 Hi, I'm Zor. Welcome to Unizor Education. I would like to continue talking about limits. There are a few problems which are actually many theorems, but I presented it as problems which I would like to talk about today. Quite important actually problems because there are certain aspects of other areas of mathematics like geometry for instance, where we are measuring lengths and areas, they do depend on these theorems, so I will definitely be referring to these particular theorems when I will talk about those geometrical properties. Alright, so first of all I would like to expand the concept of minimum and maximum towards infinite sequences. Now, obviously if you have a finite sequence, let's say 1, 2, 3, 4, you know that this particular finite sequence has the maximum element which is 4 and minimum element which is 1. Now, what if my sequence is infinite? So instead of this I will use the following sequence. One half, two thirds, three quarters, etc. and over n plus 1, etc. Now, minimum actually we can talk about one half. Now, then elements are increasing obviously and well, how about the maximum? It's infinite, it's infinitely increasing. Well, at the same time you might actually notice that it's getting closer and closer to number one. I can't really name number one as a maximum of this particular sequence. Why? Because it's not a member of this sequence. We can't really talk about maximum for something which is not an element of this particular set. So, to expand the concept of maximum and minimum mathematicians have come up with some other terminology if you wish and that's what I would like to introduce right now. It's a great new concept. Here it is. First of all, we will introduce the concept of upper bound for a sequence. What is upper bound for a sequence of an? Well, it's basically any number a which is greater than or equal to all the members of this particular sequence. If this is for all m. So, if this number a is greater or equal to any element of our sequence then a is called an upper bound. Well, that's natural. Now, are there sequences which do not have upper bounds? Yes. One, two, three, four, five, six, seven, etc. up to infinity. There is no upper bound. Alright. Because whenever we choose any particular number a there will be some kind of a member of this sequence which is greater. Okay. Now, obviously in exactly the same fashion we can introduce lower bound. Now, this is a number a which is less than or equal to a for all m. So, if there is a such a number which is less than or equal to any element of our sequence it's called lower bound. Again, this is quite obvious and there are obviously certain examples of sequences which do not have lower bound. Like for instance minus one, minus two, minus three, etc. It goes to minus infinity which means no matter how we choose the number as a potential lower bound there will be some elements of the sequence which are less than this particular number. So, these two concepts have absolutely no problems. Either they exist these bounds, upper and lower or they don't exist but if they do exist it's obvious what it is. But here is the very important axiom which is attributed to famous mathematician Dedekind. The axiom of completeness. What it says is the following. If a sequence has an upper bound it has the least upper bound. Now, what is the least upper bound? And similarly the greatest lower bound. Now, what are these? Well, the least upper bound first of all is an upper bound which is a number which is greater than or equal to all elements of our sequence. At the same time any other upper bound for the same sequence would be greater than or equal to our least upper bound. So, this is actually the minimum if you wish of all the different lower bounds. In any case, similarly, if a is the greatest lower bound, not only this is sorry, it's different, not only this is a lower bound but it's also greater, it's the greatest, right? So, it's greater than any other lower bound. So, these are correspondingly the least upper bound and the greatest lower bound. So, the axiom of completeness states that if there is an upper bound for a sequence, then there is the least upper bound. So, there is a concrete number, concrete real number. We're talking about real numbers only right now which is not only the upper bound but also the least upper bound and similar with the greatest. Now, the first one is called supremum and the second one is called infinum. Supremum infinum. So, these are mathematical terms for least upper bound and greatest lower bound. So, these are concepts which are expanding the concepts of maximum and minimum towards infinite sets. If the finite set is given, the finite sequence of elements then obviously its maximum and supremum are exactly the same and similarly its infinum and minimum are the same but if we are talking about infinite sequences they might or might not be bounded and if they are bounded then if it's bounded from above, if there is an upper bound then we can talk about supremum which is the least upper boundary and similarly for infinum. Okay. Now, why did I introduce these particular topics? Well, there is a very important theorem in the theory of limits and this particular theorem is actually dealing with supremum and infinum of infinite sequences. So, under certain circumstances certain sequences do have the limit and I am going to basically prove that this particular theorem. And this is my first problem which I wanted to present today. So, whatever I said before is just some kind of explanation of what exactly the problem is all about and it will be obvious how to solve the problem using these two new concepts, supremum and infinum. Okay. Here is the problem. Consider you have a sequence such that it's always increasing. So, every next element is greater or equal than the previous one. So, for instance if you put it on the axis of all the numbers then this is your A1, this is your