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Monotone Convergence Theorem (MCT) with Examples — Transcript

by Dr. Harish Garg · 2,537 words · 436 segments · language en · Watch on YouTube

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0:00Hello students. Welcome to the next
0:03lecture on the sequence. Today, I will
0:06explain you the concept of the monotone
0:10convergence theorem. Myself, Dr. Harish
0:12Kar, you can follow and subscribe my
0:15YouTube channel where you can find the
0:17various lectures related to the sequence
0:19and its applications such as limit of
0:22the sequence, behavior of the sequence,
0:2570 plus examples, squeeze principle,
0:28monotonic, bounded, Cauchy first
0:30theorem, second theorem, and many more
0:32lectures available in this single place.
0:35Today's, as I said you, the today's
0:37objective is the monotone convergence
0:40theorem. In this lecture, we will learn
0:43a very simple proof of the monotone
0:47convergence theorem, the idea
0:50behind this monotone convergence
0:52theorem, and various examples so that
0:56you can understand how this monotone
0:59convergence theorem applicable.
1:02First of all, if I say AN is my any
1:05sequence which is a convergent.
1:09Fine? Then, we all knows every
1:13convergent sequence must be bounded.
1:17But, we also know the converse is not
1:21true.
1:23Any example that you know?
1:25Any example where the is a bounded
1:27sequence, but it's not a convergent?
1:30Yes, minus one raised to power n.
1:34Clearly say, AN is my bounded sequence.
1:37But, it is not
1:40convergent. Why it's not a convergent?
1:43Because the limit of AN as n approaches
1:48infinity is not unique.
1:52If it's not unique, then we can say
1:53that we can say it is not convergent.
1:57Now, so what does it means? That implies
2:01the boundedness
2:03doesn't implies
2:05it to be convergent.
2:08You get it clear what I have to say? So,
2:10our target is to make something to be
2:15added
2:16bounded we have to get added curry so
2:19that it will be convergent. So, our
2:23target is to identify this question
2:27mark.
2:28What should be added in the boundedness
2:32so that the sequence will be convergent?
2:36Yeah? Now, there are the two things
2:39that's question mark, it may be
2:42monotonic sequence or not. So, our
2:46target is to find the condition on the
2:49bounded sequence which ensure us the
2:52sequence is convergent.
2:55So, we will consider the two different
2:57cases. The first case is
2:59the sequence is my monotonic.
3:03Fine? And the second case is sequence is
3:06not monotonic.
3:08Assume that firstly we will case discuss
3:11the case when the sequence is not
3:15monotonic.
3:16Fine? So, it means can you think about
3:19any example
3:21whose sequence is my bounded
3:24plus not monotonic? Can you say it's a
3:28convergent sequence?
3:30Answer is no.
3:32Why? Because if you look about that
3:35minus one raised to power n it is a
3:39bounded?
3:40Yes, it is a bounded. Again, it is not
3:44monotonic.
3:45Fine? Then, you can say it is not
3:50convergent. [clears throat]
3:52So, what does it means? If I add the
3:54condition of the non-monotonic, still we
3:57have the one counterexample
4:01which disprove our result. So, that
4:05means the non- The second case, case
4:07number two, non-monotonic sequence
4:10may be accepted.
4:12Fine? If the sequence is a
4:14non-monotonic, then the boundedness
4:16doesn't implies for the convergent.
4:20On the other hand, can you say
4:23can you say if I consider the monotonic
4:26sequence?
4:28Can you say if I take the monotonic
4:30sequence as well as the bounded sequence
4:33that implies the convergent? The answer
4:36is yes. And that is the condition we
4:40have imposed on the boundedness is any
4:43sequence
4:44which is a bounded
4:46plus monotonic, then you can say it's a
4:51convergent.
