Skip to main content
Loading Explore guides…
Algebraic Limit Laws · Acedra
Home
Explore
Mathematics
Algebraic Limit Laws
Sign in to save
Chat
If YouTube asks you to sign in, open the video on YouTube.
Watch on YouTube
Reload player
Transcript
99 segments
Copy
Hide video
0:01
In this video, we are going to continue proving results about sequences. I have two related statements here that go very well together. These are what I would call the limit laws, and they're exactly what you would predict for convergent sequences. So, in particular, as a hypothesis, imagine that the sequence little a sub n converges to the real number capital A and likewise, the sequence little b sub n converges to capital B. So here a and
0:26
b are real numbers and they are the limits of their respective sequences. The algebraic limit laws give us ways to form new sequences out of these pieces and basically do algebra on them with the corresponding algebra on the limits. So what I mean by that is the first statement tells us that if we form a new sequence by pairwise adding the terms from a n and from b_n. So that the first term in our new sequence would be a1 plus b1 and then the second term is a2 plus b2.
0:54
that new sequence would also converge and we know what its limit is. It would be the sum of a and b. That's not surprising. So the limit of a sum is the sum of the limits. Likewise, the second piece tells us that if we form a new sequence by pairwise multiplying the terms from a in and b_n, then the limit of this product is the product of the original limits. Again, not too surprising. The third one is uh just a result where if we take the original sequence and flip it,
1:22
if that's legal to do so, so none of the terms are zero, then that inversion also produces a sequence which converges and it converges to one over the original limit where again we can only do this if we're not dividing by zero. Okay, so that's like doing algebra on sequences. Our second group is the order limit laws and this is ordering how sequences behave. The first one is the only one that we actually need. So this tells us that if we have two sequences where
1:50
say a n is less than or equal to b_n for all n then the limits are also going to have the same relation between them. So in particular a will be less than or equal to b. It's like the sequence a n isn't going to jump over the sequence b_n. Again not too surprising. I claim that if you know this first line you also know the second one. So we don't actually need to prove statement 2B down here because it's a corollary. It's an immediate corollary of statement 2 A. So
2:19
let me pause for a second. See if you can just work out why 2B follows immediately from 2 A. 2 B is immediately implied by 2 A because we could take the numbers C and D to actually be constant sequences. So in other words, if we had CN is less than or equal to A N is less than or equal to DN, then we could apply 2 A to this pairing and to this pairing where imagine that CN is actually the constant number C and DN is the constant D. So it's like the constant sequence we proved
2:59
converges to its own repeated value. We we actually showed that explicitly. So if you view these constants as actual constant sequences then 2B is going to follow immediately from 2A. Similarly um I almost wrote it up here as its own result and then I decided not to but uh two sorry 1b implies through the same train of thought immediately that a constant times a sequence that converges will scale the limit accordingly. So again, once you've shown this, you've shown a
3:31
stronger result because you can just replace one of those sequences with a constant sequence. Okay? So that just saves us from having to prove extra things. We will prove 1 a, 1 b, 1 c, and 2 a. Our first proof is that the limit of the sum is the sum of the limits. This is going to be a direct proof that the sequence that we get by taking the sum satisfies the definition of a convergent sequence with limit a plus b. Now often when
4:02
you do this kind of sequential convergence proof, you do some scrap work. This one is so straightforwards that I'm going to jump right into it, but we will probably do a little bit of scrap work in the future for maybe 1b. Let me go ahead and just write this up directly right now. Let epsilon greater than 0 be given. You can include the phrase be given here.
4:22
Since the sequence a sub n converges to a for this fixed value of epsilon greater than zero, there exist a threshold index capital N sub 1 from the set of natural numbers such that whenever indices or our sequence a sub n oh I actually didn't mean to write indices. I meant to just say that to be a little bit more concise. Let me erase that. But yes, I'm talking about indices. Whenever little n is an index beyond this first threshold that we are going to identify, it follows
5:04
that the distance from the corresponding sequence term to the limit capital A is the pure definition of sequential convergence would lead me to write epsilon here.
