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Title: Reordering for Series
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Series: Real Analysis
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Chapter: Infinite Series
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YouTube-Title: Real Analysis 21 | Reordering for Series
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Quiz: Test your knowledge
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Timestamps
00:00 Intro
00:17 Meaning of reordering of series
01:13 Example: Reordering can change the value
02:32 2nd Example with a convergent series
04:58 Definition Reordering
06:10 Theorem for abs. convergent series
06:39 Proof of the Theorem
12:22 Credits
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Subtitle in English
1 00:00:00.535 –> 00:00:03.365 Hello and welcome back to Real Analysis
2 00:00:03.905 –> 00:00:06.245 and as always I want to thank all the nice people
3 00:00:06.245 –> 00:00:08.325 that support this channel on Steady or PayPal.
4 00:00:09.145 –> 00:00:13.285 In today’s part 21, we will talk about reordering series.
5 00:00:14.055 –> 00:00:15.585 Therefore, first we should talk about
6 00:00:15.775 –> 00:00:19.265 what this exactly means if we don’t have a series
7 00:00:19.485 –> 00:00:20.865 but a finite sum.
8 00:00:21.165 –> 00:00:23.865 You know, this is a short notation for the sum
9 00:00:24.705 –> 00:00:29.225 A one plus a two plus a three until we reach a N.
10 00:00:30.165 –> 00:00:32.905 Of course, we also could calculate a different sum
11 00:00:32.955 –> 00:00:37.245 where we start with a two and then we add a one.
12 00:00:37.985 –> 00:00:40.725 So you see, I just want to flip the members around
13 00:00:41.585 –> 00:00:43.085 and then we continue with this
14 00:00:43.335 –> 00:00:45.325 until we reach the end of the sum.
15 00:00:46.265 –> 00:00:49.405 And for convenience, let’s assume that n is even such
16 00:00:49.405 –> 00:00:50.845 that the end looks like this.
17 00:00:51.705 –> 00:00:53.725 Now since you know all the calculation rules,
18 00:00:53.865 –> 00:00:56.645 you immediately know this is the same sum.
19 00:00:57.515 –> 00:00:59.095 We don’t change the result
20 00:00:59.095 –> 00:01:01.895 of the sum when we just reorder the terms.
21 00:01:02.675 –> 00:01:06.445 However you already know an infinite sum a series
22 00:01:06.985 –> 00:01:08.685 is different from a finite sum.
23 00:01:09.565 –> 00:01:12.675 There a reordering can indeed change the value
24 00:01:13.815 –> 00:01:14.915 in order to see this.
25 00:01:15.045 –> 00:01:16.995 Let’s immediately look at next sample.
26 00:01:17.855 –> 00:01:19.755 So let’s take this well known series
27 00:01:19.845 –> 00:01:21.875 where we add plus and minus one.
28 00:01:22.565 –> 00:01:25.875 There we have already shown that it is not convergent.
29 00:01:26.455 –> 00:01:28.545 However, we can show that the sequence
30 00:01:28.545 –> 00:01:31.665 of partial sums has two different accumulation values
31 00:01:32.545 –> 00:01:35.605 and in this case there are zero and one.
32 00:01:36.365 –> 00:01:37.895 This is indeed easy to see
33 00:01:37.895 –> 00:01:40.775 because we start with one, then we add minus one,
34 00:01:40.875 –> 00:01:43.935 so we are at zero, then at one again, then at zero again.
35 00:01:44.115 –> 00:01:46.455 So we alternate between both values here.
36 00:01:47.375 –> 00:01:50.115 Now what we will do is we will reorder the series
37 00:01:50.495 –> 00:01:52.955 and we will get different accumulation values.
38 00:01:54.055 –> 00:01:55.795 We simply start with the one again,
39 00:01:56.055 –> 00:01:58.995 but then I want to exchange these terms here.
40 00:01:59.975 –> 00:02:03.505 Okay, so what we get is simply this new series here.
