
Title: Bounded Sequences and Unique Limits

Series: Real Analysis

Chapter: Sequences and Limits

YouTubeTitle: Real Analysis 3  Bounded Sequences and Unique Limits

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Quiz: Test your knowledge

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Subtitle on GitHub: ra03_sub_eng.srt

Timestamps
00:00 Intro
00:15 An example for showing that a sequence is divergent
03:41 Definition of a bounded sequence
04:36 A convergent sequence is also bounded
06:09 The limit of a convergent sequence is uniquely given
08:59 Outro

Subtitle in English
1 00:00:00,410 –> 00:00:03,410 Hello and welcome back to real analysis
2 00:00:04,170 –> 00:00:09,500 and as always many, many thanks to all the nice people that support this channel on Steady or Paypal.
3 00:00:09,990 –> 00:00:14,590 In todays part 3 i will tell you more things about convergence sequences.
4 00:00:15,240 –> 00:00:19,000 For this i think a good starting point would be to consider an example.
5 00:00:19,840 –> 00:00:22,040 Here i have the sequence (an)
6 00:00:22,250 –> 00:00:25,250 which is given by 1 to the power n.
7 00:00:25,640 –> 00:00:28,740 So, this is an example we had in the last video,
8 00:00:28,750 –> 00:00:32,800 but now we will show, that this sequence is not convergent.
9 00:00:33,470 –> 00:00:36,960 Or in other words: We prove that the sequence is divergent.
10 00:00:37,590 –> 00:00:39,790 Now the prove will work by contradiction.
11 00:00:39,790 –> 00:00:43,180 Which means we will assume that the sequence has a limit “a”.
12 00:00:44,090 –> 00:00:46,490 So this could be any number on the number line,
13 00:00:46,510 –> 00:00:48,800 where the sequence gets closer and closer to.
14 00:00:49,360 –> 00:00:51,440 However, if you look at the number line
15 00:00:51,450 –> 00:00:53,950 you already see that this can’t happen.
16 00:00:54,530 –> 00:00:57,430 Because the sequence jumps between 1 and 1.
17 00:00:58,420 –> 00:01:01,020 Therefore it does not matter were we set the “a”
18 00:01:01,370 –> 00:01:05,750 You always find infinitely many points that are far away from “a”
19 00:01:06,710 –> 00:01:10,000 and of course this should not happen for a convergent sequence.
20 00:01:10,990 –> 00:01:14,780 Hence for us it now remains to put this idea into a formal way.
21 00:01:15,840 –> 00:01:17,750 So a good idea at the beginning would be
22 00:01:17,770 –> 00:01:20,860 to write down the explicit definition of convergence.
23 00:01:21,510 –> 00:01:24,410 Which reads that for every small number epsilon
24 00:01:24,430 –> 00:01:26,430 we find an index “N”
25 00:01:26,460 –> 00:01:29,250 such that for all indices afterwards
26 00:01:29,270 –> 00:01:34,050 we have that the distance “an” to “a” is less than a given epsilon.
27 00:01:34,860 –> 00:01:40,640 Now what we can put in is that the distance between 1 and 1 is exactly 2.
28 00:01:41,500 –> 00:01:45,380 Ok, now the idea is that we have to choose epsilon so small
29 00:01:45,390 –> 00:01:47,860 that in the epsilonneighbourhood around “a”
30 00:01:47,870 –> 00:01:50,870 we don’t have 1 and 1 at the same time.
31 00:01:51,370 –> 00:01:54,470 This means we need a number smaller or equal than 1
32 00:01:54,510 –> 00:01:56,210 so lets choose 1.
33 00:01:57,000 –> 00:02:00,900 Now by definiton for this epsilon we find a “N”.
34 00:02:01,830 –> 00:02:07,310 Therefore we now know, the distance from “an” to “a” is less than our given epsilon.
35 00:02:07,730 –> 00:02:12,100 and of course we have also this result for the next member in the sequence.
36 00:02:12,790 –> 00:02:16,260 Now you know this holds for infinitely many sequence members,
37 00:02:16,270 –> 00:02:17,800 but we only need these two.
38 00:02:18,350 –> 00:02:22,740 Because we know one of the two indices is even and the other is odd.
39 00:02:23,600 –> 00:02:25,130 Therefore we get both outcomes
40 00:02:25,140 –> 00:02:26,829 1 and 1.
41 00:02:27,670 –> 00:02:30,000 With this you might already see the problem we have
42 00:02:30,030 –> 00:02:33,530 we measure two distances, but both are less than 1,
43 00:02:33,620 –> 00:02:36,820 but the distance between 1 and 1 is still 2.
44 00:02:37,440 –> 00:02:40,630 Therefore this should contradict the triangle inequality.
45 00:02:41,390 –> 00:02:42,800 Ok, then lets write that down.
46 00:02:43,030 –> 00:02:45,630 First we note, when we add both distances here
47 00:02:45,670 –> 00:02:47,760 we get a number that is less than 2
