absolute values of x sub n's is bigger 00:32:27.390 --> 00:32:33.650 align:middle line:84% than 1, this implies that this cannot converge a 0. 00:32:33.650 --> 00:32:43.270 align:middle line:90% 00:32:43.270 --> 00:32:48.500 align:middle line:84% So remember, I mean, let me remove this absolute value. 00:32:48.500 --> 00:32:50.190 align:middle line:90% x sub n cannot converge to 0. 00:32:50.190 --> 00:32:50.690 align:middle line:90% Why? 00:32:50.690 --> 00:32:53.570 align:middle line:84% I mean, we could go back to the basic definition 00:32:53.570 --> 00:32:57.260 align:middle line:84% of convergence, of what it means for a sequence 00:32:57.260 --> 00:33:02.480 align:middle line:84% to converge to a real number x, and what it doesn't mean, 00:33:02.480 --> 00:33:05.000 align:middle line:84% or what it means for it to not converge to x. 00:33:05.000 --> 00:33:09.660 align:middle line:84% So xn does not converge to 0 if there exists a bad epsilon. 00:33:09.660 --> 00:33:13.040 align:middle line:84% So that x sub n is outside of that interval. 00:33:13.040 --> 00:33:15.640 align:middle line:90% 00:33:15.640 --> 00:33:22.020 align:middle line:84% As long as is outside of that interval 00:33:22.020 --> 00:33:23.325 align:middle line:90% if I go far enough out. 00:33:23.325 --> 00:33:27.310 align:middle line:90% 00:33:27.310 --> 00:33:29.380 align:middle line:90% And we certainly have that here. 00:33:29.380 --> 00:33:32.960 align:middle line:90% 00:33:32.960 --> 00:33:36.150 align:middle line:90% Or you could use-- 00:33:36.150 --> 00:33:40.460 align:middle line:84% so this bigger than 1, in fact, implies 00:33:40.460 --> 00:33:48.700 align:middle line:84% that the lim sup of x of n is bigger than 1. 00:33:48.700 --> 00:33:51.130 align:middle line:84% Because if I have two sequences, one bigger 00:33:51.130 --> 00:33:54.790 align:middle line:84% than the other, then the lim sup of the bigger one 00:33:54.790 --> 00:33:56.710 align:middle line:84% is bigger than or equal to the lim sup 00:33:56.710 --> 00:33:59.260 align:middle line:90% of the other one, which is 1. 00:33:59.260 --> 00:34:09.719 align:middle line:84% And you're doing in this week's assignment, 00:34:09.719 --> 00:34:13.350 align:middle line:84% that this converges-- that xn converges to 0 if 00:34:13.350 --> 00:34:16.409 align:middle line:84% and only if the lim sup of the absolute values of xn's 00:34:16.409 --> 00:34:17.860 align:middle line:90% converge to 0. 00:34:17.860 --> 00:34:20.489 align:middle line:90% And this is not. 00:34:20.489 --> 00:34:22.020 align:middle line:90% It's bigger than or equal to 1. 00:34:22.020 --> 00:34:25.135 align:middle line:90% 00:34:25.135 --> 00:34:26.510 align:middle line:90% So we'll leave that there. 00:34:26.510 --> 00:34:37.090 align:middle line:90% 00:34:37.090 --> 00:34:43.870 align:middle line:84% So now, for the other case, that l is less than 1. 00:34:43.870 --> 00:34:46.900 align:middle line:90% 00:34:46.900 --> 00:34:56.460 align:middle line:84% Suppose l is less than 1, then let alpha 00:34:56.460 --> 00:35:00.240 align:middle line:90% be a number between l and 1. 00:35:00.240 --> 00:35:04.080 align:middle line:90% 00:35:04.080 --> 00:35:10.350 align:middle line:84% Again, since this converges to l, which is less than alpha, 00:35:10.350 --> 00:35:14.110 align:middle line:84% for all n sufficiently large, this has to be less than alpha. 00:35:14.110 --> 00:35:26.590 align:middle line:84% So there exists an integer, I have x to the n to the 1 00:35:26.590 --> 00:35:33.790 align:middle line:84% over n is less than alpha, which implies 00:35:33.790 --> 00:35:40.380 align:middle line:84% that for all n bigger than or equal to M0, 00:35:40.380 --> 00:35:45.690 align:middle line:84% x to the n in absolute value is less than alpha to the n. 00:35:45.690 --> 00:35:50.970 align:middle line:84% So here again, we're seeing this series, 00:35:50.970 --> 00:35:53.880 align:middle line:84% if it satisfies these hypotheses, 00:35:53.880 --> 00:35:58.200 align:middle line:84% is very much acting like a geometric series. 