41 00:02:04.515 –> 00:02:06.455 The first thing you see immediately is that
42 00:02:06.455 –> 00:02:09.935 with the reordering, the series is still not convergent.
43 00:02:10.445 –> 00:02:14.135 However, we still find two different accumulation values,
44 00:02:15.115 –> 00:02:18.615 but now it’s one and two as the accumulation values.
45 00:02:19.495 –> 00:02:22.875 So the first lesson we’ve learned here is always be careful
46 00:02:23.105 –> 00:02:25.555 when you reorder an infinite series.
47 00:02:26.565 –> 00:02:30.465 Indeed, such changes we have seen here can also happen when
48 00:02:30.465 –> 00:02:32.105 we look at the convergence series.
49 00:02:33.105 –> 00:02:35.515 Therefore, let’s consider this example here
50 00:02:35.725 –> 00:02:37.875 where we already know it’s convergent
51 00:02:37.875 –> 00:02:39.315 by the Leibniz criterion.
52 00:02:40.225 –> 00:02:41.765 The test here is applicable
53 00:02:41.765 –> 00:02:43.885 because we have an alternating series
54 00:02:44.385 –> 00:02:47.325 and also here a monotonically decreasing part.
55 00:02:48.135 –> 00:02:50.355 You already know the series is given
56 00:02:50.415 –> 00:02:51.595 by a positive fraction.
57 00:02:51.745 –> 00:02:53.315 Then comes a negative fraction,
58 00:02:53.315 –> 00:02:54.995 then comes a positive fraction again,
59 00:02:55.025 –> 00:02:57.115 then a negative again and so on.
60 00:02:57.695 –> 00:02:59.725 And here we know we have a unique limit,
61 00:02:59.945 –> 00:03:02.125 so we have just one accumulation value.
62 00:03:02.675 –> 00:03:04.125 Therefore, the question here is
63 00:03:04.305 –> 00:03:06.965 how can we reorder the whole thing such
64 00:03:06.965 –> 00:03:08.045 that we change the limit?
65 00:03:08.785 –> 00:03:11.965 And I can already tell you this is indeed possible.
66 00:03:11.965 –> 00:03:14.435 For this example, what we will do is
67 00:03:14.435 –> 00:03:16.475 that we take two positive numbers
68 00:03:16.815 –> 00:03:18.835 before the next negative number comes.
69 00:03:19.805 –> 00:03:22.795 Hence first we add one over one plus one third
70 00:03:23.255 –> 00:03:24.955 and then we subtract one half.
71 00:03:25.665 –> 00:03:27.075 Then in the next step we have
72 00:03:27.075 –> 00:03:28.915 to take these two positive numbers
73 00:03:29.415 –> 00:03:31.155 and then we subtract one quarter.
74 00:03:32.075 –> 00:03:34.215 And with this rule, we just continue.
75 00:03:35.075 –> 00:03:37.535 Of course, the important thing you should note here is
76 00:03:37.535 –> 00:03:40.055 that we don’t miss any numbers involved here
77 00:03:40.395 –> 00:03:42.455 and we also don’t add other ones.
78 00:03:43.205 –> 00:03:46.945 For this reason, we call this a reordering of this series.
79 00:03:47.885 –> 00:03:48.905 Of course, don’t worry,
80 00:03:49.105 –> 00:03:51.185 I give you the exact mathematical definition
81 00:03:51.185 –> 00:03:53.945 of a reordering soon. at the moment.
82 00:03:54.045 –> 00:03:57.665 For us, it’s enough to know that this C here is convergent
83 00:03:57.665 –> 00:03:59.705 with a limit C greater than zero.
84 00:04:00.845 –> 00:04:02.385 In fact, later we will see
85 00:04:02.415 –> 00:04:06.185 that this constant C is the natural logarithm of two.