4:52So, for more detail about the monotonic
4:54sequence, you can watch my previous
4:56lectures, monotonic sequence and the
4:58bounded sequence. If I quickly recall
5:01you, any sequence is said to be
5:03monotonic increasing whenever an plus
5:06one is greater than equal to an for all
5:09n. And whenever it's a less than, it is
5:12said to be monotonic decreasing.
5:15Any sequence an is said to be monotonic
5:18if it is either the monotonic increasing
5:21or decreasing.
5:23So, we can say the monotonic convergence
5:27theorem will states about every bounded
5:31sequence
5:33plus monotonic sequence
5:37that implies
5:39convergent.
5:40Any bigger sequence bounded be whole or
5:43monotonic be whole. Like you can say one
5:46by n.
5:48That sequence is my monotonic? Yes.
5:52It is a monotonic decreasing sequence.
5:55Fine. This sequence is also my bounded.
5:58Fine. That implies the sequence 1/n is
6:03my convergent.
6:05Okay,
6:06any sequence that is bounded and
6:09monotonic,
6:10that sequence is called as the
6:12convergent sequence. Now, since we are
6:15talking about the monotonic, we have the
6:17two different forms. One is a monotonic
6:20increasing and monotonic decreasing.
6:22Therefore, the monotonic convergence
6:25theorems has two different forms.
6:28The first one is, if the sequence is a
6:31monotonically increasing and bounded,
6:34then it converges to the supremum of the
6:38A N.
6:39However, if the sequence is a
6:41monotonically decreasing, then it
6:43converges to the infimum of A N.
6:47In other words, the monotonic
6:49convergence theorem says any sequence A
6:52N
6:54is increasing
6:55and bounded above, then we can say it's
6:58a convergent. On the other hand, if the
7:01sequence is my decreasing and bounded
7:04below, then we call it as the
7:06convergent.
7:07For more details about the boundedness,
7:10bounded below, and bounded above, you
7:12can watch my this lecture bounded
7:14sequence.
7:23Now,
7:24a very simple state and proof of the
7:26monotone convergence theorem. So, I can
7:29prove for the increasing sequence. You
7:32can similarly prove for the decreasing
7:34sequence. If A N is a monotonic
7:37increasing sequence and bounded from
7:40above, then A N is my convergent to the
7:44L, where L is the supremum of A.
7:49The proof is a very, very simple. Let's
7:51start with the proof.
7:54Suppose AN is a monotonic increasing and
7:56bounded from above. So, what does the
7:59meaning of the increasing? That implies
8:02AN + 1 is greater than equal to AN for
8:06all N.
8:08Clear? Now, I can suppose L
8:12is the supremum of AN.
8:16Fine? So, what is your target? Your
8:19target is to prove AN converges to the
8:24L.
8:26Now, look at that.
8:28What is the definition of the supremum?
8:30It's a matter of
8:31Yeah, we sequence A1, A2, A3, A4, A5,
8:35and so on. So, you can get a lot of
8:37bounded above here.
8:39It's a bounded from above. So, that will
8:41be my
8:43supremum. That is the L. It's a matter
8:45of
8:45Sorry, values you have all is L say to
8:47me only. That means, if I assume any
8:51epsilon which is greater than zero,
8:53by the definition of the supremum of AN,
8:58what does it means? L minus epsilon must
9:01be less than of the
9:07and
9:08by the definition of the supremum, L
9:11minus epsilon is
9:13not the upper bound.
9:16Why? Because I have got L upper bound
9:18after So, L minus epsilon is not the
9:22upper bound.
9:23So, once the L minus epsilon is not the
9:26upper bound, what does it means? There
9:28exists some natural number N
9:31such that A of N must be greater than of
9:36L minus epsilon. But love is to be made
9:38in a way that I can put ek aur sequences
9:40milengi aapko.
9:42Fine? And that will be L A N of L minus
9:46epsilon.
9:47Now, since our target is to prove A N
9:52minus L is less than epsilon for all N
9:57to yahan pe maine capital N liya hai.
9:59Fine?
10:00Now, since I can start from here.