00:35:58.200 --> 00:35:59.730 align:middle line:90% Remember, alpha's less than 1. 00:35:59.730 --> 00:36:06.040 align:middle line:90% 00:36:06.040 --> 00:36:08.680 align:middle line:84% Then for every natural number little m, 00:36:08.680 --> 00:36:14.146 align:middle line:84% if we look at the nth partial sum of the absolute value, 00:36:14.146 --> 00:36:17.070 align:middle line:90% let's say n equals 1. 00:36:17.070 --> 00:36:20.430 align:middle line:84% This is we split it up again into a part 00:36:20.430 --> 00:36:32.650 align:middle line:84% that we don't really care about, plus an interesting part. 00:36:32.650 --> 00:36:35.837 align:middle line:84% Remember, little m is the thing that's changing. 00:36:35.837 --> 00:36:38.170 align:middle line:84% This should look like a little m compared to capital M0. 00:36:38.170 --> 00:36:41.060 align:middle line:90% 00:36:41.060 --> 00:36:46.340 align:middle line:84% And this is less than or equal to sum from n equals 1 to M0, 00:36:46.340 --> 00:36:47.030 align:middle line:90% x sub n. 00:36:47.030 --> 00:36:49.130 align:middle line:84% Again, this is just a fixed number. 00:36:49.130 --> 00:36:57.980 align:middle line:84% Plus now we can put this inequality, n equals M0 plus 1, 00:36:57.980 --> 00:36:59.960 align:middle line:90% m alpha to the n. 00:36:59.960 --> 00:37:03.460 align:middle line:90% 00:37:03.460 --> 00:37:10.552 align:middle line:84% And this is less than or equal to sum from n equals M0. 00:37:10.552 --> 00:37:14.470 align:middle line:84% I'm going to go kind of fast here, because I'm running out 00:37:14.470 --> 00:37:17.020 align:middle line:90% of space on the board here. 00:37:17.020 --> 00:37:19.130 align:middle line:84% Maybe I'll write this out in just a minute. 00:37:19.130 --> 00:37:21.340 align:middle line:90% But why this is true. 00:37:21.340 --> 00:37:25.960 align:middle line:84% And this part here is less than or equal to 1 over 1 00:37:25.960 --> 00:37:28.420 align:middle line:90% minus alpha. 00:37:28.420 --> 00:37:29.830 align:middle line:90% Because what do I do? 00:37:29.830 --> 00:37:31.435 align:middle line:90% This sum is a finite sum. 00:37:31.435 --> 00:37:38.900 align:middle line:90% 00:37:38.900 --> 00:37:41.780 align:middle line:84% And it's certainly bigger than if I 00:37:41.780 --> 00:37:46.790 align:middle line:84% make the lower bound smaller and the upper bound larger. 00:37:46.790 --> 00:37:49.840 align:middle line:90% 00:37:49.840 --> 00:37:52.210 align:middle line:84% And this is just the nth partial sum corresponding 00:37:52.210 --> 00:37:54.460 align:middle line:84% to a geometric series with non-negative terms, 00:37:54.460 --> 00:37:56.000 align:middle line:90% the alpha to the n. 00:37:56.000 --> 00:37:58.270 align:middle line:84% And so, that's less than or equal to sum from n 00:37:58.270 --> 00:38:03.460 align:middle line:84% equals 0 to infinity of alpha to the n, which equals 1 over 1 00:38:03.460 --> 00:38:04.210 align:middle line:90% minus alpha. 00:38:04.210 --> 00:38:07.718 align:middle line:90% 00:38:07.718 --> 00:38:09.010 align:middle line:90% So that's how we got this term. 00:38:09.010 --> 00:38:11.370 align:middle line:84% And therefore, the partial sums corresponding 00:38:11.370 --> 00:38:16.440 align:middle line:84% to the series with absolute values is bounded. 00:38:16.440 --> 00:38:21.120 align:middle line:84% And therefore, that series with the absolute values converges. 00:38:21.120 --> 00:38:22.920 align:middle line:84% And we have absolute convergence. 00:38:22.920 --> 00:38:28.960 align:middle line:90% 00:38:28.960 --> 00:38:35.920 align:middle line:84% So now, let me state a theorem about alternating series. 00:38:35.920 --> 00:38:37.000 align:middle line:90% It is not-- 00:38:37.000 --> 00:38:40.180 align:middle line:84% I prefer not to call it an alternating series test, 00:38:40.180 --> 00:38:43.660 align:middle line:84% because there's nothing really to test. 00:38:43.660 --> 00:38:46.180 align:middle line:84% I mean, at least with the ratio and root test, 00:38:46.180 --> 00:38:49.240 align:middle line:84% you have to recompute a limit, which 00:38:49.240 --> 00:38:50.815 align:middle line:90% might require some work to do. 00:38:50.815 --> 00:39:02.920 align:middle line:90% 00:39:02.920 --> 00:39:06.370 align:middle line:84% And therefore, at least to me, that's kind of a real thing, 00:39:06.370 --> 00:39:09.820 align:middle line:84% that you have to do a little work to test 00:39:09.820 --> 00:39:11.590 align:middle line:90% whether a series converges. 