5:20
And maybe I'm going to do that actually. So I'm going to write epsilon, but I'm going to write it a little bit high. And I know that I'm going to come back and annotate that. I kind of want to show you what happens with these proofs pretty often. So leave yourself room to scoot a little denominator underneath this epsilon. But here I'm really just vocalizing our knowledge that the sequence a n converges to a. We were given that. And this is an interpretation of
5:43
what that hypothesis tells us. Let's do the same thing for the other sequence. Likewise, I'm not going to restate this because I'm saying likewise, there exists a threshold index which I will call capital N sub 2 from the set of natural numbers such that whenever a sequence index little n is beyond that threshold, it follows that
6:17
the distance from uh I'm going to write it here to squeeze it in, but it's going to be the exact same construction. Again, we are just stating what we know when we say that little a sub n converges to a and little b sub n converges to b. We want to now add these together and compare it to the limit a + b. I want to know that both of these statements are happening at the same time. So if I'm going
6:47
to take a random index little n and say that it's large enough, I want it to be larger than both n2 so that this happens and n1 so that this happens. That means that we want to take the maximum of the two. So this is going to be the threshold that we are identifying for our new sequence. The sum of the two that we were given. Let capital N be the maximum of the two thresholds that we have already identified. And suppose I now grab for the new sequence an index beyond this threshold.
7:31
If little n is greater than or equal to capital N, then by the way that we made capital N, it's also greater than or equal to n1 and n2. You want to be sure that you're going out farther and farther in convergent sequences. And so typically taking a maximum when you have multiple thresholds is the correct approach. Uh let me continue it here. Then observe. I want to show that when I compute the distance between a sub n plus b sub n, this is a term from
8:08
the sequence. By putting it in parentheses, I made it look like the entire sequence and it doesn't need to be in parentheses. So let me just erase that. But the idea is that this is a a single term from the new sequence. this term with this index, what is its distance to our proposed limit? I need to break this up with the triangle inequality. I'm going to come down a little bit to the right. Normally, I would write this to the
8:34
right. With the triangle inequality, I'm going to pair up a n minus a. Actually, before I do that, because we haven't done too many of these yet, let me leave it inside of one absolute value. So, I'm just going to rearrange terms. we have a n minus a plus b_n minus b. Okay, so that's just rearranging the terms inside of this first absolute value here. Now with the triangle inequality, this is something that we see a lot. We have this pairing which is really
9:04
nice because it reminds us of this statement and we have this pairing which reminds us of this one. So with the triangle inequality, we break this into the sum of a n minus a plus b_n minus b. Now if little n is greater than or equal to n1, this distance here is less than this weird epsilon that I wrote, which I'm going to again write kind of high. And similarly little n is greater than or equal to n2 so that this distance is less than this
9:44
other strange epsilon that I wrote. Okay, let me pause here. This is a very common technique here of you know we have this comparison that we want to make comparing our sequence terms to the limit and these pair up these pair up. So you rearrange things and then you split them up with the triangle inequality. This comes up all the time. So the triangle inequality is a phrase that we will say frequently together. Now if you look at this I have two epsilon that
10:16
doesn't feel like a big deal but especially in first semester real analysis I want you to land on epsilon. So 2 epsilon is actually like in practice it's good enough but let's really write these correctly so that it sounds exactly like the definition of sequential convergence. It's good to start with that level of requirement on yourself. So for first semester real analysis, I want your computation here to end with epsilon, not a modification of epsilon. I
10:42
want you to get to epsilon exactly. How would we see that? Uh well, if we had epsilon / two, then we would have two copies of epsilon over two leading us to epsilon. And at this point, we're almost done with the proof. You can do this when your hypothesis is that the sequence converged. The sequence here being a sub n. Okay, we were told that a sub n converges to a. Which means that for any choice of positive number you make, you can then go find
11:17
a threshold so that the distance from some term beyond the threshold to the limit is less than that positive number which you picked. So the idea with this modification is that I've already fixed epsilon. When I now talk about the sequence a sub n, the positive number I'm going to pick when I state the definition of sequential convergence here and I apply it to this problem. I'm going
11:39
to pick my positive number to be epsilon / 2 which is also a positive number. So it's like for this choice of positive number which happens to have the form of this number divided by two go for this number. So for this positive number in one is a suitable threshold beyond which the distance from any term in the sequence to the sequential limit is less than this number. Same thing down here. So
12:05
so I pick n1 and n2 for these bounds over here on the right hand side. And as long as they're positive and we know that our sequences converge, we can do this kind of operation. And it's not unusual because because we split things up with the triangle inequality so many times. It comes up all the time in real analysis that you break things up with the triangle inequality. You often have things like epsilon over two or epsilon over three if you have three pairings or something like
12:29
that. So you should get in the habit of of kind of looking for this step. Okay, at this point we are basically done. So you know we have the statement up here. Let me just write as desired in a box. And that wraps up our first result. Let's look now at the second one.