86 00:04:06.655 –> 00:04:09.145 However, also not so important right now
87 00:04:09.615 –> 00:04:12.825 because here you just have to do some suitable calculations
88 00:04:12.825 –> 00:04:13.865 with the two series
89 00:04:14.445 –> 00:04:17.665 and we get out that this series is also convergent,
90 00:04:18.535 –> 00:04:22.035 but the result, the limit is three halves times C,
91 00:04:22.695 –> 00:04:25.275 so strictly greater than the limit we had before.
92 00:04:26.075 –> 00:04:27.485 Therefore, the important result
93 00:04:27.485 –> 00:04:31.205 for you is here we have a convergence series, we reorder it
94 00:04:31.425 –> 00:04:33.165 and we get another limit out.
95 00:04:33.925 –> 00:04:36.665 The whole calculation for this is a nice exercise,
96 00:04:36.725 –> 00:04:38.225 but I don’t discuss it now
97 00:04:38.655 –> 00:04:40.865 because I want to focus on the result
98 00:04:40.975 –> 00:04:43.785 that this strange thing here cannot happen
99 00:04:43.885 –> 00:04:45.945 for an absolutely convergent series.
100 00:04:46.335 –> 00:04:49.105 Therefore, for such a series, it’s always allowed
101 00:04:49.105 –> 00:04:52.665 to reorder the whole series without changing the limit.
102 00:04:53.565 –> 00:04:56.945 But before we do this, let’s finally fix the definition
103 00:04:56.945 –> 00:04:59.065 of a reordering for this.
104 00:04:59.115 –> 00:05:00.305 Let’s take a series
105 00:05:00.525 –> 00:05:01.745 and a map tau
106 00:05:02.785 –> 00:05:05.755 that goes from the natural numbers into the natural numbers
107 00:05:06.095 –> 00:05:07.275 and is byjective.
108 00:05:08.295 –> 00:05:11.475 Of course, if you have another index set for the series,
109 00:05:11.615 –> 00:05:13.595 you would change the natural numbers here
110 00:05:13.785 –> 00:05:15.035 into this index set.
111 00:05:15.775 –> 00:05:17.995 The visualization for this map tau is
112 00:05:17.995 –> 00:05:19.555 that it is our reordering
113 00:05:19.605 –> 00:05:22.235 where we have the old indices on the right hand side
114 00:05:22.335 –> 00:05:24.355 and the new indices on the left hand side.
115 00:05:25.085 –> 00:05:27.985 For example, the new index one could correspond
116 00:05:28.005 –> 00:05:29.865 to the old index three.
117 00:05:31.005 –> 00:05:33.185 In the same way we have similar relations
118 00:05:33.245 –> 00:05:34.465 for the other indices.
119 00:05:35.135 –> 00:05:38.225 However, the only important thing here is that we have a one
120 00:05:38.245 –> 00:05:41.625 to one correspondence between the old and the new indices,
121 00:05:42.345 –> 00:05:45.245 and this is exactly given by the term by bijective.
122 00:05:46.425 –> 00:05:48.885 Now, you might already guess the new series
123 00:05:48.885 –> 00:05:52.845 where we put in tau K as the index is called a reordering
124 00:05:52.845 –> 00:05:54.285 of the original series.
125 00:05:54.965 –> 00:05:56.785 So we have the same terms involved
126 00:05:56.845 –> 00:05:58.765 but possibly in a different order,
127 00:05:59.545 –> 00:06:02.125 and this simply works by shuffling the natural numbers
128 00:06:02.315 –> 00:06:04.605 with the help of a bijective map tau.
129 00:06:05.395 –> 00:06:08.205 Okay, by having this, let’s formulate the theorem
130 00:06:08.265 –> 00:06:09.525 for this video today.
131 00:06:10.585 –> 00:06:13.605 As promised, it’s about absolutely convergence series.
132 00:06:14.585 –> 00:06:17.165 If we have such a series, then we know
133 00:06:17.165 –> 00:06:21.045 that any reordering is also absolutely convergent
134 00:06:21.725 –> 00:06:25.425 and more importantly, the values, the limits are the same.