00:39:11.590 --> 00:39:17.000 align:middle line:84% And for alternating series, the test is, you look at it. 00:39:17.000 --> 00:39:19.200 align:middle line:90% And that's it. 00:39:19.200 --> 00:39:20.820 align:middle line:90% You don't compute anything. 00:39:20.820 --> 00:39:22.110 align:middle line:90% You look at it. 00:39:22.110 --> 00:39:25.560 align:middle line:84% So I prefer not to call this theorem 00:39:25.560 --> 00:39:28.230 align:middle line:84% about alternating series an alternating series test. 00:39:28.230 --> 00:39:40.480 align:middle line:90% 00:39:40.480 --> 00:39:42.945 align:middle line:84% So the theorem is, for alternating series. 00:39:42.945 --> 00:39:46.440 align:middle line:90% 00:39:46.440 --> 00:39:47.970 align:middle line:84% And the statement is the following. 00:39:47.970 --> 00:39:52.990 align:middle line:90% 00:39:52.990 --> 00:40:03.870 align:middle line:84% Let x sub n be a monotone decreasing 00:40:03.870 --> 00:40:13.710 align:middle line:90% sequence converging to 0. 00:40:13.710 --> 00:40:16.700 align:middle line:90% 00:40:16.700 --> 00:40:19.340 align:middle line:84% So because this thing is monotone, and monotone 00:40:19.340 --> 00:40:23.900 align:middle line:84% decreasing and converging to 0, it's all-- 00:40:23.900 --> 00:40:26.400 align:middle line:90% so let me-- 00:40:26.400 --> 00:40:28.320 align:middle line:90% I'll put this in parentheses. 00:40:28.320 --> 00:40:33.200 align:middle line:90% 00:40:33.200 --> 00:40:36.410 align:middle line:84% Therefore xn is bigger than or equal to 0 for all n. 00:40:36.410 --> 00:40:44.600 align:middle line:90% 00:40:44.600 --> 00:40:47.660 align:middle line:84% I cannot have a monotone decreasing sequence converging 00:40:47.660 --> 00:40:49.678 align:middle line:84% to 0 if one of the xn's is less than 0, 00:40:49.678 --> 00:40:51.095 align:middle line:84% because they keep getting smaller. 00:40:51.095 --> 00:40:57.670 align:middle line:90% 00:40:57.670 --> 00:40:59.410 align:middle line:90% This is it. 00:40:59.410 --> 00:41:01.990 align:middle line:84% This is not-- this is all of the hypotheses. 00:41:01.990 --> 00:41:04.120 align:middle line:84% You don't have to compute anything. 00:41:04.120 --> 00:41:12.490 align:middle line:84% Then the series minus 1 to the n, xn, n equals-- 00:41:12.490 --> 00:41:19.090 align:middle line:84% let's of course, we don't have to start at 1 in particular. 00:41:19.090 --> 00:41:23.020 align:middle line:84% But at least for this statement, let's make it precise. 00:41:23.020 --> 00:41:25.540 align:middle line:84% Sum from n equals 1 to infinity of minus 1 to the n, 00:41:25.540 --> 00:41:28.870 align:middle line:90% x to the n-- x sub n converges. 00:41:28.870 --> 00:41:31.590 align:middle line:90% 00:41:31.590 --> 00:41:33.690 align:middle line:84% And we can just say convergence, not necessarily 00:41:33.690 --> 00:41:34.800 align:middle line:90% absolute convergence. 00:41:34.800 --> 00:41:37.770 align:middle line:84% Because again, if we look at minus 1 to the n, 00:41:37.770 --> 00:41:41.850 align:middle line:84% 1 over n, which is 1 over n is a monotone decreasing sequence 00:41:41.850 --> 00:41:43.050 align:middle line:90% converging to 0. 00:41:43.050 --> 00:41:45.120 align:middle line:84% That converges, but not absolutely. 00:41:45.120 --> 00:41:48.380 align:middle line:90% 00:41:48.380 --> 00:41:51.530 align:middle line:84% And if we have x sub n equal 1 over n squared, 00:41:51.530 --> 00:41:54.950 align:middle line:84% again, 1 over n squared is a monotone decreasing sequence 00:41:54.950 --> 00:41:56.030 align:middle line:90% converging to 0. 00:41:56.030 --> 00:41:57.530 align:middle line:90% That would converge absolutely. 