12:54
For our second result, we want to show that the limit of the product is the product of the limits. The proof is going to follow the same train of thought as the first exercise that we just did, but there's a little bit more work here. So, let's start with scrap work. With scrap work, what I like to do is is jump straight to the the quantity that we start simplifying at the very end of the proof. So what I mean by that is it's going to look like a and b_n minus ab.
13:27
The first example was easier because we had a n by itself with a. We paired that together and then we had b_n and b. So it was an immediate application of the triangle inequality. Here they're kind of stuck together. So what you want to do in this situation is realize that we are making a change from A into A and from B in to B with just one term. So from this term to this term, I'm
13:52
changing two quantities. The the step the the goal here is to only change one at a time. So whenever you're like this change to A and this change to B, freeze one of them and only change the other. So what I mean by that is let's do a in b_n and then I'm going to leave b_n alone and only change a. So the change would be to go from a in to capital A the limit leave b_n as it is. Now I can't just randomly bring this term in out of thin air. So what we're really
14:26
doing is adding a fancy version of the number zero. So add it back and then subtract off a b.
14:36
Now notice that I've only made a change here for this first pairing and I've only made a change here for the second pairing. A is a b_n is b_n. And now we have two pairs. So again with the triangle inequality we're going to say uh a n b_n minus a b_n. And let me do a simplification at the same time. Notice that b_n factors out of both of them. So what if we also took this and pulled it to the right. I'm just trying to save a little bit of room here. So I could take this
15:11
first term and write a n minus a * b the absolute value of b_n like that. And then likewise here a is common to both expressions. So let's pull the absolute value of a in front and then b minus b. This is just scrap work. So what I like to do is put here what I would like to see. We have the sum of two terms. So what I would like to have is epsilon / 2 plus epsilon / two like we had in the first example. So that overall this would sum up to epsilon.
15:48
So if I could get this to epsilon over two and this to epsilon over two that would be great. Um well here you know this could be made smaller than epsilon / 2 certainly we did that in the first exercise but then I'm scaling it by b_n the thing is b sub n is a convergent sequence and we have already proved that convergent sequences are bounded which means that for this sequence there's some number capital m which is positive so that this absolute value is literally by definition
16:21
less then cap capital M m will be positive. So it's safe to put down here in the denominator. It's a fixed positive number that's simply a bound for the convergent sequence B sub n which means that this is just a positive number. And the rationale that we just talked through with the first example allows me to find a threshold for the sequence a sub n so that if I take an index beyond that threshold, this distance is less than this fixed positive number. Notice there's no
16:49
dependence on the index here. epsilon is fixed in the first sentence of our proof. Two is a number. Capital M will be a fixed number. So this is just a fixed positive number. It will work for the definition of sequential convergence. Capital A is also just a fixed number. So this is just going to be absolute value of A. And then here b_n is converging to B. So again we think epsilon / 2.
17:17
Little danger here. A lot of students will write 2 * the absolute value of a because they want it to cancel out exactly as it did, but a could be zero. And then I've seen people go into proof by cases and that's overkill. Actually, a little trick is just to take this number which is greater than or equal to zero and make it positive by adding one. Notice um epsilon / 2 times the absolute value of
17:48
a over the absolute value of a + 1. This number is less than or equal to 1. So this is less than epsilon / two which is good. You want to go up in size. You know if a is zero this is just zero. That's less than epsilon less than epsilon over two. If a is four this is four fifths. So you see how this plus one works. So it gives us a number which is less than one actually. So that this number is actually less than epsilon / two and that's fine. That's the good direction to go
18:17
with that inequality. Okay. So we have a lot to do in the preamble of this proof. First thing we're going to do is let epsilon be greater than zero. Then we should create m before using it. So I will mention b_n is a convergent sequence. So there's a bound capital M so that this size of all the terms in the b_n sequence are less than M. I'll write this down if I'm not saying it very well.