10:03And sequence aapke paas kya hai is a
10:06increasing sequence to agar sequence
10:08increasing hai that means A N must be
10:12greater than of the A of capital N.
10:15Yahi hoga. Fine? I can write also
10:19A of capital N is less than equal to
10:23A N. Fine? Now, because A N is greater
10:28than of
10:30L minus epsilon.
10:32Fine? And A N because L is my supremum,
10:36so I can say that number is less than of
10:39the L and L is always be less than of
10:42the L plus epsilon because epsilon is my
10:45positive number. So, what does it means?
10:49L minus epsilon is less than of the A N
10:52is less than of the L plus epsilon. What
10:55does it means?
10:57A N minus L is less than epsilon
11:01whenever N is greater than equal to
11:04capital N. What does it means? A N
11:07converges to the L is the required proof
11:11of the result.
11:13Fine? Therefore, the sequence A N is my
11:18convergent sequence. To aap dekho yahan
11:21pe ek this is the concept [laughter] of
11:23the supremum. And this is the concept of
11:26the increasing sequence.
11:30Similarly, you can prove for the
11:32monotonically decreasing sequence. So,
11:34what is that? Instead of the infimum
11:36supremum, you can take it as the
11:38infimum.
11:39Instead of the greater than sign, you
11:40can take it as a less than sign. Instead
11:43of the L minus epsilon, you can take it
11:45as a L plus epsilon. You have a less
11:47than sign I think.
11:48You can change the inequality and get
11:50the proof in a simple manner.
11:53But, a very, very important warning to
11:56all of you is
11:57by using the monotone convergence
11:59theorem, we
12:01can't conclude the sequence converges to
12:04the bound.
12:06For example,
12:08I don't make sequence little one by N.
12:11Fine. We know
12:13what are their bounds?
12:15We know the sequence is AN.
12:19Fine. Is it strictly greater than
12:20equality than you have you have you have
12:21to pass. Also,
12:24also we know minus of one, minus of
12:28four, minus four by three, fine, minus
12:3117, etc. are also the lower bound
12:35because the sequence is decreasing.
12:39You are decreasing sequence and you have
12:41to pass. So, are also the lower bounds.
12:44Fine. That is you can say it is bounded
12:48below.
12:50So, it's kind of a matter of care.
12:52If I say the sequence AN is bounded
12:54below by minus one, minus 14, minus four
12:57by three, minus 17, it doesn't implies
13:01that the sequence AN converges to minus
13:05four by three or minus 17. You don't
13:09need to carry on.
13:11Fine. That's the important warning.
13:13Warning that you have to pass. Sequence
13:15will converges to the infimum of the AN.
13:19So, if you have to stop by infimum they
13:20go to get out of there. Zero. Although
13:23equality you have to include in here.
13:25So, AN
13:27converges to the zero.
13:29converges to the zero.
13:32So, because AN is a monotonically
13:34decreasing sequence, which is bounded
13:36below by minus 17 as well, but it does
13:39not converges to the minus 17. Make sure
13:43the sequence converges to the infimum
13:47for monotonically decreasing sequence,
13:50while converges to the supremum of the
13:53AN for the monotonically increasing
13:56sequence.
13:57For example,
13:59consider the sequence. First of all, you
14:01can check whether it's a increasing or
14:03decreasing.
14:04Clearly say, if I found the AN + 1 - AN,
14:08that will be 1 over
14:102n + 1
14:12into n + 1 whole square. You can watch
14:15this lecture, monotonic sequence 15 plus
14:18examples, and clearly say this is
14:20greater than zero. Therefore, the
14:23sequence is my monotonically increasing
14:26sequence.
14:28Fine? Now, you can also check whether
14:31it's a bounded or not. We know the value
14:34of the AN.
14:36This is less because it's a 2 raised to
14:39power K. So, I also say K is equal to
14:42three, 1 over
14:452 raised to power K.