00:41:57.530 --> 00:41:59.780 align:middle line:84% So we just have a statement about convergence. 00:41:59.780 --> 00:42:07.870 align:middle line:90% 00:42:07.870 --> 00:42:18.260 align:middle line:84% So how we're going to do this is, we are-- 00:42:18.260 --> 00:42:21.440 align:middle line:84% it's kind of like how we proved convergence 00:42:21.440 --> 00:42:27.080 align:middle line:84% for p series, in that we're going 00:42:27.080 --> 00:42:31.790 align:middle line:84% to show that a certain subsequence of partial sums 00:42:31.790 --> 00:42:32.720 align:middle line:90% converges. 00:42:32.720 --> 00:42:36.110 align:middle line:84% And then, we're going to use that to show 00:42:36.110 --> 00:42:38.990 align:middle line:84% that the full sequence of partial sums converge. 00:42:38.990 --> 00:42:45.630 align:middle line:90% 00:42:45.630 --> 00:42:52.080 align:middle line:84% So let me state this as claim one. 00:42:52.080 --> 00:43:02.160 align:middle line:84% So the subsequence of partial sums, 00:43:02.160 --> 00:43:09.625 align:middle line:84% S sub 2k, so this is just the sum from n equals 1 to 2k. 00:43:09.625 --> 00:43:18.890 align:middle line:90% 00:43:18.890 --> 00:43:23.140 align:middle line:84% We're going to show this converges. 00:43:23.140 --> 00:43:28.600 align:middle line:84% So again, just to be complete, S sub m, 00:43:28.600 --> 00:43:30.910 align:middle line:84% this is the nth partial sum. it is the sum from m 00:43:30.910 --> 00:43:32.800 align:middle line:90% equals 1 to m. 00:43:32.800 --> 00:43:45.315 align:middle line:90% 00:43:45.315 --> 00:43:46.690 align:middle line:84% So how we're going to do this is, 00:43:46.690 --> 00:43:51.220 align:middle line:84% we're going to show that these partial sums are 00:43:51.220 --> 00:43:55.030 align:middle line:84% in fact monotone decreasing and bounded from below. 00:43:55.030 --> 00:44:00.940 align:middle line:84% Monotone decreasing, basically because these guys 00:44:00.940 --> 00:44:02.710 align:middle line:90% are monotone decreasing. 00:44:02.710 --> 00:44:10.070 align:middle line:84% And bounded from below by the same reasoning. 00:44:10.070 --> 00:44:12.620 align:middle line:90% So let's show that. 00:44:12.620 --> 00:44:20.440 align:middle line:90% 00:44:20.440 --> 00:44:31.340 align:middle line:84% For k, a natural number, if I look at S 2k, so this is a sum. 00:44:31.340 --> 00:44:38.320 align:middle line:84% It equals 1 2k, minus 1 to the n x sub n. 00:44:38.320 --> 00:44:46.110 align:middle line:84% Now I have the n equals 1 term, which I can write as minus x-- 00:44:46.110 --> 00:44:50.170 align:middle line:84% so just writing this out in a certain way. 00:44:50.170 --> 00:44:56.760 align:middle line:84% So this is minus x1 plus x2 minus x3 plus-- 00:44:56.760 --> 00:44:59.430 align:middle line:90% and then n equals 2k is even. 00:44:59.430 --> 00:45:01.860 align:middle line:90% So then I get plus x 2k. 00:45:01.860 --> 00:45:04.880 align:middle line:90% 00:45:04.880 --> 00:45:07.910 align:middle line:84% This is equal to x sub 2 minus x sub 1, 00:45:07.910 --> 00:45:16.880 align:middle line:84% plus x sub 4 minus x sub 3, plus all the way to x sub 00:45:16.880 --> 00:45:20.810 align:middle line:90% 2k minus x sub 2k, minus 1. 00:45:20.810 --> 00:45:27.010 align:middle line:90% 00:45:27.010 --> 00:45:31.080 align:middle line:84% Now, this sequence is monotone decreasing. 00:45:31.080 --> 00:45:36.330 align:middle line:84% So x sub 2 going to be smaller than x sub 1. 00:45:36.330 --> 00:45:40.890 align:middle line:84% x sub 4 is going to be less than x sub 3. 00:45:40.890 --> 00:45:45.750 align:middle line:84% x sub 2k minus x sub 2k minus 1 is less than 0. 00:45:45.750 --> 00:45:49.800 align:middle line:84% So since this is monotone decreasing, 00:45:49.800 --> 00:45:52.605 align:middle line:90% this is bigger than or equal to. 00:45:52.605 --> 00:45:57.330 align:middle line:90% 00:45:57.330 --> 00:46:01.110 align:middle line:84% So what I just said has nothing to do with this inequality 00:46:01.110 --> 00:46:03.870 align:middle line:90% I'm writing here-- 00:46:03.870 --> 00:46:04.370 align:middle line:90% yet. 00:46:04.370 --> 00:46:07.960 align:middle line:90% 00:46:07.960 --> 00:46:09.610 align:middle line:90% So I kept everything here. 00:46:09.610 --> 00:46:18.490 align:middle line:84% And now I add x sub 2k plus 2 minus x sub 2k plus 1. 