18:44
And then we'll pick a threshold for the sequence A so that this happens. And then we'll pick the threshold for the sequence B so that this happens. We'll pick the maximum of those thresholds and the rest of it will be exactly this. So we'll see this algebra again. All right, let me erase this scrap work and we will write up the proof. I'm not going to talk through all the logic as much as uh I might if I were writing the proof from the get-go because we did so much of it here. So,
19:06
I will pretty much just write up the proof without stopping and explaining every single step. Let's now translate all those ideas that we saw in the scrap work into a formal proof. I'm going to open with let epsilon be greater than zero. At this point, it is a fixed positive number. It's a secret. Our proof needs to work for whatever epsilon is, but it's now given to us. Since the sequence b_n converges to the number B,
19:36
b_n is bounded by a prior result that we proved in an earlier video. We don't need to redo that proof. We already know this. Choose capital M greater than zero so that the size of b_n is less than M for all N in the set of natural numbers less than or less than or equal to both would would work here. Since a n converges to capital A oops little a n converges to capital A there exist a threshold n sub one from the set of
20:21
natural numbers so that for all indices little n beyond this threshold the distance from some nth term to the limit a is less than epsilon / the positive number capital M. And again, I don't want to talk through this too much because I said I wasn't going to, but this is a safe thing to do for a convergent sequence so long as your right hand side here is a fixed positive number. It doesn't have any index dependence. So if I had little n over here,
20:54
that would look real weird. But here it's okay. And then b_n converges to b. Similarly, there exists a capital N2 from the set of natural numbers, excuse me, natural numbers such that for all indices little n beyond this threshold, the distance which I will write here from the nth term in the sequence B sub n to the limit of that sequence B is less than epsilon / Capital A + 1 where again this plus one is protecting me in case A is zero. All right,
21:43
we have two thresholds. We're working with sequences. If anything, we want to go up farther in the sequences. So let capital N, which will be the threshold for our new sequence formed by taking this product, be the maximum of this one and this one.
22:04
Okay, so we've picked our threshold for our new sequence and suppose little n is beyond that threshold. I would like to do the full computation formatted nicely for you. So I'm going to erase all of this. Let's pause for a moment. Make sure that you have all of this written down because when I come back it will be gone. And then also uh you have the scrap work presumably and maybe you feel good about these kinds of proofs. See if you
22:31
can just go ahead and take it all the way to its conclusion before I return. Moving into the final step of our proof for that index little n that we just identified. We find I just want to format this nicely a n b_n minus capital A * b is less than or equal to a n b_n minus a b n + a n minus a. So I have brought in this fancy zero. Triangle inequality happens so often in these proofs that you don't need to cite it. At least not for me. And likewise doing things
23:16
like pulling out a shared term. You don't need an extra step for that. So let's go straight to less than or equal to the product of the distance from a n to capital A times the size of the b_n term plus the size of a time the distance from the term b_n to the number capital B. So this is such a common thing to see in these kinds of proofs. You don't need too much explanation here. it's not obscure what you've done. Uh do bring this in explicitly
23:53
though. I think that's nice for the reader. And now we have chosen our index far enough out that we know that this first term is less than epsilon / twice that positive number that is larger than all of these terms b sub n or their absolute values. And then the number a is just what it is. So take its absolute value and that we've gone far enough out in our sequences that this is less than epsilon / 2 * that number + 1.
24:29
And then I like to see that this is epsilon / 2 plus actually this is strictly less than epsilon / two. So let me switch this to less than equals epsilon. And on paper I would go down to the left and write as desired in line with text. But let me write it here so that I don't run out of room. Okay. So that is our proof that the limit of a product is the product of the limits. I think it's a really
25:01
good important exercise for several reasons. It's more interesting than that first one. One is to recognize how to bring in this fancy zero and then how to use the knowledge that we have here to construct a way to cancel these out to epsilon over two. Also, please feel free to use this little trick down here which saves you from having to write a proof by cases which is more than is necessary for this exercise. Let's look at our third statement in the algebraic climate laws.