135 00:06:26.325 –> 00:06:28.505 In other words, such a strange thing as
136 00:06:28.505 –> 00:06:30.105 before can’t happen here
137 00:06:31.035 –> 00:06:35.175 and indeed this nice result holds for any reordering.
138 00:06:35.925 –> 00:06:39.295 Okay, now I think we are able to prove such a statement,
139 00:06:40.225 –> 00:06:43.115 therefore this is what we do for the rest of the video.
140 00:06:44.055 –> 00:06:45.515 Now first, please recall
141 00:06:45.585 –> 00:06:47.875 that being absolutely convergent means
142 00:06:47.875 –> 00:06:51.595 that the series here is still convergent when you put the
143 00:06:51.835 –> 00:06:53.675 absolute value bars around a K
144 00:06:54.505 –> 00:06:56.805 and for this convergence series, I want
145 00:06:56.805 –> 00:06:58.445 to use the Cauchy criterion.
146 00:06:59.235 –> 00:07:02.085 This means that for a given epsilon greater than zero,
147 00:07:02.465 –> 00:07:06.085 the Cauchy criterion implies there exists at capital N,
148 00:07:06.175 –> 00:07:09.085 which we call N one for the moment, such that
149 00:07:09.085 –> 00:07:12.685 for all indices n, m afterwards we have
150 00:07:12.685 –> 00:07:15.885 that this finite sum here is less than epsilon.
151 00:07:16.615 –> 00:07:19.205 Using this, we can show on the one hand
152 00:07:19.555 –> 00:07:22.485 that this series is also absolutely convergent,
153 00:07:22.665 –> 00:07:23.725 and on the other hand
154 00:07:23.995 –> 00:07:27.445 that this equation holds both proofs work similarly.
155 00:07:27.475 –> 00:07:30.605 Therefore, I just show you that the limits coincide
156 00:07:31.075 –> 00:07:33.685 because I think that is just more interesting.
157 00:07:34.355 –> 00:07:36.445 More concretely, this means we want to show
158 00:07:36.445 –> 00:07:39.885 that the distance from the limit to the partial sums
159 00:07:40.495 –> 00:07:43.845 where we have tau K here is less than epsilon.
160 00:07:44.375 –> 00:07:46.965 Maybe it’s a good idea to use an abbreviation here.
161 00:07:47.015 –> 00:07:49.285 Let’s call this limit capital A.
162 00:07:49.985 –> 00:07:51.365 So in the next step I want
163 00:07:51.365 –> 00:07:53.725 to bring in an expression we can control
164 00:07:54.145 –> 00:07:56.645 and then I want to use the triangle inequality,
165 00:07:57.385 –> 00:08:01.165 and the term we use is the finite sum of a K where we start
166 00:08:01.165 –> 00:08:03.605 with one and go to N one minus one.
167 00:08:04.355 –> 00:08:07.855 We do this because we already know by the Cauchy criterion
168 00:08:08.045 –> 00:08:11.575 that this term is already close to A, okay?
169 00:08:11.605 –> 00:08:14.575 Then the triangle inequality makes this less
170 00:08:14.675 –> 00:08:16.015 or equal than two terms.
171 00:08:16.725 –> 00:08:17.775 Okay? I’ve told you
172 00:08:17.775 –> 00:08:19.175 before for the first term,
173 00:08:19.355 –> 00:08:21.935 we already know it’s less than epsilon
174 00:08:21.935 –> 00:08:23.175 by the Cauchy criterion.
175 00:08:23.885 –> 00:08:25.175 More concretely, we see
176 00:08:25.175 –> 00:08:28.535 that difference here gives us this infinite sum,
177 00:08:29.155 –> 00:08:31.015 and if you write the limit in front,
178 00:08:31.155 –> 00:08:35.095 you can see here we can use the triangle inequality again.
