• Title: Comparison Test

  • Series: Real Analysis

  • Chapter: Infinite Series

  • YouTube-Title: Real Analysis 19 | Comparison Test

  • Bright video: https://youtu.be/mI40-tAtP58

  • Dark video: https://youtu.be/KH8fwgy6qg4

  • Quiz: Test your knowledge

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  • Subtitle on GitHub: ra19_sub_eng.srt

  • Timestamps

    00:00 Intro

    00:21 Absolutely convergent

    00:47 Absolutely convergent implies convergent

    02:06 Implication doesn’t work in the other direction

    02:45 Majorant criterion

    04:16 Proof of majorant criterion

    05:31 Minorant criterion

    06:30 Example for the minorant criterion

    07:44 Credits

  • Subtitle in English

    1 00:00:00,400 –> 00:00:03,644 Hello and welcome back to real analysis

    2 00:00:04,129 –> 00:00:09,366 and as always first I want to thank all the nice people that support this channel on Steady or Paypal.

    3 00:00:09,686 –> 00:00:15,318 In today’s part 19 we will talk about the so called comparison test for series.

    4 00:00:15,943 –> 00:00:20,924 It’s a criterion we can use to decide if a given series is convergent,

    5 00:00:21,124 –> 00:00:24,592 but first we need a new definition about convergence.

    6 00:00:24,792 –> 00:00:30,686 Namely we call a series given by a sequence a_k, absolutely convergent

    7 00:00:31,229 –> 00:00:36,112 if the series given by the absolute value of a_k is simply convergent.

    8 00:00:36,686 –> 00:00:43,857 So what you see immediately is that for a series that has only positive a_k’s, this is not a new definition at all.

    9 00:00:44,386 –> 00:00:47,494 However in general it’s a stronger definition.

    10 00:00:48,214 –> 00:00:52,645 In other words being absolutely convergent implies being convergent.

    11 00:00:53,100 –> 00:00:58,175 Indeed this is not hard to show, we can just use the Cauchy criterion.

    12 00:00:58,375 –> 00:01:07,196 Applying this for our series here just means: For all epsilon there is an N, such that for all n, m we have

    13 00:01:07,314 –> 00:01:10,740 that this finite sum here is less than epsilon.

    14 00:01:11,271 –> 00:01:18,961 Now, in this special case here, we know that the sum is never negative. Therefore we can omit the absolute values around it.

    15 00:01:19,757 –> 00:01:26,986 Ok, but know we want to say something about the original series with a_k. So without the absolute value here inside.

    16 00:01:27,643 –> 00:01:33,415 Hence what we really want is that this sum with the absolute value outside is less than epsilon.

    17 00:01:33,829 –> 00:01:38,825 In order to get this we can just use the triangle inequality for the absolute value.

    18 00:01:39,129 –> 00:01:45,315 It just tells us that we can push the absolute value inside the sum, but then we have an inequality.

    19 00:01:45,515 –> 00:01:51,597 Now, here we already know from the line above that the right-hand side here is less than epsilon.

    20 00:01:52,114 –> 00:01:58,237 So this is our implication and we just get the Cauchy criterion for our original series with a_k.

    21 00:01:59,057 –> 00:02:02,185 Hence the series is convergent as well.

    22 00:02:02,385 –> 00:02:06,551 Ok, now please remember that we have always this implication here.

    23 00:02:06,751 –> 00:02:10,101 However it does not work the other way around.

    24 00:02:10,414 –> 00:02:14,172 Indeed that’s what we already learned in the last video.

    25 00:02:14,514 –> 00:02:19,217 Namely this series here is convergent by the Leibniz criterion.

    26 00:02:19,757 –> 00:02:23,127 However it’s not absolutely convergent.

    27 00:02:23,327 –> 00:02:29,313 We see that immediately, because when we take the absolute value here, we get out the harmonic series.

    28 00:02:29,700 –> 00:02:34,696 Ok, now in summary what we have is a stronger notion of convergence

    29 00:02:35,314 –> 00:02:39,127 and indeed this is exactly what we need for the comparison test.

    30 00:02:39,886 –> 00:02:44,793 Now in fact, this comparison test consists of 2 separate criteria.