25:34
This third one is one I normally skip in the classroom because I'm not sure it adds enough to the discussion. But since we're making a video, let's go ahead and record this. We want to show that when we form this quotient sequence, it converges accordingly. And here we know that none of the terms nor the limit are zero. So that's nice and convenient. Again, let's start with some scrap work. This one is actually more obscure how we're going to
25:59
approach it than the previous exercise. So we've leveled up with each statement in this algebraic limit laws. Again, what we want to do is compare 1 / a n to 1 / capital A. The natural thing to do here is put this into one fraction with a common denominator. So we have a - a n / a n a denominator here is totally safe because everything is non zero. Now, this expression is exactly what you want to see whenever you're working with sequential
26:31
convergence because we're comparing here a sequence term to the limit. So, this is the kind of thing that we can make small using the definition of convergence. Let's go ahead and write this as a n minus a. That's the same absolute value as a minus a n. It just looks better. And then divided by the absolute value of a n times the absolute value of a. And here some things look good, some things look bad. Okay, so this is great because this is the
27:04
kind of expression we want to see in our scrap work. This is what we control with our threshold. This is also good because capital A is just a nonzero number. It's a fixed number. This term down here is kind of bad as it is because it's going to change for different values of n. So I can't um I'm think what would be a mistake I might see what I'm about to write down is a mistake. It would be bad to set this less than epsilon and say okay we're going to have a n minus
27:36
a less than epsilon time I guess just time a n * a. Um so that when we replace this with the epsilon statement we get everything cancels out and leaves us with epsilon. You do not ever want your right hand side here to depend on the index. So this is fine. This is fine but this is critically wrong. So we have to do something else. Here's how we will handle this term. I want to again replace it with a quantity that maybe just involves capital A because that's a fixed number.
28:13
I don't want to necessarily bring in anything to do with epsilon because look how it's living down in the denominator. I want epsilon, you know, I want this. So I don't want to have epsilon down here in the denominator. That would be problematic. Let's see if we can just come up with an algebraic statement giving us a way to replace the quantity. Let me write one more line here. 1 over the absolute value of a n times 1 over the absolute value of a time a n minus a. Okay,
28:44
so this is what I I want to replace and I would love to replace it with just a plain old positive number like any expression involving the absolute value of capital A because then we'll be able to say this is less than epsilon divided by whatever we need to cancel that out. How we do that here? It's a little bit of a trick. I'm going to take capital A's size and bring in a sub n with a fancy zero. So let's take this and say it's equal to a minus
29:12
a n + a n. As soon as you see that now you should start thinking triangle inequality. And as I do that I'm actually going to flip this ordering here just so it looks a little nicer. Now this is what I have here. Hang on. Okay. This expression here is what I would like to use to handle this one here. This is a number. When you see this, you might think, oh, epsilon over two or something like that. But I
29:55
actually here just want an expression with capital A. I don't want to risk canceling out an epsilon or something like that. I've got my epsilon. It's going to live here. I really just want a number, a fixed number independent of epsilon even. Let's pick a threshold large enough so that this distance here is less than capital A, the absolute value of that divided by two, which is completely fair to do because this is a positive number. And it might not even be hard to do. I mean,
30:30
capital A could be like 100. And so we're saying go out in the sequence far enough so that the distance between sequence terms and the limits is less than 50. which might happen right away. It's probably less restrictive than what we'll do here. Okay, I'm worried I'm losing you, but it will make more sense when we write it up formally actually. Uh what is happening now? Well, notice that we can
30:52
say the size of a sub n this term I don't like here is greater than or equal to the absolute value of a minus the absolute value of a / 2 which is actually equal to the absolute value of a / 2. Now we can take the flip to get to 1 over the absolute value of a sub n. That's what we want to replace. So 1 over the absolute value of a sub n is going to be less than or equal to because we're inverting 2 / the absolute value of a. Okay, this is scrap work. It's going to be really messy.