00:46:18.490 --> 00:46:22.540 align:middle line:90% 00:46:22.540 --> 00:46:25.465 align:middle line:84% So what I was saying was that the x sub n's 00:46:25.465 --> 00:46:26.890 align:middle line:90% are monotone decreasing. 00:46:26.890 --> 00:46:30.560 align:middle line:84% This term is right after this term. 00:46:30.560 --> 00:46:33.410 align:middle line:90% So it's smaller than this term. 00:46:33.410 --> 00:46:37.990 align:middle line:84% So this term here is less than or equal to 0. 00:46:37.990 --> 00:46:42.910 align:middle line:90% 00:46:42.910 --> 00:46:46.990 align:middle line:84% And therefore, this thing plus itself plus something 00:46:46.990 --> 00:46:49.990 align:middle line:84% that's negative is less than or equal to this thing. 00:46:49.990 --> 00:47:01.150 align:middle line:90% 00:47:01.150 --> 00:47:03.920 align:middle line:84% Now, what is this thing on the right hand side? 00:47:03.920 --> 00:47:21.730 align:middle line:84% This is just a x 2k minus plus x 2k, minus x 2k plus 1, 00:47:21.730 --> 00:47:24.940 align:middle line:90% plus x sub 2, k plus 1. 00:47:24.940 --> 00:47:32.230 align:middle line:84% And this is the k plus 1 2k plus 1 partial sum. 00:47:32.230 --> 00:47:43.270 align:middle line:90% 00:47:43.270 --> 00:47:57.050 align:middle line:84% Thus this series is monotone decreasing. 00:47:57.050 --> 00:48:06.910 align:middle line:90% 00:48:06.910 --> 00:48:10.270 align:middle line:84% So to show that the subsequence of partial sums converges, 00:48:10.270 --> 00:48:14.350 align:middle line:84% we just have to show that this subsequence is 00:48:14.350 --> 00:48:18.460 align:middle line:90% bounded, or bounded below. 00:48:18.460 --> 00:48:20.530 align:middle line:84% If it's monotone decreasing, it's already 00:48:20.530 --> 00:48:22.640 align:middle line:90% bounded above by something. 00:48:22.640 --> 00:48:24.640 align:middle line:84% So we just need to show it's bounded from below. 00:48:24.640 --> 00:48:27.970 align:middle line:90% 00:48:27.970 --> 00:48:37.760 align:middle line:84% Now, for all k, a natural number, let's look at S 00:48:37.760 --> 00:48:44.080 align:middle line:84% sub 2k and group term slightly differently. 00:48:44.080 --> 00:48:46.060 align:middle line:90% So this is equal now to-- 00:48:46.060 --> 00:48:53.370 align:middle line:84% I'm going to write it as minus x1 plus x sub 3. 00:48:53.370 --> 00:48:59.290 align:middle line:90% 00:48:59.290 --> 00:49:04.780 align:middle line:84% x sub 2 minus x sub 3, plus x sub 4 minus x sub 5, 00:49:04.780 --> 00:49:14.740 align:middle line:84% plus so on, plus x 2k minus x 2k minus 1. 00:49:14.740 --> 00:49:28.090 align:middle line:84% OK, I think-- ah, no, that's not right. 00:49:28.090 --> 00:49:31.600 align:middle line:90% 00:49:31.600 --> 00:49:37.490 align:middle line:90% So this is minus 2 plus x of 2k. 00:49:37.490 --> 00:49:37.990 align:middle line:90% All right. 00:49:37.990 --> 00:49:41.546 align:middle line:90% 00:49:41.546 --> 00:49:44.720 align:middle line:90% Now, this is right. 00:49:44.720 --> 00:49:46.760 align:middle line:90% So this is equal to minus-- 00:49:46.760 --> 00:49:51.770 align:middle line:84% so again, the x sub n's are monotone decreasing. 00:49:51.770 --> 00:49:56.240 align:middle line:84% So x sub 3 is smaller than x sub 2. 00:49:56.240 --> 00:49:57.830 align:middle line:90% So this is non-negative. 00:49:57.830 --> 00:50:00.400 align:middle line:90% 00:50:00.400 --> 00:50:04.480 align:middle line:84% This is also non-negative, because x sub 5 is smaller 00:50:04.480 --> 00:50:07.160 align:middle line:90% than x sub 4, so on and so on. 00:50:07.160 --> 00:50:08.980 align:middle line:90% So this is also non-negative. 00:50:08.980 --> 00:50:15.640 align:middle line:84% So S sub 2k is bigger than or equal to minus x sub 1. 