48 00:02:48,320 –> 00:02:50,110 and on the other hand we already know
49 00:02:50,110 –> 00:02:52,910 the distance between 1 and 1 is exactly 2.
50 00:02:54,010 –> 00:02:57,900 Ok and now the idea is to go the detour with “a” in the middle.
51 00:02:58,650 –> 00:03:02,050 In other words i want to get “a” into the absolute value here.
52 00:03:02,900 –> 00:03:07,800 So lets write “ a + a” here and use the triangle inequality
53 00:03:08,670 –> 00:03:09,560 and there you should see
54 00:03:09,560 –> 00:03:13,160 this is exactly the same we have written at the right hand side before.
55 00:03:13,900 –> 00:03:18,900 Which immediately leads to our contradiction, because we have 2 is less than 2.
56 00:03:19,730 –> 00:03:23,720 So this is not correct, therefore the assumption has to be false.
57 00:03:24,400 –> 00:03:27,600 Or in other words. This sequence is indeed divergent.
58 00:03:28,500 –> 00:03:33,900 Ok, so here we have seen a sequence which does not have the nice property of convergence,
59 00:03:33,940 –> 00:03:35,990 but still it has some properties.
60 00:03:36,730 –> 00:03:40,860 and one important one is that this sequence is a bounded sequence.
61 00:03:41,510 –> 00:03:44,600 Therefore now lets define what a bounded sequence is.
62 00:03:45,720 –> 00:03:48,910 For this let “an” be any sequence of real numbers
63 00:03:48,940 –> 00:03:50,940 and then we call it bounded
64 00:03:51,830 –> 00:03:54,450 If there exists a real number “C”
65 00:03:55,290 –> 00:03:57,990 such that for all sequence members we have
66 00:03:58,540 –> 00:04:03,000 that the distance “an” to 0 is less or equal than “C”
67 00:04:03,760 –> 00:04:05,760 You can visualize this on a number line
68 00:04:05,770 –> 00:04:08,970 by putting in two bounds: “ C” and “C”
69 00:04:09,530 –> 00:04:13,330 and then the whole sequence “an” just lives inside.
70 00:04:14,030 –> 00:04:16,420 and of course if we don’t find such a “C”
71 00:04:16,420 –> 00:04:18,820 we just call the sequence unbounded.
72 00:04:19,410 –> 00:04:22,480 An example of a bounded sequence you have already seen
73 00:04:22,540 –> 00:04:24,240 in the example from the beginning.
74 00:04:25,100 –> 00:04:27,440 There we can just set the “C” to be 1.
75 00:04:28,120 –> 00:04:30,120 and then you see, the picture fits in.
76 00:04:30,990 –> 00:04:33,590 However, we have even seen more examples,
77 00:04:33,650 –> 00:04:35,940 because we have the following important fact.
78 00:04:36,600 –> 00:04:41,600 If we have a convergent sequence we can conclude it’s also bounded.
79 00:04:41,970 –> 00:04:44,040 However, not the other way around
80 00:04:44,050 –> 00:04:46,440 as we have seen in the first example today.
81 00:04:47,020 –> 00:04:50,650 Another important fact you should remember here is the contraposition.
82 00:04:51,460 –> 00:04:54,000 Which means if we have an unbounded sequence
83 00:04:54,020 –> 00:04:56,560 we already know it can’t be convergent
84 00:04:57,290 –> 00:04:59,290 So lets prove this important fact here.
85 00:04:59,290 –> 00:05:01,000 Which is indeed not so hard.
86 00:05:01,520 –> 00:05:05,100 For this just recall again the definition of convergence.
87 00:05:05,900 –> 00:05:08,480 So you know there exists a limit we call “a”
88 00:05:08,480 –> 00:05:12,180 such that we have something we can visualize with this picture here.
89 00:05:12,690 –> 00:05:17,280 We have that almost all points lie in the epsilonneighbourhood around “a”
90 00:05:17,920 –> 00:05:20,290 Therefore to get something like in the picture above
91 00:05:20,290 –> 00:05:25,120 we simply have to define “C” or “ C” to be “a + epsilon”
92 00:05:25,140 –> 00:05:29,000 or to be the member that is the farthest away from this epsilonneighbourhood.
93 00:05:29,560 –> 00:05:30,860 To put it into a formular
94 00:05:30,860 –> 00:05:35,580 we just define “C” to be the maximum out of these finitely many numbers.
95 00:05:36,380 –> 00:05:38,360 And if we take the absolute values
96 00:05:38,370 –> 00:05:40,370 we don’t have a problem with the sign.
97 00:05:40,390 –> 00:05:43,070 We just get out “C” as a positive number.
98 00:05:43,650 –> 00:05:46,640 and if we write “a + epsilon”
99 00:05:46,820 –> 00:05:50,510 we get out a bound here, that is the farthest away from 0.
100 00:05:50,970 –> 00:05:52,970 It could be this one or this one.
101 00:05:53,060 –> 00:05:55,659 Depending where 0 lies here on the number line.