    31 00:02:44,993 –> 00:02:48,563 The first one is the so called majorant criterion.

    32 00:02:48,871 –> 00:02:54,384 If we are able to apply this, we get the absolute convergence out for a series.

    33 00:02:54,584 –> 00:02:59,568 Ok, therefore let’s fix a series and now I will tell you what we have to check.

    34 00:02:59,768 –> 00:03:05,428 We just need another sequence b_k that is larger than the sequence a_k

    35 00:03:05,986 –> 00:03:09,399 and of course we put a_k in the absolute value

    36 00:03:09,599 –> 00:03:14,433 and we don’t need the inequality for all k, but we need it eventually.

    37 00:03:14,714 –> 00:03:18,561 Hence for all k after a given index n_0.

    38 00:03:18,971 –> 00:03:24,342 So the idea here is that we have another sequence b_k that is well behaved.

    39 00:03:24,542 –> 00:03:28,163 In the sense that corresponding series is convergent

    40 00:03:28,500 –> 00:03:36,031 and then we can conclude with this inequality that the series corresponding to the absolute value a_k is also convergent.

    41 00:03:36,371 –> 00:03:46,263 Therefore our premise is that we find an index n_0 and a convergent series and this series is then often called the majorant

    42 00:03:46,815 –> 00:03:52,461 and of course our inequality only makes sense if the corresponding b_k’s are non-negative.

    43 00:03:52,661 –> 00:03:58,660 Ok and then our conclusion is that our series over a_k is absolutely convergent.

    44 00:03:58,860 –> 00:04:06,600 So the idea is here that we can control the whole sequence of partial sums. We simply keep the sequence members small enough.

    45 00:04:07,143 –> 00:04:11,657 Now, applications for this nice theorem I show you in the next video.

    46 00:04:12,071 –> 00:04:19,460 Here we first should convince ourselves that this is indeed correct and in fact this proof is very short.

    47 00:04:19,660 –> 00:04:25,256 We just have to apply our already known Cauchy criterion to the series over b_k.

    48 00:04:25,456 –> 00:04:31,951 So this is the same as before. We find epsilon and N and get that this sum is less than epsilon.

    49 00:04:32,314 –> 00:04:40,144 However of course we want to say something about this sum. Now you might already know, we can just apply our inequality here.

    50 00:04:40,443 –> 00:04:47,360 In other words if we substitute a_k here with b_k, we get a value that is larger or stays the same.

    51 00:04:47,857 –> 00:04:54,186 The only small restriction for this calculation is that we have to choose the capital N larger than n_0

    52 00:04:54,771 –> 00:04:57,585 and of course it’s always possible to do that.

    53 00:04:57,871 –> 00:05:02,698 Ok and then on the right-hand side these 2 things here are obviously the same.

    54 00:05:03,157 –> 00:05:07,584 In summary now, this number is indeed less than epsilon.

    55 00:05:07,784 –> 00:05:14,167 This simply means that the Cauchy criterion holds for the series over the absolute value of a_k.

    56 00:05:14,367 –> 00:05:21,890 So we have a convergent series there or to put differently the series over a_k is absolutely convergent.

    57 00:05:22,243 –> 00:05:25,453 Ok and with this the whole proof is finished.

    58 00:05:25,829 –> 00:05:31,433 Ok, so this was the majorant criterion, but we also have the minorant criterion.

    59 00:05:31,943 –> 00:05:37,196 It works similarly, but now we don’t check for convergence but for divergence.

    60 00:05:37,486 –> 00:05:45,457 Therefore we almost have the same assumptions, but now we only check for series that have a sequence a_k that is non-negative.

    61 00:05:45,857 –> 00:05:52,300 So now we can copy almost all things from above, but now we need a divergent series b_k

    62 00:05:52,500 –> 00:05:56,344 and then of course the inequality goes the other direction.

    63 00:05:56,686 –> 00:06:00,955 Hence now the sequence a_k is simply too large.

    64 00:06:01,155 –> 00:06:06,285 Therefore the idea here is, if this series has already the limit infinity,

    65 00:06:06,485 –> 00:06:11,402 then the series over a_k, which is larger, should also have the limit infinity.

