WEBVTT
00:00:00.960 --> 00:00:07.910
In this video, we will learn how to determine if a series is absolutely convergent, conditionally convergent, or divergent.
00:00:08.770 --> 00:00:14.700
We know that if a series is called convergent, it means that the partial sums approach a specific limit.
00:00:15.070 --> 00:00:19.430
But what does it mean for a series to be called absolutely convergent?
00:00:20.340 --> 00:00:27.020
We say a series is called absolutely convergent if the series of absolute values is convergent.
00:00:28.040 --> 00:00:35.780
One thing to notice is that if π π is a series with positive terms, then the absolute value of π π is equal to π π.
00:00:36.120 --> 00:00:39.780
So absolute convergence implies convergence.
00:00:40.480 --> 00:00:46.040
First, letβs look at an example where we have to determine whether a series is absolutely convergent.
00:00:46.710 --> 00:00:54.820
Is the series the sum from π equals one to β of negative one to the power of π add one over π squared absolutely convergent?
00:00:55.330 --> 00:01:11.200
Remember to test for absolute convergence, we need to check whether the series of absolute values is convergent, in other words, is the sum from π equals one to β of the absolute value of negative one to the power of π add one over π squared convergent.
00:01:12.100 --> 00:01:20.660
First of all, notice that negative one to the power of π add one is always going to be either one or negative one, depending on whether the power is even or odd.
00:01:21.500 --> 00:01:27.900
So if we take the absolute value of negative one raised to the power of π add one, this is always going to be one.
00:01:28.820 --> 00:01:31.250
We know the π runs from one to β.
00:01:31.460 --> 00:01:34.070
So π squared is always going to be positive.
00:01:35.130 --> 00:01:39.080
So we can actually rewrite this as one over π squared.
00:01:39.850 --> 00:01:43.710
Remember that weβre trying to determine whether this converges or diverges.
00:01:44.460 --> 00:01:51.400
But we actually recognize the sum from π equals one to β of one over π squared to be a series that we know.
00:01:51.600 --> 00:01:52.880
Itβs a π-series.
00:01:53.530 --> 00:02:00.490
So we use the fact that a π-series converges if π is greater than one and diverges if π is less than or equal to one.
00:02:00.950 --> 00:02:05.640
So for our question, we see that this is a π-series with π equal to two.
00:02:06.600 --> 00:02:13.860
Because this is greater than one, we can say that the sum from π equals one to β of one over π squared is convergent.
00:02:15.100 --> 00:02:24.420
So because we found the series of absolute values to be convergent, then the series is absolutely convergent.
00:02:25.560 --> 00:02:32.620
Interestingly, if we find a series, which is not absolutely convergent, it may still be convergent.
00:02:33.190 --> 00:02:36.040
We call this conditional convergence.
00:02:36.860 --> 00:02:44.200
A series is conditionally convergent if the series is convergent but not absolutely convergent.
00:02:44.950 --> 00:02:50.660
In other words, the sum from π equals one to β of the absolute value of π π diverges.
00:02:50.660 --> 00:02:54.240
But the sum from π equals one to β of π π converges.
00:02:55.050 --> 00:03:01.590
And if a series is not absolutely convergent and itβs not conditionally convergent, then itβs divergent.
00:03:02.370 --> 00:03:05.250
Letβs see an example of conditional convergence.
00:03:05.970 --> 00:03:18.180
Is the alternating harmonic series the sum from π equals one to β of negative one to the power of π add one multiplied by one over π absolutely convergent, conditionally convergent, or divergent?
00:03:19.140 --> 00:03:25.740
Letβs firstly remember that a series is absolutely convergent if the series of absolute values is convergent.
00:03:26.350 --> 00:03:32.770
And a series is conditionally convergent if the series of absolute values diverges but the series converges.
00:03:33.680 --> 00:03:36.010
And otherwise, the series is divergent.
00:03:36.540 --> 00:03:39.980
So letβs start by testing for absolute convergence.
00:03:40.440 --> 00:03:46.770
We can see that negative one raised to the power of π add one is always going to give us one when we take the absolute value.
00:03:47.260 --> 00:03:51.520
So this is in fact the same as the sum from π equals one to β of one over π.