102 00:05:56,460 –> 00:05:58,720 An important thing to note here is that “C”
103 00:05:58,730 –> 00:06:00,660 obviously has to be positive,
104 00:06:00,670 –> 00:06:03,670 but we don’t have to find the minimal possible “C”
105 00:06:04,000 –> 00:06:05,400 We just have to find a “C”.
106 00:06:05,920 –> 00:06:08,320 and that’s what we get with this definition here.
107 00:06:09,090 –> 00:06:12,970 Knowing this lets get to another important fact about sequences.
108 00:06:13,460 –> 00:06:17,740 You know if you consider a convergent sequence, you find such a limit “a”,
109 00:06:17,780 –> 00:06:20,460 but this “a” is indeed uniquely given.
110 00:06:21,260 –> 00:06:25,150 Therefore we are able to speak of THE limit of the convergent sequence.
111 00:06:25,950 –> 00:06:29,430 Moreover the notation “limit an” makes sense.
112 00:06:30,050 –> 00:06:33,350 So lets end this video by proving this important fact.
113 00:06:33,980 –> 00:06:38,280 Lets assume that we have two different limits “a” and “ã”
114 00:06:38,610 –> 00:06:43,210 This means the distance between both of them is greater than 0.
115 00:06:43,750 –> 00:06:46,110 Now, if you visualize this on the number line
116 00:06:46,130 –> 00:06:48,520 you already see, this can’t happen
117 00:06:49,470 –> 00:06:51,990 because you know, for an arbitrary epsilon
118 00:06:52,000 –> 00:06:56,300 the sequence members lie inside the epsilonneighbourhood eventually.
119 00:06:56,830 –> 00:06:59,820 and this works for “a” and " ã " as well.
120 00:07:00,490 –> 00:07:04,000 Which is clearly a contradiction when epsilon is small enough.
121 00:07:04,300 –> 00:07:06,910 Ignoring finitely many sequence members
122 00:07:06,920 –> 00:07:10,520 you can’t have all sequence members inside this epsilonneighbourhood
123 00:07:10,530 –> 00:07:11,560 and in this one.
124 00:07:12,380 –> 00:07:15,570 This clearly contradicts the definition of convergence.
125 00:07:16,210 –> 00:07:18,210 In this sense this picture says it all.
126 00:07:18,220 –> 00:07:20,770 Now we just have to put this into formulars.
127 00:07:21,260 –> 00:07:22,630 Or in other words
128 00:07:22,650 –> 00:07:26,100 choose a small enough epsilon such that this picture works.
129 00:07:26,740 –> 00:07:30,240 You see it has to be less than the distance between “a” and " ã "
130 00:07:30,260 –> 00:07:32,750 so lets choose a quarter of this distance.
131 00:07:33,300 –> 00:07:36,100 Therefore this is our epsilon now.
132 00:07:37,020 –> 00:07:39,409 Now by the definition of the convergence
133 00:07:39,410 –> 00:07:42,700 we know for this epsilon we find a “N”.
134 00:07:43,230 –> 00:07:47,830 and then we have that all sequence members after “N” fulfill this.
135 00:07:48,550 –> 00:07:52,350 However, of course by assumption we have the same for “ã”
136 00:07:53,040 –> 00:07:57,540 and the “N” we find for “ã” we can just call “Ñ”
137 00:07:58,270 –> 00:08:01,410 Therefore now we can just choose a “n”
138 00:08:01,420 –> 00:08:04,220 which is greater than “N” and “Ñ”
139 00:08:04,810 –> 00:08:08,210 Hence we simply set it as greater or equal than the maximum.
140 00:08:08,740 –> 00:08:12,440 and then of course we want to use the triangle inequality again.
141 00:08:12,890 –> 00:08:17,280 This means between “a” and " ã " we put in our “an”
142 00:08:17,900 –> 00:08:20,680 and then we just split that into our two parts.
143 00:08:21,520 –> 00:08:25,150 and at this point we know by the assumption of the two limits
144 00:08:25,160 –> 00:08:27,700 that this one is less than epsilon
145 00:08:27,760 –> 00:08:29,760 and this one
146 00:08:29,960 –> 00:08:33,549 Now using the definition of epsilon we now get
147 00:08:34,110 –> 00:08:37,409 that the left hand side is less than the right hand side.
148 00:08:37,539 –> 00:08:39,730 which is one half of the left hand side.
149 00:08:40,260 –> 00:08:42,620 However, we have positive numbers here.
150 00:08:42,640 –> 00:08:44,130 So this can’t happen.
151 00:08:45,060 –> 00:08:46,640 and with this we have the whole proof.
152 00:08:46,830 –> 00:08:49,620 Which was indeed just the picture i showed you before.
153 00:08:50,080 –> 00:08:54,180 Of course this fact here is what we will use for the whole course.
154 00:08:55,010 –> 00:08:58,800 Therefore please remember a convergent sequence has a unique limit.
155 00:08:59,670 –> 00:09:01,760 With this i hope to see you in the next video.
156 00:09:01,850 –> 00:09:03,850 Have a nice day and Bye!