31:40
If we've done this replacement, what do we now have? We have that this goes up in size to 2 over the absolute value of a. This is just a number. That's fine. In fact, well, I'll leave it like that. I want to have just plain old epsilon down here. So, what should I put here? We're going to take our threshold for this term in the sequence to be large enough so that this distance from some term beyond that threshold in the limit is less than uh absolute value
32:16
of a squar * epsilon / 2 so that this will cancel with that these will cancel with that and leave us with epsilon. Okay. Okay, so at this point in the proof writing, here's the conversation I have with myself. How many precisions do I need to make for the threshold? So basically, how many thresholds do I need to pick that I then maximize? I need one threshold so that this will happen. So I need to
32:47
say choose capital N sub one so that if little n is any index beyond n sub one, this distance is less than this positive number. And then I also need a second threshold so that this happens. So I will say pick capital N sub 2. So that if little n is any index beyond n sub 2, this distance is less than the number capital A squared. Actually you don't even need absolute values there but that's okay. Capital a^ squ* epsilon / 2. So that's just a positive number. and we'll pick
33:25
our second threshold for that. Then we will say, okay, we had two selections, two thresholds to make two inequalities true. Let's make them both true by picking the maximum and then we'll be able to just kind of do this. So I will repeat what I did in the previous proof where I will try to do it without talking too much because I think we worked through all the algebra here. So I will try to be fairly concise in our write up in just one moment. As I write up this proof,
33:53
I'm going to break convention a little bit just to show you that you can. Namely, I'm not going to start with the phrase let epsilon be greater than zero because I don't need epsilon right away. So, let me hold off on that for just one moment. Since the sequence a n is known to converge to a, there exist a threshold capital N sub 1 from the natural numbers such that for all little n greater
34:23
than or equal to capital N, we know that the size of the distance from a sub n to a, the absolute value of a sub n minus a is less than the size of a / 2. Nothing to do with epsilon. Observe. I think it's kind to the reader to do that algebra that we did with uh this expression here. It's not obvious right away when you first see it. So the size of a is equal to the size of a minus the term a sub n plus that term back is less than or equal to a sub n minus a plus the size of a sub n
35:13
i.e a sub n. Let me flip this inequality is greater than or equal to I'm flipping it. Actually, you know what? I'm changing my mind. No, I'm not. The size of a sub n is greater than or equal to the size of a minus this distance like that.
35:58
Consequently, for all indices beyond this threshold that we have identified and I should have a subscript one here, we find that the size of a sub n is greater than or equal to the size of a minus the size / 2 / 1 over the size of that nth term from our sequence is less than or equal to 2 / the size of a. Notice nothing about this had to do with any choice of epsilon out there. I kind of made a a choice
36:56
of a different epsilon here, but I gave it a very specific name. So, this is nothing to do with the epsilon I'm going to bring in in just a moment. In fact, I'm going to bring it in now. Uh, let me see if I can add it. Uh, I'm going to add it down here. Let epsilon be greater than zero. Since a n converges to a as we know very well at this point there exists capital N sub 2 a second threshold such that and now going to continue the proof here whenever
37:39
a sequence index little n is beyond that threshold we find that The distance from the nth term in our sequence to the limit a is less than oh shoot what was it? I think it was A squared or the size of A squared times this epsilon which we just introduced divided by two. Yes, that looks right.
38:05
Let capital N be the maximum of the two thresholds that we have named so that both of these behaviors will be true beyond the threshold capital N. And this will allow us in just one moment to study the new sequence that we have defined by taking the reciprocal of the one that we were given. Let me pause here in a moment. I will erase all of this and we will just do the ending. So all of this
38:30
will be gone in just a moment. Let me say a little bit before I do this erasing. I needed all of this and it has nothing to do with epsilon. So I didn't want to confuse the reader by saying let epsilon greater than zero be given and then here's a whole bunch of stuff that has nothing to do with epsilon because I think as a reader you're like looking for the connection to the epsilon that was just
38:52
introduced in the first sentence. So I wanted to do this algebra here before introducing epsilon. Then I said let epsilon be greater than zero. We're going to bring that in and use it with a second threshold. So since we have a convergent sequence, there's a threshold. So that whenever we're beyond that threshold, we can say that the distance from corresponding sequence terms to the limit is less than this fixed positive number which has nothing to do with the index. That's
39:17
why we did all this work. This is a positive number. This is a positive number that we've just introduced. And this is the number two. You know, as a professor who teaches real analysis, whenever I'm reading student work and it doesn't begin with let epsilon be greater than zero and they're proving something like a sequence converges, I always get a little bit nervous, right? So, it's
39:38
something that it's going to worry your instructor if you see that um if they see this. But as long as you introduce epsilon in before you use it, you're okay. So, it was fair to do this. It's just it's a little bit scary when you're a professor and you don't see that right away. However, this is legitimate. Okay, so this proof is correct so far. In a moment I will have erased all of this. We will start with the sentence let little actually let me go ahead and do it here because
40:03
that's how we've been doing it. And suppose that now I'm just going to grab a random index beyond this new threshold for our new sequence. Suppose we're beyond that threshold. I will start with we find and then the nice triangle inequality argument that will take us through to the end.