00:50:15.640 --> 00:50:20.560 align:middle line:90% Plus non-negative terms, so 2k. 00:50:20.560 --> 00:50:24.610 align:middle line:84% And remember, x sub n's, these are monotone 00:50:24.610 --> 00:50:26.110 align:middle line:90% decreasing converging to 0. 00:50:26.110 --> 00:50:28.270 align:middle line:84% And therefore, that they're all non-negative. 00:50:28.270 --> 00:50:32.020 align:middle line:84% So this is bigger than or equal to minus x sub 1. 00:50:32.020 --> 00:50:41.260 align:middle line:90% 00:50:41.260 --> 00:50:43.340 align:middle line:90% Thus, what do we get? 00:50:43.340 --> 00:50:49.780 align:middle line:84% So for all natural numbers k, I have minus x1 00:50:49.780 --> 00:50:51.670 align:middle line:84% is less than or equal to S sub 2k. 00:50:51.670 --> 00:50:53.710 align:middle line:84% And because these are monotone decreasing, 00:50:53.710 --> 00:50:57.740 align:middle line:84% this is always less than or equal to S k equals 1. 00:50:57.740 --> 00:51:01.860 align:middle line:90% 00:51:01.860 --> 00:51:07.440 align:middle line:84% And this is-- so if you like S sub 2, is just minus x sub 1 00:51:07.440 --> 00:51:09.210 align:middle line:90% plus x sub 2. 00:51:09.210 --> 00:51:12.540 align:middle line:84% So this sequence of numbers is bounded 00:51:12.540 --> 00:51:14.490 align:middle line:84% between this real number and this real number. 00:51:14.490 --> 00:51:16.440 align:middle line:90% And therefore, it's bounded. 00:51:16.440 --> 00:51:17.912 align:middle line:84% And since it's monotone decreasing, 00:51:17.912 --> 00:51:18.870 align:middle line:90% it has to have a limit. 00:51:18.870 --> 00:51:35.780 align:middle line:90% 00:51:35.780 --> 00:51:39.610 align:middle line:90% So S sub 2k converges. 00:51:39.610 --> 00:51:47.730 align:middle line:90% 00:51:47.730 --> 00:51:50.780 align:middle line:90% Let's call this limit something. 00:51:50.780 --> 00:52:00.420 align:middle line:84% Let's call it S. And we'll show that the sequence 00:52:00.420 --> 00:52:02.370 align:middle line:84% of partial sums converge to S now. 00:52:02.370 --> 00:52:05.280 align:middle line:84% We just showed this for along a sequence of-- 00:52:05.280 --> 00:52:07.080 align:middle line:84% along a subsequence of partial sums. 00:52:07.080 --> 00:52:09.750 align:middle line:84% Let's show it converges to S along the full sequence. 00:52:09.750 --> 00:52:18.080 align:middle line:90% 00:52:18.080 --> 00:52:23.700 align:middle line:84% So this is, if you like, claim 2, 00:52:23.700 --> 00:52:36.710 align:middle line:84% which is that the full sequence of partial sums converges to S. 00:52:36.710 --> 00:52:42.450 align:middle line:84% And we're going to do this by brute force using epsilon M 00:52:42.450 --> 00:52:44.760 align:middle line:90% argument. 00:52:44.760 --> 00:52:46.840 align:middle line:90% So remember, to show something-- 00:52:46.840 --> 00:52:50.260 align:middle line:84% a limit of something equals S means for all epsilon, 00:52:50.260 --> 00:52:53.220 align:middle line:84% there exists a capital M, so that for all M bigger than 00:52:53.220 --> 00:52:57.120 align:middle line:84% or equal to capital M, S sub M minus S in absolute value 00:52:57.120 --> 00:52:58.830 align:middle line:90% is less than epsilon. 00:52:58.830 --> 00:53:00.600 align:middle line:90% So let epsilon be positive. 00:53:00.600 --> 00:53:04.710 align:middle line:90% 00:53:04.710 --> 00:53:13.410 align:middle line:84% Since S sub 2k convergence to S, there exists M0 natural number. 00:53:13.410 --> 00:53:25.280 align:middle line:84% So that all k bigger than or equal to M0 S sub 2k 00:53:25.280 --> 00:53:28.025 align:middle line:84% minus S in absolute value is less than epsilon. 00:53:28.025 --> 00:53:33.800 align:middle line:90% 00:53:33.800 --> 00:53:37.370 align:middle line:84% Now, we haven't used at all in this proof 00:53:37.370 --> 00:53:39.230 align:middle line:84% that the xn's really converge to 0. 00:53:39.230 --> 00:53:41.810 align:middle line:84% We did use that they were non-negative at one point. 