14:46Fine? Because 1 K squared is always a
14:48positive number. Because it is up to the
14:51end, I can also say the sum will be less
14:54than equal to
14:571 by 2 raised to power K. Infinity to
14:59the earlier minute.
15:00Fine? Just infinity earlier now it's a
15:03GP. The GP give up
15:05first of A over 1 minus R. A will be 1
15:08over 2 raised to power 3, 1 minus 1 over
15:112. So, that will be 1 over
15:144. It's a common
15:16sequence after pass KO key.
15:18Bounded be okay.
15:20Fine? Sequence is a monotonically
15:22increasing. Sequence the monotonic
15:25convergence theorem, the sequence an
15:29convergent.
15:32But, here you all must be noted about
15:36one most important thing.
15:39I mean that I got your sequence is a
15:40convergent, but I'm you need to get
15:43something. The sequence is convergent to
15:46the 1 by 4.
15:481 by 4 is Is it 1 by 4? Because it's a
15:52increasing. Is 1 by 4 is a supremum of
15:55an?
15:56Have you proved that?
15:58No. So, we can never say that an
16:02converges to the 1 by 4.
16:04Until you prove that 1 by 4 is a
16:07supremum of the an. So, we can only say
16:11that that by the MCD monotone
16:14convergence theorem, the sequence is
16:16convergent.
16:18Look at the other one.
16:20First of all, check whether it's a
16:21increasing or decreasing. I can start
16:23from the an minus one
16:25minus of an.
16:27That will be n plus one over n plus two
16:31minus n over n plus one. You can take
16:34the LCM.
16:36n plus two
16:38n plus one. If you solve them, it's a n
16:41squared plus one plus
16:432n minus n squared minus of 2n. n
16:48squared will be cancelled out. 2n and 2n
16:51will be cancelled out. So, the result
16:52will be 1 over
16:55n plus one n plus two. Because n is my
16:58natural number that is greater than zero
17:01for all n. Therefore, your sequence is
17:06monotonically
17:08Yeah, okay. Yeah. Because an plus one
17:10minus an is a greater than zero. So,
17:12it's a monotonically increasing
17:14sequence.
17:16Fine. Now, you can check for the
17:18boundedness.
17:20Because a n is my n over n plus one. As
17:26n belongs to the natural number, that is
17:28always greater than zero. Strictly
17:31greater than zero because n can never be
17:33zero.
17:36Also,
17:38also
17:43So, that will be
17:45Fine? And we know n plus one is always
17:49be greater than equal to zero.
17:51Fine?
17:53That implies one over n plus one is also
17:55greater than zero. So, that means that
17:57number is strictly less than one.
18:01Yeah, one always equality one to come in
18:04the end. So, that means the sequence is
18:07lies between
18:09a one up to infinity, but infinity is
18:11not a natural number. So, that means the
18:14number is lies
18:16open one
18:17open zero.
18:20So, therefore the sequence is bounded as
18:22well. Sequence is a monotonic
18:24increasing. So, by using the monotone
18:26convergence theorem, the sequence a n is
18:30converges to the L, where
18:34where
18:37L is my supremum of the a n. And since
18:41we already found the bound of the a a n,
18:45can you find the supremum of this bound?
18:48Supremum of this is one. Upper that's a
18:50one.
18:51So, therefore the sequence a n is
18:54converges to the one is the right
18:57answer.
18:58What you check we got
19:00n over n plus one is equal to
19:03as n approaches infinity. That will be
19:06one. So, therefore the sequence is
19:10convergent.
19:11But make sure monotone convergence
19:13theorem is for the purpose
19:16is that if a sequence is monotonic
19:19and
19:20bounded
19:21then it can say you that it is a
19:23convergent.
19:29I hope you have understood this lecture
19:31in a very very simple manner. We will
19:33see the important lecture called as a
19:35subsequence in our next part. Till then
19:37you can share this video with your
19:39friends, like and comment on the videos.
19:41Best of luck students. Happy learning.

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