00:03:52.350 --> 00:03:55.200
But this is actually a series that weβre familiar with.
00:03:55.200 --> 00:03:56.770
Itβs the harmonic series.
00:03:57.500 --> 00:04:00.310
And we know that the harmonic series diverges.
00:04:01.060 --> 00:04:05.710
So the alternating harmonic series is not absolutely convergent.
00:04:06.140 --> 00:04:09.480
But is it conditionally convergent or divergent?
00:04:10.160 --> 00:04:14.660
So our next step is to test the alternating harmonic series for convergence.
00:04:15.340 --> 00:04:19.900
Because thatβs an alternating series, we can do this with the alternating series test.
00:04:20.440 --> 00:04:34.910
Recall this says that for an alternating series, the sum of negative one to the power of π add one multiplied by π π if π π is decreasing and the limit as π approaches β of π π is equal to zero, then π π is convergent.
00:04:35.590 --> 00:04:40.810
So for the alternating harmonic series, we can say that π π equals one over π.
00:04:41.800 --> 00:04:43.910
So is π π decreasing?
00:04:44.520 --> 00:04:47.790
Well, as π increases, one over π does decrease.
00:04:48.410 --> 00:04:50.160
So that condition is satisfied.
00:04:50.820 --> 00:04:54.580
But does the limit as π approaches β of π π equal zero?
00:04:55.190 --> 00:05:02.330
Well, the limit as π approaches β of one over π is going to be one over β, which we know is zero.
00:05:03.010 --> 00:05:04.720
So that condition is satisfied.
00:05:05.680 --> 00:05:17.330
So because we found that the alternating harmonic series is not absolutely convergent, but it is convergent, we can conclude that the alternating harmonic series is conditionally convergent.
00:05:18.450 --> 00:05:24.760
We can summarize the check for absolute convergence, conditional convergence, and divergence in a helpful diagram.
00:05:25.630 --> 00:05:32.270
Letβs say we want to find out whether the series π π is absolutely convergent, conditionally convergent, or divergent.
00:05:32.910 --> 00:05:38.550
We begin by testing whether the series of absolute values is convergent or divergent.
00:05:39.170 --> 00:05:43.030
Letβs say that we find that the series of absolute values is convergent.
00:05:43.610 --> 00:05:46.590
Then, the series π π is absolutely convergent.
00:05:47.310 --> 00:05:55.230
But if we find that the series of absolute values is divergent, then the series π π is not absolutely convergent.
00:05:55.530 --> 00:05:58.150
But it may still be conditionally convergent.
00:05:58.630 --> 00:06:06.370
So we try a different test on the series π π to check for convergence, for example, the alternating series test.
00:06:07.010 --> 00:06:13.690
And if we find that the series π π converges, then we say that the series π π is conditionally convergent.
00:06:14.260 --> 00:06:21.340
But if we find that the series π π diverges, then we conclude that the series π π is divergent.
00:06:22.300 --> 00:06:25.720
So these are the three possible conclusions that we can draw.
00:06:26.760 --> 00:06:28.960
So letβs now see some more examples.
00:06:30.090 --> 00:06:35.320
Consider the series the sum from π equals one to β of sin of π over π cubed.
00:06:35.670 --> 00:06:41.320
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
00:06:42.060 --> 00:06:49.100
Recall that a series π π is absolutely convergent if the series of absolute values is convergent.
00:06:49.720 --> 00:06:55.800
And if we find that the series is not absolutely convergent, it may still be conditionally convergent.
00:06:56.050 --> 00:06:59.380
So we then test the series for convergence or divergence.
00:07:00.330 --> 00:07:04.650
So letβs begin by testing this series for absolute convergence.
00:07:05.510 --> 00:07:13.640
So we want to find out whether the sum from π equals one to β of the absolute value of sin of π over π cubed is convergent or divergent.
00:07:14.270 --> 00:07:21.140
Well, because π only runs through positive values from one to β, π cubed is always going to be positive.
00:07:21.680 --> 00:07:27.370
So this is just the sum from π equals one to β of the absolute value of sin of π over π cubed.
00:07:28.030 --> 00:07:32.390
Now we know that sin of π will always be between negative one and one.
