• Title: Cauchy Criterion

  • Series: Real Analysis

  • Chapter: Infinite Series

  • YouTube-Title: Real Analysis 17 | Cauchy Criterion

  • Bright video: https://youtu.be/lr12HwXQ3uw

  • Dark video: https://youtu.be/NuYO5il1Dko

  • Quiz: Test your knowledge

  • PDF: Download PDF version of the bright video

  • Dark-PDF: Download PDF version of the dark video

  • Print-PDF: Download printable PDF version

  • Thumbnail (bright): Download PNG

  • Thumbnail (dark): Download PNG

  • Subtitle on GitHub: ra17_sub_eng.srt

  • Timestamps

    00:00 Intro

    00:36 Properties of series

    02:20 Cauchy Criterion

    03:46 Proof

    05:29 Example

    07:15 Conclusion

    08:47 Credits

  • Subtitle in English

    1 00:00:00,329 –> 00:00:03,100 Hello and welcome back to real analysis

    2 00:00:03,500 –> 00:00:08,933 and as always first I want to thank all the nice people that support this channel on Steady or Paypal.

    3 00:00:09,133 –> 00:00:13,214 In today’s part 17 we still talk about series.

    4 00:00:13,386 –> 00:00:19,643 More concretely we will use this video and the next ones to talk about criteria we have to check for convergence.

    5 00:00:19,971 –> 00:00:24,573 For this please recall that a series is just a sequence of partial sums

    6 00:00:25,129 –> 00:00:31,063 and in the case that this sequence is convergent we also use this symbol to denote the limit.

    7 00:00:31,263 –> 00:00:36,247 Therefore by applying the well known limit theorems we can write down some properties.

    8 00:00:36,900 –> 00:00:40,485 Therefore we take 2 series that need to be convergent

    9 00:00:40,843 –> 00:00:47,028 and then we can say something about the series that is given by the sequence a_k + b_k.

    10 00:00:47,657 –> 00:00:51,338 So here keep in mind we have a new sequence of partial sums.

    11 00:00:51,943 –> 00:00:57,293 Now the result, which is not so surprising, is that this series is also convergent.

    12 00:00:57,600 –> 00:01:04,881 We know even more. We can immediately calculate the limit. It’s just given by the sum of the two limits.

    13 00:01:05,200 –> 00:01:11,064 Of course all of this is just an application of the limit theorems you already learned for sequences

    14 00:01:11,529 –> 00:01:19,230 However in this case, when we have convergent series, it tells us that we can pull the addition out of this symbol.

    15 00:01:19,430 –> 00:01:26,108 Of course for a normal sum this always works but please keep in mind this symbol here is not a normal sum.

    16 00:01:26,600 –> 00:01:32,171 Ok and another property we get from ordinary sums is given by the multiplication.

    17 00:01:32,555 –> 00:01:38,474 Namely we have the distributive law when we multiply all the terms in the sum with the same factor.

    18 00:01:38,674 –> 00:01:44,821 For a series this means, if the original series is convergent then also this one.

    19 00:01:45,021 –> 00:01:52,446 No matter which real number we choose for lambda. Also here this follows immediately from the limit theorems

    20 00:01:52,646 –> 00:01:56,779 and we get the calculation rule that we can pull out lambda.

    21 00:01:57,614 –> 00:02:04,002 Ok these two properties you really should remember and you should remember that they hold for convergent series.

    22 00:02:04,202 –> 00:02:09,649 Therefore a natural question would be: how can we check that a given series is indeed convergent?

    23 00:02:09,849 –> 00:02:14,970 In fact there are a lot of different criteria one can apply to answer this question.

    24 00:02:15,170 –> 00:02:20,333 The first one I want to show you just uses the completeness axiom of the real numbers.

    25 00:02:20,533 –> 00:02:23,500 Therefore it’s called the Cauchy criterion.

    26 00:02:24,143 –> 00:02:30,899 Please recall in the real numbers being a convergent sequence is equivalent for being a Cauchy sequence

    27 00:02:31,271 –> 00:02:37,044 and there the important part was that for a Cauchy sequence we don’t have to know the limit.

    28 00:02:37,557 –> 00:02:42,532 Therefore this criterion will check for convergence but it does not tell you the limit.