179 00:08:35.765 –> 00:08:37.695 Therefore, in fact, for this term,
180 00:08:37.995 –> 00:08:40.975 we already know it’s less than epsilon no matter
181 00:08:40.975 –> 00:08:42.175 how large N is.
182 00:08:43.085 –> 00:08:46.795 Hence in summary, the first term here is no problem at all.
183 00:08:47.145 –> 00:08:50.115 However, for the second term, we have to work a little bit
184 00:08:50.185 –> 00:08:52.475 because here tau is involved.
185 00:08:53.265 –> 00:08:54.515 Okay, we need some space,
186 00:08:54.515 –> 00:08:57.285 therefore let’s call this part just star.
187 00:08:58.325 –> 00:09:00.265 In order to deal with this, we have
188 00:09:00.265 –> 00:09:02.625 to define a suitable capital N.
189 00:09:03.365 –> 00:09:06.545 Why that there the picture for tau might help Again,
190 00:09:07.285 –> 00:09:10.505 so you should see here in the second sum, we go
191 00:09:10.505 –> 00:09:13.905 through all the tau K, where K goes from one
192 00:09:13.925 –> 00:09:15.065 to an arbitrary N.
193 00:09:15.415 –> 00:09:17.025 However, in the first sum,
194 00:09:17.285 –> 00:09:20.545 we always stop at the index N one minus one.
195 00:09:21.195 –> 00:09:24.305 Hence, it would be nice if we would subtract all the terms
196 00:09:24.375 –> 00:09:25.905 here with the second sum,
197 00:09:26.685 –> 00:09:29.265 but this means with the tau map here, we have
198 00:09:29.265 –> 00:09:34.025 to hit all the indices until we reach N one minus one, and
199 00:09:34.025 –> 00:09:35.625 because the map is bijective,
200 00:09:35.965 –> 00:09:38.425 we already know this will eventually happen.
201 00:09:39.375 –> 00:09:42.985 Also, it’s no problem at all that maybe the four is mapped
202 00:09:42.985 –> 00:09:45.745 to a number that comes after N one minus one.
203 00:09:46.555 –> 00:09:47.855 We just continue further
204 00:09:48.185 –> 00:09:51.815 until all the points on the right hand side we want are hit,
205 00:09:52.435 –> 00:09:54.495 and the last index on the left hand side,
206 00:09:54.755 –> 00:09:56.015 we need for that to happen.
207 00:09:56.395 –> 00:09:57.975 We just call capital N.
208 00:09:58.675 –> 00:10:01.805 Therefore, for all indices n, which are greater
209 00:10:01.865 –> 00:10:05.245 or equal than this capital N, we can consider the set
210 00:10:05.245 –> 00:10:10.165 of the images tau one, tau two until we reach tau N,
211 00:10:10.985 –> 00:10:14.565 and this is always a super set of the set given
212 00:10:14.625 –> 00:10:18.845 by the indices one, two until N one minus one.
213 00:10:19.465 –> 00:10:21.645 In other words, at least we hit all
214 00:10:21.645 –> 00:10:25.545 of these indices on the right hand side for our part star.
215 00:10:25.695 –> 00:10:29.185 This means we completely eliminate this sum here,
216 00:10:29.885 –> 00:10:32.815 therefore only terms in this sum remain.
217 00:10:33.325 –> 00:10:37.255 However, there we have the question, what are the indices
218 00:10:37.255 –> 00:10:39.935 that remain before answering
219 00:10:39.965 –> 00:10:42.495 that we should immediately simplify the whole expression
220 00:10:42.715 –> 00:10:44.695 by using the triangle inequality.
221 00:10:44.785 –> 00:10:47.165 Again, this is indeed simpler
222 00:10:47.165 –> 00:10:50.285 because now we just add up non-negative terms.
223 00:10:51.025 –> 00:10:53.045 We still don’t know the indices here,
224 00:10:53.345 –> 00:10:56.565 but now you should see we don’t need to know them exactly
225 00:10:57.115 –> 00:11:01.005 because we already know that this A does not have any index
226 00:11:01.075 –> 00:11:02.805 that is less than N one.
