
Title: Leibniz Criterion

Series: Real Analysis

Chapter: Infinite Series

YouTubeTitle: Real Analysis 18  Leibniz Criterion

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Quiz: Test your knowledge

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Subtitle on GitHub: ra18_sub_eng.srt

Timestamps
00:00 Intro
00:18 Recalling the harmonic series
01:24 Leibniz Criterion
03:07 Proof
07:52 Example
08:35 Credits

Subtitle in English
1 00:00:00,257 –> 00:00:03,669 Hello and welcome back to real analysis
2 00:00:03,869 –> 00:00:09,247 and as always many, many thanks for all the nice people that support this channel on Steady or Paypal.
3 00:00:09,447 –> 00:00:13,385 In today’s part 18 we will talk about the Leibniz criterion.
4 00:00:13,585 –> 00:00:19,486 In short it’s a theorem that gives us the sufficient condition for a series to be convergent.
5 00:00:19,686 –> 00:00:24,922 As a good introduction for this, let’s recall a series you already know. The Harmonic series.
6 00:00:25,686 –> 00:00:29,952 It’s the sequence of partial sums, where we sum over 1/k.
7 00:00:30,152 –> 00:00:33,938 A nice visualization for this would be to look at rectangles.
8 00:00:34,900 –> 00:00:40,146 Here all the rectangles should have width 1 and the height should be given by 1/k.
9 00:00:40,671 –> 00:00:46,102 This means that the area of the rectangles are given by our numbers we want to sum up.
10 00:00:46,671 –> 00:00:51,996 Of course here the area of one rectangle gets smaller and smaller when we go to the right.
11 00:00:52,196 –> 00:00:58,457 However we have already shown that the whole infinite sum, the whole area is infinity.
12 00:00:58,814 –> 00:01:04,518 In other words the Harmonic series is not convergent but divergent to infinity.
13 00:01:04,843 –> 00:01:12,168 Therefore having a sequence inside the sum that goes to 0 is not sufficient for convergence of the series.
14 00:01:12,368 –> 00:01:15,911 However we know from the last video it’s necessary.
15 00:01:16,357 –> 00:01:20,720 Ok now the Leibniz criterion will fix this missing sufficient condition.
16 00:01:21,071 –> 00:01:24,476 For this we also can use the same visualization.
17 00:01:24,943 –> 00:01:27,645 Let’s again look at the rectangles from before.
18 00:01:27,943 –> 00:01:31,909 However now we want to alternate the signs of the areas.
19 00:01:32,686 –> 00:01:39,466 Hence the first area is positive, the second is negative, the third is positive again and so on.
20 00:01:39,714 –> 00:01:48,779 Therefore we don’t look at the Harmonic series here, but rather at a sequence of partial sums , where we have (1) to the power (k+1) in front.
21 00:01:49,471 –> 00:01:53,959 So we have the same sequence involved, but know with alternating signs
22 00:01:54,300 –> 00:02:00,377 and now the Leibniz criterion tells us this is indeed enough to make the whole series convergent.
23 00:02:00,577 –> 00:02:08,044 Therefore please remember adding, subtracting, adding, subtracting and so on, gives us a different behavior than just adding.
24 00:02:08,643 –> 00:02:12,388 Ok, now we are ready to formulate the general Leibniz criterion.
25 00:02:13,086 –> 00:02:17,726 For obvious reason this is also known as the alternating series test.
26 00:02:17,926 –> 00:02:22,133 Of course there are also other names some people just call it Leibniz’s test.
27 00:02:22,829 –> 00:02:27,281 Ok for the formulation let’s take a sequence of real numbers called a_k.
28 00:02:27,481 –> 00:02:33,893 Now, we already know if we want a convergent series we need this sequence to be convergent with limit 0.
29 00:02:34,643 –> 00:02:39,133 However now we also put in that a_k is monotonically decreasing.
30 00:02:39,414 –> 00:02:43,304 So please keep in mind this is exactly what we have for 1/k
31 00:02:43,857 –> 00:02:48,338 and now the conclusion is that the alternating series here is convergent.