    29 00:02:43,257 –> 00:02:50,402 Of course often this is good enough. For example if you just want to justify that you are allowed to use these calculation rules here.

    30 00:02:51,071 –> 00:02:57,258 Hence the criterion tells us that the series is convergent if and only if

    31 00:02:57,458 –> 00:03:03,875 for all epsilon greater than 0 there exists a capital N,

    32 00:03:04,075 –> 00:03:11,820 such that for all indices n and m greater than this capital N, where also n is greater or equal than m,

    33 00:03:12,229 –> 00:03:21,589 we have that the absolute value of the partial sum, starting with m and ending with n is less than epsilon.

    34 00:03:22,314 –> 00:03:32,157 In other words, if you can show that this expression here is arbitrarily small eventually, then you know the series is convergent

    35 00:03:32,286 –> 00:03:38,565 and on the other hand, if you can show that this expression stays large or larger than a given constant,

    36 00:03:38,765 –> 00:03:41,724 then you know the series is not convergent.

    37 00:03:42,114 –> 00:03:46,874 Ok, before I show you an example let’s do the proof of the statement here.

    38 00:03:47,074 –> 00:03:51,676 This is not so complicated, because we just have to use the completeness axiom.

    39 00:03:52,043 –> 00:03:56,606 First let’s define the partial sums as always denoted by s_n.

    40 00:03:56,806 –> 00:04:03,757 Then we know the sequence s_n is convergent if and only if it is a Cauchy sequence.

    41 00:04:04,500 –> 00:04:08,481 That is exactly what we know as the completeness of the real numbers.

    42 00:04:08,681 –> 00:04:13,783 If you don’t know anymore what a Cauchy sequence is, let’s write down the definition.

    43 00:04:14,486 –> 00:04:23,775 It just means that for all epsilon greater 0 there exists a capital N such that for all indices afterwards we can calculate the difference

    44 00:04:23,975 –> 00:04:27,870 and then in the absolute value it should be less than epsilon.

    45 00:04:28,471 –> 00:04:35,654 Here you see the indices are called n-tilde and m-tilde, just because I want to redefine them in the next step.

    46 00:04:35,854 –> 00:04:39,322 You already know, what we want is an expression like this.

    47 00:04:39,943 –> 00:04:47,768 Hence now you can see n should be always the larger index here and m-tilde I want to shift by 1.

    48 00:04:47,968 –> 00:04:51,932 Hence here we get s_n - s_(m-1).

    49 00:04:52,132 –> 00:04:56,856 First you should see this is indeed a special case of the property we had here

    50 00:04:57,056 –> 00:05:04,779 and on the other hand you should see it’s not a restriction at all, because the only case we really miss is when we have the same index here.

    51 00:05:04,979 –> 00:05:11,880 But of course in the case that we have the same index there we always get 0 out, which is always less than epsilon.

    52 00:05:12,080 –> 00:05:16,577 Therefore this reformulation for a Cauchy sequence is always allowed.

    53 00:05:16,777 –> 00:05:21,850 Ok, now we only have to see that this difference is exactly this sum here.

    54 00:05:22,050 –> 00:05:25,772 Hence this is the whole proof of the Cauchy criterion.

    55 00:05:26,214 –> 00:05:29,244 Now, as promised let’s apply it to an example.

    56 00:05:29,814 –> 00:05:34,159 Ok, I would say let’s look at the example we had at the beginning of this course.

    57 00:05:34,529 –> 00:05:40,986 Of course we already know it’s not convergent, but the question would be how can we show it with the Cauchy criterion.

    58 00:05:41,414 –> 00:05:44,554 For this we have to calculate this partial sum here.

    59 00:05:45,086 –> 00:05:52,232 Now you already know we only need to calculate this for indices n, m that are greater or equal than some capital N.

    60 00:05:52,614 –> 00:05:56,603 Therefore maybe let’s substitute the m with the capital N.

    61 00:05:56,803 –> 00:06:01,612 In other words we look at the special case for m and we can do the same for n.

    62 00:06:01,812 –> 00:06:11,971 For example it would be allowed to choose N+2. Therefore the sum consists of exactly three numbers and of course we can calculate them

    63 00:06:12,300 –> 00:06:19,056 We just have to distinguish two different cases. The first case would be N is even.