227 00:11:03.735 –> 00:11:07.125 Hence, when we write a J, we know that we start
228 00:11:07.125 –> 00:11:08.965 with J is equal to N one.
229 00:11:09.585 –> 00:11:12.485 Of course, the start index could be larger than N one,
230 00:11:12.665 –> 00:11:14.525 but we just want an estimate here,
231 00:11:14.945 –> 00:11:17.885 and for this reason we can also put infinity
232 00:11:17.985 –> 00:11:19.005 at the end index.
233 00:11:19.705 –> 00:11:21.765 Of course, this is just a rough estimate
234 00:11:21.825 –> 00:11:26.125 for the real number, but we already know this part here is
235 00:11:26.125 –> 00:11:30.085 less than epsilon and obviously we don’t need more than that
236 00:11:30.975 –> 00:11:34.905 because now we have both terms here, less than epsilon,
237 00:11:35.615 –> 00:11:38.525 hence the whole sum is less than two epsilon,
238 00:11:39.105 –> 00:11:41.365 and as usual, we can call this two times
239 00:11:41.635 –> 00:11:43.365 epsilon epsilon prime.
240 00:11:44.165 –> 00:11:45.255 Therefore, in summary,
241 00:11:45.645 –> 00:11:48.535 this whole calculation shows the convergence property.
242 00:11:49.305 –> 00:11:52.435 This means that for all epsilon Prime created than zero,
243 00:11:52.435 –> 00:11:56.085 we find a capital N such that for all indices afterwards,
244 00:11:56.395 –> 00:11:58.965 this difference is less than epsilon prime.
245 00:11:59.625 –> 00:12:03.245 So here we have the limit for this sequence of partial sums,
246 00:12:03.905 –> 00:12:05.445 and this is what we wanted to show.
247 00:12:05.745 –> 00:12:08.805 The reordering has the same limit as the series before.
248 00:12:09.635 –> 00:12:13.165 Okay, so this closes our proof, okay,
249 00:12:13.225 –> 00:12:15.085 and then in the next video I show you
250 00:12:15.105 –> 00:12:17.605 how we can immediately apply this nice fact.
251 00:12:18.255 –> 00:12:21.665 Therefore, I hope I see you there and have a nice day. Bye.
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Quiz Content
Q1: Is the series $\displaystyle \sum_{k = 1}^\infty (-1)^{k+1} \frac{1}{k}$ absolutely convergent?
A1: Yes, it is.
A2: No, it isn’t.
Q2: When do we call a series $\displaystyle \sum_{k = 1}^\infty a_{\tau(k)}$ a reordering of the series $\displaystyle \sum_{k = 1}^\infty a_k$?
A1: It’s called a reordering if $\tau : \mathbb{N} \rightarrow \mathbb{N}$ is map.
A2: It’s called a reordering if $\tau :\mathbb{N} \rightarrow \mathbb{N}$ is a injective map.
A3: It’s called a reordering if $\tau :\mathbb{N} \rightarrow \mathbb{N}$ is a surjective map.
A4: It’s called a reordering if $\tau :\mathbb{N} \rightarrow \mathbb{N}$ is a bijective map.
A5: It’s called a reordering if $\tau :\mathbb{N} \rightarrow \mathbb{N}$ is a bijective map with $\tau(1) = 1$.
Q3: Can one change the value of a convergent series with a reordering.
A1: Yes, that is always possible.
A2: No, this is never possible.
A3: Yes, this is possible but only if the series is not absolutely convergent.
Q4: Let $q \in \mathbb{R}$ with $|q| < 1$. Can one change the limit of the series $\displaystyle \sum_{k = 0}^\infty q^k$ with a reordering?
A1: Yes, this is always possible.
A2: No, this is not possible.
A3: It possible for $q < 0$ but not for $q \geq 0$.
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Last update: 2025-09