00:53:41.810 --> 00:53:44.300 align:middle line:84% But now, this is where we'll use a conversion to 0, 00:53:44.300 --> 00:53:48.110 align:middle line:90% because here's the intuition. 00:53:48.110 --> 00:53:52.890 align:middle line:84% The S sub 2k's are converging to something. 00:53:52.890 --> 00:53:55.500 align:middle line:84% So now, I just have to look at the odd ones. 00:53:55.500 --> 00:54:00.240 align:middle line:84% I've shown all the even partial sums converge to S. 00:54:00.240 --> 00:54:04.260 align:middle line:84% So if I can show the odd partial sums also converge to S, 00:54:04.260 --> 00:54:05.580 align:middle line:90% then essentially, I'm done. 00:54:05.580 --> 00:54:07.920 align:middle line:84% What's the difference between an even partial sum 00:54:07.920 --> 00:54:09.540 align:middle line:90% and an odd partial sum? 00:54:09.540 --> 00:54:14.700 align:middle line:84% Well, it's just x to the 2k plus x sub 2k plus 1, 00:54:14.700 --> 00:54:16.080 align:middle line:90% which is converging to 0. 00:54:16.080 --> 00:54:20.280 align:middle line:90% So they don't differ by much. 00:54:20.280 --> 00:54:26.680 align:middle line:84% And that's essentially the whole argument right there. 00:54:26.680 --> 00:54:35.300 align:middle line:84% So since xn's converge to 0, there exists natural number M 00:54:35.300 --> 00:54:43.030 align:middle line:84% sub 1 so that for all n bigger than or equal to M sub 1, 00:54:43.030 --> 00:54:45.860 align:middle line:84% x of n is less than epsilon over 2. 00:54:45.860 --> 00:54:47.110 align:middle line:90% I should have written Epsilon. 00:54:47.110 --> 00:54:47.950 align:middle line:90% Over 2 here. 00:54:47.950 --> 00:54:55.070 align:middle line:90% 00:54:55.070 --> 00:55:01.850 align:middle line:84% Choose M to be the maximum of two numbers 2M0 0 00:55:01.850 --> 00:55:04.700 align:middle line:90% plus 1 and M1. 00:55:04.700 --> 00:55:07.265 align:middle line:84% So you'll see why I made these choices in just a minute. 00:55:07.265 --> 00:55:11.740 align:middle line:90% 00:55:11.740 --> 00:55:15.820 align:middle line:84% So suppose m is bigger than or equal to M. 00:55:15.820 --> 00:55:20.350 align:middle line:84% We now want to show that S sub m minus S is less than epsilon 00:55:20.350 --> 00:55:23.630 align:middle line:90% in absolute value. 00:55:23.630 --> 00:55:24.610 align:middle line:90% So there's two cases. 00:55:24.610 --> 00:55:33.560 align:middle line:84% If m is even, then m over 2 is bigger than 00:55:33.560 --> 00:55:38.360 align:middle line:84% or equal to capital M over 2, which is the max of these. 00:55:38.360 --> 00:55:40.250 align:middle line:84% If I divide by 2, that's certainly 00:55:40.250 --> 00:55:44.030 align:middle line:90% bigger than or equal to M0. 00:55:44.030 --> 00:55:50.360 align:middle line:84% And therefore, I use this first inequality, S sub m minus s, 00:55:50.360 --> 00:55:57.440 align:middle line:84% which is equal to S2 times m over 2, minus S. Now, m over 2 00:55:57.440 --> 00:56:01.240 align:middle line:84% is an integer bigger than or equal to m sub 0. 00:56:01.240 --> 00:56:02.963 align:middle line:90% So I can use this inequality. 00:56:02.963 --> 00:56:05.380 align:middle line:84% It's less than epsilon over 2, which is less than epsilon. 00:56:05.380 --> 00:56:08.920 align:middle line:90% 00:56:08.920 --> 00:56:13.700 align:middle line:90% And now, we do odd. 00:56:13.700 --> 00:56:28.578 align:middle line:90% 00:56:28.578 --> 00:56:30.620 align:middle line:84% So again, m is bigger than or equal to capital M. 00:56:30.620 --> 00:56:32.630 align:middle line:84% And there's two cases, even and odd. 00:56:32.630 --> 00:56:38.900 align:middle line:84% If m is odd, let k be this integer now. 00:56:38.900 --> 00:56:41.570 align:middle line:84% Since m is odd, m minus 1 is even, 00:56:41.570 --> 00:56:42.980 align:middle line:90% divided by 2, that's an integer. 00:56:42.980 --> 00:56:47.470 align:middle line:90% 00:56:47.470 --> 00:56:49.300 align:middle line:90% And so, this is-- 00:56:49.300 --> 00:56:52.390 align:middle line:90% 00:56:52.390 --> 00:56:56.200 align:middle line:90% so m equals 2k plus 1. 