00:07:33.130 --> 00:07:47.130
So we can say that the absolute value of sin of π will always be less than or equal to one, which means that we can write the absolute value of sin of π over π cubed is less than or equal to one over π cubed.
00:07:47.740 --> 00:07:51.430
Writing it this way allows us to do a direct comparison.
00:07:52.110 --> 00:08:03.210
Recall that this means if π π is less than π π and the sum from π equals one to β of π π converges, then the sum from π equals one to β of π π also converges.
00:08:03.960 --> 00:08:07.770
And one over π cubed is actually a series we recognize.
00:08:08.340 --> 00:08:14.510
Recall that a π-series is a series of the form the sum for π equals one to β of one over π to the π power.
00:08:14.730 --> 00:08:19.800
And this converges if π is greater than one and diverges if π is less than or equal to one.
00:08:20.600 --> 00:08:24.430
So one over π cubed is a π-series with π equal to three.
00:08:24.860 --> 00:08:27.300
So one over π cubed converges.
00:08:27.860 --> 00:08:33.510
So by direct comparison, the absolute value of sin of π over π cubed also converges.
00:08:34.030 --> 00:08:45.700
Then, because we found the series of absolute values to be convergent, then our series the sum from π equals one to β of sin of π over π cubed is absolutely convergent.
00:08:47.090 --> 00:08:59.570
State whether the series the sum from π equals one to β of negative one to the power of π add one multiplied by two over the square root of π add one converges absolutely, conditionally, or not at all.
00:09:00.210 --> 00:09:07.730
Firstly, recall that for a series π π, this is absolutely convergent if the series of absolute values converges.
00:09:08.330 --> 00:09:13.100
And itβs conditionally convergent if the series of absolute values diverges.
00:09:13.310 --> 00:09:15.440
But the series itself converges.
00:09:15.960 --> 00:09:20.780
So letβs first of all find out whether this series is absolutely convergent or not.
00:09:21.540 --> 00:09:33.410
This means testing whether the series from π equals one to β of the absolute value of negative one to the power of π add one multiplied by two over the square root of π add one is convergent or divergent.
00:09:34.280 --> 00:09:39.160
Well, negative one to the power of π add one is always going to be one or negative one.
00:09:39.390 --> 00:09:42.280
But if we take the absolute value, it will always be one.
00:09:42.990 --> 00:09:49.400
Whereas two over the square root of π add one will always be positive because π runs through positive values.
00:09:49.800 --> 00:09:55.050
So we can write this as the sum from π equals one to β of two over the square root of π add one.
00:09:55.470 --> 00:10:00.670
Then, we can use the constant multiplication rule to bring the two to the front of the sum.
00:10:01.360 --> 00:10:06.180
From here, we need to work out whether this series converges or diverges.
00:10:07.070 --> 00:10:11.860
One way we can actually do this is with a direct comparison with the harmonic series.
00:10:12.510 --> 00:10:19.110
Because for π is greater than two, we have that one over π is less than one over the square root of π add one.
00:10:19.590 --> 00:10:26.090
And we know that if we have π π less than π π where π π diverges, then π π also diverges.
00:10:26.670 --> 00:10:32.870
And we know that the sum from π equals one to β of one over π is the harmonic series which diverges.
00:10:33.690 --> 00:10:38.960
Then, the sum from π equals one to β of one over the square root of π add one also diverges.
00:10:39.400 --> 00:10:48.020
So we found that the series of absolute values diverges, which means that this series is not absolutely convergent.
00:10:48.500 --> 00:10:51.740
But it could still be conditionally convergent.
00:10:52.180 --> 00:10:55.560
So weβre going to test the series itself for convergence.
00:10:56.100 --> 00:10:57.550
So letβs clear some space.
00:10:58.250 --> 00:11:01.450
We can firstly bring the constant two to the front of the sum.
00:11:01.830 --> 00:11:11.710
And then, if we look at this negative one to the power of π add one, this creates an alternating series because it makes the terms alternate between positive and negative.
00:11:12.390 --> 00:11:18.550
So we can decide whether this series is convergent or divergent using the alternating series test.
00:11:19.210 --> 00:11:33.790
Remember that this says for a series π π, where π π is equal to negative one to the power of π add one multiplied by π π, if π π is decreasing and the limit as π approaches β of π π is equal to zero, then the series π π is convergent.