    66 00:06:11,957 –> 00:06:15,125 In conclusion the series is divergent.

    67 00:06:15,457 –> 00:06:21,557 Ok, the proof for this theorem is also not so hard. You can just use the majorant criterion.

    68 00:06:21,757 –> 00:06:26,216 So you can do a proof by contraposition or if you want by contradiction.

    69 00:06:26,614 –> 00:06:30,394 Ok, because it’s a very short proof, I think it’s a good exercise for you

    70 00:06:30,957 –> 00:06:35,701 and then we can close this video with an example for the minorant criterion.

    71 00:06:35,901 –> 00:06:41,030 We will show that the series with 1 over square root of k is divergent.

    72 00:06:41,486 –> 00:06:47,400 The first thing you should note here is that the square root of k is indeed less or equal than k itself.

    73 00:06:48,143 –> 00:06:51,737 Hence the reciprocals fulfill the other inequality.

    74 00:06:52,171 –> 00:06:56,999 So 1 over the square root of k is greater or equal than 1 over k

    75 00:06:57,257 –> 00:07:01,055 and of course this holds for all natural numbers k

    76 00:07:01,400 –> 00:07:06,602 and at this point you see, here we have our minorant, the sequence b_k,

    77 00:07:06,802 –> 00:07:14,557 because this gives us the harmonic series and we already showed that this is a divergent series.

    78 00:07:15,014 –> 00:07:21,959 Hence if we are able to write down such an inequality, we immediately know that we have a divergent series as well.

    79 00:07:22,343 –> 00:07:29,060 The only problem this might give you is that you don’t know at the beginning, which minorant or majorant you should choose.

    80 00:07:29,686 –> 00:07:34,635 Therefore in the next video I’ll show you some criteria that help you exactly there,

    81 00:07:34,835 –> 00:07:40,536 but still if you know what to do, the minorant and the majorant criterion can be very helpful.

    82 00:07:40,857 –> 00:07:44,757 Ok, then I hope I see you in the next video. Bye!

  • Quiz Content

    Q1: What is the definition of saying that a series $\displaystyle \sum_{k = 1}^\infty a_k$ is absolutely convergent?

    A1: $\displaystyle \sum_{k = 1}^\infty a_k$ is convergent

    A2: $\displaystyle \sum_{k = 1}^\infty |a_k|$ is convergent

    A3: $\displaystyle \sum_{k = 1}^\infty (-1)^{k} a_k$ is convergent

    A4: $\displaystyle \sum_{k = 1}^\infty (-1)^{k} |a_k|$ is convergent

    Q2: Is the series $\displaystyle \sum_{k = 1}^\infty (-1)^{k} \frac{1}{k}$ absolutely convergent?

    A1: Yes, it is.

    A2: No, it isn’t.

    Q3: Is the series $\displaystyle \sum_{k = 1}^\infty \frac{1}{\sqrt{k}} $ absolutely convergent?

    A1: Yes, it is.

    A2: No, it isn’t.

    Q4: What is the correct formulation of the majorant criterion, also known as the comparison test, for a series $\displaystyle \sum_{k = 1}^\infty a_k$ ?

    A1: If there is $n_0$ and a convergent series $\displaystyle \sum_{k = 1}^\infty b_k$ with $|a_k| \leq b_k$ for all $k \geq n_0$, then the series $\displaystyle \sum_{k = 1}^\infty a_k$ is divergent.

    A2: If there is $n_0$ and a convergent series $\displaystyle \sum_{k = 1}^\infty b_k$ with $|a_k| \leq b_k$ for all $k \geq n_0$, then the series $\displaystyle \sum_{k = 1}^\infty a_k$ is absolutely convergent.

    A3: If there is $n_0$ and a convergent series $\displaystyle \sum_{k = 1}^\infty b_k$ with $|a_k| \leq b_k$ for all $k \leq n_0$, then the series $\displaystyle \sum_{k = 1}^\infty a_k$ is convergent.

    A4: If there is $n_0$ and a divergent series $\displaystyle \sum_{k = 1}^\infty b_k$ with $|a_k| \leq b_k$ for all $k \geq n_0$, then the series $\displaystyle \sum_{k = 1}^\infty a_k$ is divergent.

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