32 00:02:49,114 –> 00:02:55,461 This means whenever you see you have alternating signs you can always check if the Leibniz criterion is applicable.
33 00:02:55,661 –> 00:03:03,340 You just ignore the signs and look if the remaining part defines a sequence with limit 0 and is monotonically decreasing.
34 00:03:03,540 –> 00:03:07,505 Ok, now I guess you are interested in the proof of this statement.
35 00:03:07,800 –> 00:03:15,771 Indeed this is not so hard. We just defined the partial sums as always and then we will look at suitable subsequences.
36 00:03:16,386 –> 00:03:22,414 So here you see the formulation is a little bit different than before, because here we shifted the index by 1,
37 00:03:22,614 –> 00:03:29,249 which is of course no problem at all, but the picture changes a little bit because the first member here gets a minus sign.
38 00:03:29,871 –> 00:03:33,763 Hence now you see here the correct visualization for our case.
39 00:03:34,057 –> 00:03:38,854 Now the first thing you should note here is that all the a_k’s are nonnegative,
40 00:03:38,886 –> 00:03:43,337 because the limit is 0 and the sequence is monotonically decreasing.
41 00:03:43,537 –> 00:03:45,903 Therefore this picture is indeed correct.
42 00:03:45,971 –> 00:03:51,624 Maybe the sizes of the rectangles are different, but we have that they don’t get larger when we go to the right
43 00:03:51,686 –> 00:03:54,159 and we have the alternating sign.
44 00:03:54,359 –> 00:03:59,257 Now the first idea, the first subsequence is that we only consider even ends here,
45 00:03:59,457 –> 00:04:03,408 because then you see here, the negative part dominates.
46 00:04:03,914 –> 00:04:12,858 So let’s take a natural number “l”, then this index is even and then we subtract the predecessor given by an even number as well.
47 00:04:13,486 –> 00:04:17,701 Therefore in the picture what remains are just 2 rectangles, maybe here.
48 00:04:18,371 –> 00:04:26,728 and then we know by assumption that the right area here, can’t be bigger than the left area here. Of course measured with the absolute value.
49 00:04:27,314 –> 00:04:29,575 On the other hand in the formula this looks like this.
50 00:04:29,671 –> 00:04:37,429 With the same reasoning, when we subtract the 2 partial sums only 2 terms remain and only the odd number gets the minus sign.
51 00:04:38,043 –> 00:04:43,000 Now, since the sequence is monotonically decreasing this is less or equal than 0.
52 00:04:43,700 –> 00:04:46,769 That’s what I said, the negative part here dominates.
53 00:04:47,400 –> 00:04:51,625 Ok, maybe not so surprising we can do the same for odd indices.
54 00:04:51,871 –> 00:04:58,585 This works completely the same but only shifted by one index, which means now the positive part dominates.
55 00:04:58,785 –> 00:05:05,087 Of course we see that in the formula as well. These 2 members remain and the odd index gets the minus sign.
56 00:05:05,514 –> 00:05:09,411 Therefore this is again by the monotonicity greater or equal than 0.
57 00:05:10,114 –> 00:05:14,805 Hence the odd indices give us a monotonically increasing subsequence
58 00:05:15,005 –> 00:05:20,704 and on the other hand we have a monotonically decreasing subsequence for the even indices.
59 00:05:20,904 –> 00:05:24,651 Ok, then in the next step let’s compare both subsequences.
60 00:05:25,329 –> 00:05:31,089 In other words: what is the difference between the members with indices 2l+1 and 2l?
61 00:05:31,571 –> 00:05:37,040 Also this is easy to calculate. It’s simply a with index 2l+1.
62 00:05:37,800 –> 00:05:44,914 Which is less or equal than 0. Hence we get this inequality when we want to compare both subsequences
63 00:05:45,486 –> 00:05:53,471 and now we can put in that we know that this sequence is monotonically decreasing and this sequence is monotonically increasing.