00:11:34.330 --> 00:11:38.700
So for our series π π is equal to one over the square root of π add one.
00:11:39.050 --> 00:11:40.430
But is this decreasing?
00:11:40.920 --> 00:11:45.100
Well, for π π to be decreasing, we need π π to be greater than π π add one.
00:11:45.750 --> 00:11:51.870
Well, we can see that as π increases by one, the square root of π add one is going to get bigger.
00:11:52.150 --> 00:11:55.560
So one over the square root of π add one is going to decrease.
00:11:56.160 --> 00:11:57.850
So π π is decreasing.
00:11:58.410 --> 00:12:02.830
Then, we need to check whether the limit as π approaches β of π π is equal to zero.
00:12:03.480 --> 00:12:08.620
In other words, is the limit as π approaches β of one over the square root of π add one zero?
00:12:09.620 --> 00:12:13.220
Well, the square root of π add one is increasing as π gets bigger.
00:12:13.250 --> 00:12:15.460
So this will be one over β.
00:12:16.000 --> 00:12:21.750
So as π approaches β, then one over the square root of π add one approaches one over β.
00:12:21.960 --> 00:12:25.080
And so the limit as π approaches β is zero.
00:12:25.590 --> 00:12:28.030
So both of these conditions are satisfied.
00:12:28.400 --> 00:12:30.760
So this series is convergent.
00:12:31.390 --> 00:12:40.190
Remember that we said that a series is conditionally convergent if the series of absolute values diverges but the series converges.
00:12:40.790 --> 00:12:42.950
And thatβs exactly what weβve found here.
00:12:43.200 --> 00:12:46.120
The series of absolute values was divergent.
00:12:46.420 --> 00:12:49.250
But we found the series itself to be convergent.
00:12:49.750 --> 00:12:53.340
So we can conclude that this series converges conditionally.
00:12:55.030 --> 00:13:01.270
But why is that helpful to differentiate between absolutely convergent and conditionally convergent series?
00:13:02.550 --> 00:13:08.450
Well, absolutely convergent infinite series holds some of the same properties as finite sums.
00:13:08.860 --> 00:13:15.950
For example, if we have a finite sum, then any rearrangement of the terms still gives the same sum.
00:13:16.760 --> 00:13:20.770
And this also holds true for an absolutely convergent series.
00:13:21.200 --> 00:13:23.820
Any rearrangement yields the same sum.
00:13:24.990 --> 00:13:28.910
But this is not true for a conditionally convergent series.
00:13:29.280 --> 00:13:33.480
This is because rearranging the terms of a series changes the partial sums.
00:13:33.660 --> 00:13:37.680
So this can change the limit of the partial sums when some of the terms are negative.
00:13:38.070 --> 00:13:41.280
So we donβt have that issue with absolutely convergent series.
00:13:41.700 --> 00:13:45.370
But of course, this doesnβt apply to conditionally convergent series.
00:13:45.910 --> 00:13:52.970
As an example, the alternating harmonic series, which weβve seen is convergent, can be shown to converge to the natural log of two.
00:13:53.510 --> 00:14:02.260
But if the terms in the series are rearranged so that every positive term is followed by two negative terms, this does change the value of the sum.
00:14:02.840 --> 00:14:12.700
So with a conditionally convergent series, rearrangement changes the relative rate at which positive and negative terms are used and in turn changes the sum of the series.
00:14:13.170 --> 00:14:20.300
In fact, we can actually use this to rearrange a conditionally convergent series to converge to any value we want.
00:14:20.760 --> 00:14:24.470
But itβs beyond the scope of this video to go through these proofs in detail.
00:14:24.890 --> 00:14:27.400
Letβs now summarize some of the main points.
00:14:28.330 --> 00:14:34.450
A series π π is absolutely convergent if the series of absolute values is convergent.
00:14:35.140 --> 00:14:43.290
And a series π π is conditionally convergent if the series of absolute values is divergent, but the series itself is convergent.
00:14:44.130 --> 00:14:53.070
And finally, if π π is an absolutely convergent series with sum π , then any rearrangement of π π yields the same sum π .