    64 00:06:19,256 –> 00:06:31,910 Then we have 1+(-1)+1. Therefore for an even N we get out 1 for the sum. Of course then the absolute value does not change anything

    65 00:06:32,571 –> 00:06:41,135 Ok, then the other case would be then an odd N, which means we start with -1, then comes 1 and then we add -1 again.

    66 00:06:41,814 –> 00:06:52,768 Hence the result is -1, but with the absolute value it’s 1 as well. Hence no matter what N is this expression is always exactly 1.

    67 00:06:52,968 –> 00:06:58,483 Which is of course not less than epsilon, if we choose epsilon less than 1.

    68 00:06:58,683 –> 00:07:03,457 In conclusion this property here can’t be fulfilled for all epsilon.

    69 00:07:03,657 –> 00:07:09,657 Indeed now this is all we have to do to show that this series is not convergent.

    70 00:07:09,857 –> 00:07:15,547 So here you see the Cauchy criterion as simple as it is, can be very helpful in calculations

    71 00:07:16,329 –> 00:07:20,379 and it is also very helpful for showing the next important fact.

    72 00:07:20,579 –> 00:07:27,263 Maybe in this example you have already seen, there is a necessary condition for being a convergent series.

    73 00:07:27,463 –> 00:07:32,558 Just take any sequence of real numbers a_k and look at the series

    74 00:07:32,758 –> 00:07:40,576 and if this is convergent you know the sequence a_k has to be a sequence that converges to 0.

    75 00:07:40,714 –> 00:07:43,822 Ok, here you see the contraposition to this is,

    76 00:07:43,823 –> 00:07:51,085 if you start with a sequence a_k that is not convergent to 0, the series we get out is divergent.

    77 00:07:51,285 –> 00:07:55,769 and of course here you see immediately one example where you can apply this fact.

    78 00:07:56,300 –> 00:08:00,147 -1 to the power k is not a convergent sequence at all.

    79 00:08:00,347 –> 00:08:05,294 However one important thing to note here is, this is not an equivalence.

    80 00:08:05,494 –> 00:08:10,803 Please remind yourself in the last video we have already seen a counter example for this.

    81 00:08:11,003 –> 00:08:16,471 Now I can tell you this proof works similarly to the things we did in the example above.

    82 00:08:16,671 –> 00:08:21,094 Therefore I think this is a good exercise for you. Try to prove this fact.

    83 00:08:21,514 –> 00:08:29,466 Afterwards, then you know every time someone asks you about series and the convergence of it, you should check this thing here first.

    84 00:08:29,666 –> 00:08:35,453 However of course most of the time you still don’t know if the series is convergent then.

    85 00:08:35,653 –> 00:08:42,435 In this case you have to apply another criterion and such another one I show you in the next video.

    86 00:08:42,635 –> 00:08:46,429 Therefore I hope I see you there and have a nice day. Bye!

  • Quiz Content

    Q1: Is the following implication correct? If $ \displaystyle \sum_{k = 1}^\infty a_k $ and $ \displaystyle \sum_{k = 1}^\infty b_k $ are convergent, then also $ \displaystyle \sum_{k = 1}^\infty ( a_k + b_k )$ is convergent.

    A1: Yes, it is.

    A2: No, there are counterexample.

    Q2: Is the following implication correct? If $ \displaystyle \sum_{k = 1}^\infty ( a_k + b_k )$ is convergent, then also $ \displaystyle \sum_{k = 1}^\infty a_k $ and $ \displaystyle \sum_{k = 1}^\infty b_k $ are convergent.

    A1: Yes, they are.

    A2: No, there are counterexample.

    Q3: Let $(a_k){k \in \mathbb{N}}$ be a convergent sequence with limit $1$. What can one say about $ \displaystyle \sum{k = 1}^\infty a_k $ in this case?

    A1: The series cannot be divergent to $\infty$.

    A2: The series cannot be convergent.

    A3: The sequence of partial sums $(s_n)_{n \in \mathbb{N}}$ satisfies $s_n \geq 0$ for all $n \in \mathbb{N}$.

    A4: The sequence of partial sums $(s_n)_{n \in \mathbb{N}}$ satisfies $s_n \leq 0$ for all $n \in \mathbb{N}$.

  • Back to overview page