40:27
As promised, we find the distance from 1 / little a sub n to what we propose is the limit 1 / a. Again, it's going to repeat a lot of our scrap work. That's okay. This is a - a n over a n a and then keep going with equality. This is the same thing as a n minus a time the size of a n times the size of a. We have two replacements to make and we're going to make them at the same time. So we want to replace
41:10
this. We want to replace this. just break it up kind of nicely for the reader. So this is less than 2 over the size of a for the first one times 1 over the size of a time the size of A squared or a 2 epsilon / 2. Okay, so if I'm a reader think I could follow that. And then of course we have a lot of beautiful cancellation here. This is going to be equal to epsilon. Now I actually have room to write a nice sentence at the end. So I will say thus 1 / a n converges to 1 / a as
41:53
desired. Oops. Okay. Kind of a fun little proof. Um again I always skip this in class but I think the thought process behind passing from this term to this one is a worthy endeavor. So hopefully you were able to see why it was necessary for us to replace this term in the denominator which varies for different ends with this nice concrete fixed positive number bound that that uses the fixed number capital A. So I think that is something worth lifting up. We have just one more proof
42:28
to do the order limit statement. Uh just the first one because again as we discussed in the introduction the first one implied the second one. Here is our final statement for this video. If two sequences a n and b_n satisfy that for every index little n a sub n is less than or equal to the corresponding b sub n, then their limits have to preserve that inequality. So the limit capital A is less than or equal to the limit capital B. Basically a n can't leapfrog over b_n somehow if
43:05
this ordering is always true. This proof is going to be a little bit different because it's going to be a proof by contradiction and our scrap work is just going to be a picture. What would be the contradiction? Actually, you know what? It's going to be a proof by contrapositive. So, what I'm going to do is contrapositive if B is less than A. So, that would be negating the conclusion. So,
43:29
somehow A was strictly greater than B. then it's not the case or in other words there exist some index let's call it capital N such that uh b_n is less than a n. So this is a contrapositive statement or it's the contrapositive for the conjecture that we are trying to prove. The reason why I like contradiction here is I like the strict inequality that we get. It tells us that we can draw capital A and capital B on a number line with distance between them.
44:08
So I'm checking which of these markers is still working and probably go through several in just a moment. So this inequality allows me to situate these limits with space between them because they are strictly we have capital B is strictly less than capital A. Okay. What does it mean to converge? Well, if I plot the sequence terms a sub n and b sub n on this number line after some threshold, it would make sense to kind of cluster the a ends here and the b ends down
44:37
here. Right? So the sequence terms trending to a should be have values close to a. And likewise the sequence which trends to capital B should have values close to b. But look at my picture. This would put a lot of b sub n's less than a sub n's. And that contradicts the hypothesis that we started with. Here's how we're going to formalize this. Let's chop this interval in half. The midpoint here is a + b / 2. It's just the average of those two numbers. And then this
45:10
is going to define our distance epsilon. It'll be the same for each, which is nice. So this epsilon I'm going to define will be half the distance between them. So that's going to be larger minus smaller / 2. Notice that because we have strict a strict inequality here. This is positive. So a is strictly greater than b. This numerator is positive divided by two. It's positive. So this
45:39
is a valid choice of epsilon to use when we study these sequences. And I think this is all the scrap work we need. What we're going to do just in words kind of mapping out the rest of the proof is we're going to say suppose this for the you know we'll prove the contrapositive so suppose this define this epsilon choose some capital N so that for all terms beyond that threshold uh the corresponding sequence terms are close to their respective limits we'll do it in two stages
46:10
because this is real analysis one so I'll pick two thresholds we'll take the max as we've been doing and then we'll be able to have a contradiction So I think it it kind of falls out naturally with these explicit expressions, but I'll say pick some um b. Notice it's to the left of this pink line, but then we go up to the right to get to a end. And then we've got this relationship. Okay, so I think we are ready to go. This is the key idea here is to recognize that with the negation