Quiz Content
Q1: Which of the following sequences is convergent?
A1: $( (1)^n )_{n \in \mathbb{N}}$
A2: $( (2)^n )_{n \in \mathbb{N}}$
A3: $( (1)^n \frac{1}{n} )_{n \in \mathbb{N}}$
A4: $( 2^n )_{n \in \mathbb{N}}$
Q2: What is the correct definition of a bounded sequence $( a_n )_{n \in \mathbb{N}}$?
A1: $\forall \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \leq N ~:~ a_n  a < \varepsilon$
A2: $\exists C \in \mathbb{R} ~~ \forall n \in \mathbb{N} ~:~ a_n \leq C$
A3: $\exists C \in \mathbb{R} ~~ \forall n \in \mathbb{N} ~:~ a_n \geq C$
A4: $\exists \varepsilon > 0 ~~ \exists N \in \mathbb{N} ~~ \forall n \leq N ~:~ a_n  a < \varepsilon$
Q3: Which of the sequences is unbounded?
A1: $( (1)^n )_{n \in \mathbb{N}}$
A2: $( n^2 )_{n \in \mathbb{N}}$
A3: $( \frac{1}{n})_{n \in \mathbb{N}}$
A4: $( 1 )_{n \in \mathbb{N}}$
Q4: Which of these implications is false?
A1: $( a_n ){n \in \mathbb{N}}$ convergent $\Rightarrow$ $( a_n ){n \in \mathbb{N}}$ bounded.
A2: $( a_n ){n \in \mathbb{N}}$ unbounded $\Rightarrow$ $( a_n ){n \in \mathbb{N}}$ divergent.
A3: $( a_n ){n \in \mathbb{N}}$ bounded $\Rightarrow$ $( a_n ){n \in \mathbb{N}}$ convergent.
A4: $( a_n ){n \in \mathbb{N}}$ not bounded $\Rightarrow$ $( a_n ){n \in \mathbb{N}}$ not convergent.
A5: $( a_n ){n \in \mathbb{N}}$ not divergent $\Rightarrow$ $( a_n ){n \in \mathbb{N}}$ bounded.