00:56:56.200 --> 00:57:03.840 align:middle line:90% 00:57:03.840 --> 00:57:10.600 align:middle line:84% And since m is bigger than or equal to M0, 00:57:10.600 --> 00:57:12.430 align:middle line:90% this implies a couple of things. 00:57:12.430 --> 00:57:14.320 align:middle line:90% This implies that this integer-- 00:57:14.320 --> 00:57:20.500 align:middle line:84% so our m, this implies that 2k plus 1 00:57:20.500 --> 00:57:21.970 align:middle line:90% is bigger than or equal to 2. 00:57:21.970 --> 00:57:24.700 align:middle line:90% 00:57:24.700 --> 00:57:26.810 align:middle line:90% So m is just equal to 2k plus 1. 00:57:26.810 --> 00:57:28.310 align:middle line:84% So that's bigger than or equal to m, 00:57:28.310 --> 00:57:31.480 align:middle line:84% which is bigger than or equal to 2 m sub 0 plus 1. 00:57:31.480 --> 00:57:34.950 align:middle line:90% 00:57:34.950 --> 00:57:38.940 align:middle line:84% And therefore, k, this integer here, is bigger than M0. 00:57:38.940 --> 00:57:44.900 align:middle line:90% 00:57:44.900 --> 00:57:52.250 align:middle line:84% Also, m is bigger than or equal to m sub 1 as well. 00:57:52.250 --> 00:57:53.960 align:middle line:84% I mean, m is bigger than or equal to m. 00:57:53.960 --> 00:57:56.640 align:middle line:84% And m is bigger than or equal to the max of these two things. 00:57:56.640 --> 00:57:58.160 align:middle line:84% So it's bigger than or equal to 1. 00:57:58.160 --> 00:58:02.630 align:middle line:90% 00:58:02.630 --> 00:58:11.080 align:middle line:84% Then if I look at S of n minus S, this is equal to S sub m-- 00:58:11.080 --> 00:58:15.880 align:middle line:84% so the nth partial sum is equal to the m 00:58:15.880 --> 00:58:29.420 align:middle line:84% minus first partial sum plus the next term minus S. Now 00:58:29.420 --> 00:58:32.030 align:middle line:84% m minus 1, in terms of the integer k, 00:58:32.030 --> 00:58:34.580 align:middle line:90% is equal to 2 times k. 00:58:34.580 --> 00:58:41.130 align:middle line:90% 00:58:41.130 --> 00:58:44.400 align:middle line:84% And now, I'm going to take this S and group it with this guy. 00:58:44.400 --> 00:58:55.130 align:middle line:90% 00:58:55.130 --> 00:59:01.720 align:middle line:84% And so, since k is bigger than or equal to m sub 0, 00:59:01.720 --> 00:59:06.160 align:middle line:84% we can use this inequality here after we 00:59:06.160 --> 00:59:07.640 align:middle line:90% do the triangle inequality. 00:59:07.640 --> 00:59:09.520 align:middle line:84% So this is less than or equal to S 2k 00:59:09.520 --> 00:59:12.700 align:middle line:84% minus S, plus the absolute value of this thing. 00:59:12.700 --> 00:59:15.670 align:middle line:90% 00:59:15.670 --> 00:59:19.240 align:middle line:84% So again, since k is bigger than or equal to m sub 0, 00:59:19.240 --> 00:59:24.070 align:middle line:84% I can use this inequality to get this is less than epsilon 00:59:24.070 --> 00:59:29.530 align:middle line:90% over 2, plus this guy. 00:59:29.530 --> 00:59:32.950 align:middle line:84% And since m is bigger than or equal to, capital M, which 00:59:32.950 --> 00:59:35.800 align:middle line:84% is bigger than or equal to capital M sub 1, 00:59:35.800 --> 00:59:37.480 align:middle line:90% I can use this inequality. 00:59:37.480 --> 00:59:41.930 align:middle line:90% 00:59:41.930 --> 00:59:45.130 align:middle line:90% And this is less than epsilon. 00:59:45.130 --> 00:59:51.240 align:middle line:84% So we've done the case of m even or m odd, m bigger than 00:59:51.240 --> 00:59:51.960 align:middle line:90% or equal to m. 00:59:51.960 --> 00:59:57.430 align:middle line:84% So in summary, we've shown that if m 00:59:57.430 --> 01:00:00.880 align:middle line:84% is bigger than or equal to m, S sub m minus S 01:00:00.880 --> 01:00:02.570 align:middle line:90% is less than epsilon. 01:00:02.570 --> 01:00:07.210 align:middle line:84% And therefore, the S sub n's converge to S. 01:00:07.210 --> 01:00:11.560 align:middle line:84% And I don't think I have enough time to do this next theorem. 01:00:11.560 --> 01:00:14.490 align:middle line:90% So we'll stop there. 01:00:14.490 --> 01:00:20.000 align:middle line:90%