46:38
of the conclusion, we have this number line that we can actually sketch by putting these numbers far apart. Cut it in half and then realize that the sequence terms are going to be like on either side of this dividing learn dividing line beyond some threshold. Whenever you do an indirect proof, it's kind to the reader to let them know that right away. So we will show the contrapositive. Yep. new marker. We will show the contrapositive
47:11
contrapositive which I will abbreviate like that namely going to tell them the statement if a is greater than b so not the conclusion then not the original hypothesis. Notice with the original hypothesis we have a statement about for all n. So negating that means there exist an n for which we don't observe that behavior. Okay. So if not the original conclusion then not the original hypothesis is if a is greater
47:45
than b then there exist an index which I will call capital N from the set of natural numbers such that b sub capital N is less than a sub capital N. So that's contrapositive. Let epsilon be the average. Um, no, sorry, it was a minus b divided by two. You know, I feel like I want to mention my jumping off point one more time. Sorry, I'm kind of doing these proofs on the spot and experimenting with all these markers, but I have
48:28
these rolling markers. Okay, this is what we're going to show. So, let's show it. Thus, suppose a is greater than b and define epsilon to b a minus b / 2 and I might add to that. Notice epsilon is greater than zero. Okay. So now I've stated our new goal in the proof is to show the contrapositive and then I'm saying okay so let's start with the hypothesis of the contrapositive and now it's quite clear why this epsilon is positive because a is larger than b since a
49:13
n converges to a there exists a capital N1 threshold such that for all indices beyond this first threshold the distance from A into A is less than epsilon. Likewise, there exist capital N2 such that for all indices beyond N2, the distance from b_n to B is also less than epsilon. Okay, there we go. I think this one is good. Yes, I actually in my contrapositive statement I'm just working with one index. Notice it's really
50:04
saying there's an index for which the original inequality fails. You could if you wanted to make it all indices beyond capital N, right? But it's not necessary. So you can if you want to, but I'm just going to stick with this idea of one index. Let Capital N be the maximum of the two. Notice this is true which means let's see I want to start with B is on the left. I want to start here. I'm not going to spell this out in the proof but I will do it off to
50:44
the side to remind you of how this inequality works. to say that this absolute value is less than epsilon. This is equivalent to the statement negative epsilon is less than b_n minus capital B is less than epsilon. Right? So these are two versions of the exact same statement. Hope you can read that. This marker is also kind of on its last legs. All right. So let me apply this statement to B sub capital N. And what I'm going to
51:19
do is move that B over. B sub capital N is less than epsilon + B. And I think I will spell out this algebra. So this is epsilon plus sorry I'm going to replace epsilon and I'm going to replace it with its explicit definition which we have given here. a - b / 2 + b. But then that's equal to (a - b + 2 b) / 2. Or in other words, this simplifies to a + b / 2 just like in the picture. So you step to the right epsilon and you
51:59
land on that midpoint. And then this is actually the same thing as a minus a minus b / 2. partly because I I have the picture in mind but you can work out algebraically that this is equivalent. So this is a minus a over 2 that's just a 2 minus negative so plus b over 2 that's a plus b over 2. So going back here which is equal to a minus epsilon. Now imagine you know the same thing but for the a ends we could say you if I go through and add the limit it would be a minus epsilon
52:38
is less than a n is less than a plus epsilon so now going to be here is less than a sub capital N and that's exactly what we wanted to show b sub capital N was less than a sub capital N and again I think the picture is really key there to seeing exactly why the choice of epsilon made sense and also to help you work through this algebra so that everything checks out correctly. Thus, going back to our original statement because this was an indirect proof if
53:16
a n is less than or equal to b_n for all indices little n a is less than b or sorry less than or equal to that's what I meant less than or equal to. Okay, that wraps up this fourth and final proof in this video. Thank you for your attention.
Here to help you learn
Sign in to keep a private conversation with your course tutor.
Ask me anything about the material…
Sign in to chat