64 00:05:54,000 –> 00:06:00,756 Therefore for all natural numbers “l”, we have that s_2l is less or equal than s_2
65 00:06:00,956 –> 00:06:03,933 and the same on the left hand side with S_3.
66 00:06:04,133 –> 00:06:08,405 So from this we conclude that both subsequences are bounded
67 00:06:08,700 –> 00:06:14,086 and now you might already recognize one important theorem we had for sequences.
68 00:06:14,429 –> 00:06:17,999 We simply called it the monotone convergence criterion.
69 00:06:18,800 –> 00:06:27,346 Here in our case we have a sequence given by the even indices, which is monotonically decreasing and also bounded from below.
70 00:06:28,129 –> 00:06:31,021 Hence we know it’s a convergent sequence.
71 00:06:31,771 –> 00:06:39,529 On the other hand we have a sequence which is monotonically increasing and also bounded, so it’s also convergent.
72 00:06:40,243 –> 00:06:46,495 We can even show that the limit of both sequences is the same, because we can look at the difference again
73 00:06:46,914 –> 00:06:50,737 We already know it’s simply given by a sequence member “a”.
74 00:06:51,400 –> 00:06:55,595 Therefore when “l” goes to Infinity this sequence goes to 0.
75 00:06:56,000 –> 00:07:00,658 So here we put in our assumption about the limit of the sequence a_k.
76 00:07:00,858 –> 00:07:04,629 Ok, so there’s only one limit and we can call it s.
77 00:07:05,329 –> 00:07:11,003 So at this point please recall that we wanted to show that the sequence s_n is convergent.
78 00:07:11,714 –> 00:07:20,363 Here we just considered 2 subsequences. However both subsequences together hit all the sequence members of s_n.
79 00:07:21,014 –> 00:07:26,795 Therefore we can conclude that the limit of s_n is also given by s.
80 00:07:27,286 –> 00:07:32,251 If you have never seen this step here, then it’s a good exercise for you to show it.
81 00:07:32,871 –> 00:07:39,232 Nevertheless for us the result here is important. The sequence of partial sums s_n is convergent.
82 00:07:39,714 –> 00:07:48,107 and this closes our proof. So we have proven the Leibniz criterion and now you are allowed to use it whenever you can.
83 00:07:48,786 –> 00:07:52,354 Therefore let’s close the video with one last example.
84 00:07:52,857 –> 00:07:59,770 Let’s simply consider a series that has 1 to the power k over the square root of k.
85 00:08:00,386 –> 00:08:03,732 So here we have everything we need for the Leibniz criterion.
86 00:08:03,932 –> 00:08:11,048 We have alternating signs and a sequence which is monotonically decreasing and is convergent to 0.
87 00:08:11,686 –> 00:08:15,309 Therefore we know this whole series here is convergent.
88 00:08:15,700 –> 00:08:20,647 We cannot immediately calculate the limit, but we know the limit is a real number.
89 00:08:21,214 –> 00:08:24,692 Ok, so that’s how you apply the Leibniz criterion.
90 00:08:25,286 –> 00:08:35,286 Ok, then in the next video we will talk about other criteria as well. Therefore I hope I see you there and have a nice day. bye!

Quiz Content
Q1: What is not an assumption one needs for applying the Leibniz criterion to a series $\displaystyle \sum_{k = 1}^\infty (1)^{k} a_k$ ?
A1: The sequence $(a_k)_{k \in \mathbb{N}}$ needs to be convergent.
A2: The sequence $(a_k)_{k \in \mathbb{N}}$ needs to be monotonically decreasing.
A3: The sequence $(a_k)_{k \in \mathbb{N}}$ needs to be nonnegative.
A4: The sequence $(a_k)_{k \in \mathbb{N}}$ needs to be alternating.
A5: The sequence $(a_k)_{k \in \mathbb{N}}$ needs to have limit 0.
Q2: Is the series $\displaystyle \sum_{k = 1}^\infty (1)^{k} \frac{1}{k}$ convergent?
A1: Yes, it is.
A2: No, it isn’t.
Q3: Is the series $\displaystyle \sum_{k = 1}^\infty (1)^{k} a_k$ with $a_k = (1)^{k} \frac{1}{k}$ convergent?
A1: Yes, it is.
A2: No, it isn’t.
Q4: Is the series $\displaystyle \sum_{k = 1}^\infty (1)^{k},k$ convergent?
A1: Yes, it is.
A2: No, it isn’t.