A2, this is your A3, etc. So, all the points are directed towards increasing numbers. Now, let's also assume that everything is bounded from above. So, all members for all n, this is true. So, not only we have the sequence which goes to the right, but also there is a bound. There is a boundary which they do not exceed. So, they are moving to the right but there is a wall here. Now, this type of sequence is called monotonic or monotonous, monotonously or monotonically increasing in this particular case, because obviously there is a decreasing to the opposite side. So, the theory in which I would like to prove, and this is my first problem actually in this particular lecture is that monotonically increasing sequence bounded from above has the limit. It's actually very important because this particular theory will be used in some very important cases. So, let's just think about how to prove it. It's not difficult. I mean, it's a lot of different explanations of the concept, concepts, etc. But first of all intuitively it's obvious, right? So, if you're moving towards the wall infinite number of steps, you never go across that wall, but you're always moving towards it. There should be something that you are converging to, right? Now, so what exactly are you converging to? Well, my statement is that the limit of this particular sequence when n goes to infinity is equal to its supremum. Let's use capital letters, supremum of an. So, this is the least upper bound for this particular set. First of all it exists supremum because of the axiom of completeness for real numbers. So, the sequence is bounded from above and the axiom says that if it's bounded, there is a least bound, upper bound in this particular case. So, this thing does exist. Now I'm going to prove that this particular number, supremum of this sequence, which does exist according to an axiom which we have accepted, from this we can prove that this particular supremum is the limit of this sequence. Now, how can we do it? Alright, actually it's not very difficult. Let's think about this. Well, for obvious reasons the limit cannot be something which is above this boundary because everything is on the left side and if you put the limit on the right side, considering the sequence will be infinitely close to this limit, it will be definitely above the boundary. But that's kind of an obvious. I'm talking about intuitive understanding of this. Also, again, intuitively if the limit is somewhere on this side and again considering the sequence would be infinitely close to this particular limit then most likely this is not the least upper limit. Something like this would also be an upper limit. So, that's how I plan to prove that this is the limit of this particular sequence because it cannot be on the left, it cannot be on the right, etc. Alright, so let's do it from a more rigorous standpoint. How can I prove that this is a limit of this? Well, naturally let's choose any distance g greater than 0 and let's find the number n such that if n greater than capital N, then the distance between let's call it a, then the distance between a and a n less than d. So, basically that's what we have to do. So, we choose a number d which is the distance. Now we have to find the number after which this is closer, the elements will be closer to this particular number than d. Let's assume that there is no such number that whenever we choose a number, it will not actually satisfy this particular thing. First of all, let's use the fact that our sequence is monotonous. If there is one particular number n after which this is true, then all other numbers which are exceeding this would be definitely true because we are moving towards the right side of this, on this axis, right? Which means every time a n plus first, a n plus second, etc., will be even closer to that particular least upper limit supremum than the previous one. So, we actually have to find just one. You don't have to find the number n after which all the numbers, all the elements would satisfy this particular inequality. It's enough if we will find just one. So, if we will find one, then this particular inequality will be definitely true for all subsequent ones because we are increasing and we are moving towards eight without obviously overlapping because it's a boundary, it's an upper boundary, right? So, how can we prove that there is such a number n which satisfies this? Well, let's assume that there is no such number. Which means that all elements, whatever number we choose, all elements are further from the a than d. So, if this is a and this is d, all elements if they are further from a than d, what does it mean? It means that any number between a and a minus d would also be the upper boundary and which contradicts the fact that a is the least upper boundary. So, we are actually using two things, the monotonousness of our sequence and that's why we really have to find just one number n for any d. We don't have to worry about the subsequent one. And then we were using the fact that a is the least upper limit because if we consider that there is no such number n, if this is not true, if the distance is greater than d, then obviously there is another upper limit which is less than the one which we have claimed to be the least. So, absolutely symmetrical proof can be made for an opposite case. When sequence is decreasing and it's limited from below, then it has a limit and the limit is infinite. I don't want to do it right now. The proof actually is presented in the lecture's notes on Unisor.com so please refer to this to have a complete proof. So, this is just a purely logical exercise of what actually is the upper limit, the least upper limit and how they are related to limits of monotonic sequences. Now, what I will do next is I will just use this for a couple of other problems. So, next problem is related to series. Now, let's consider the series. The series is something, as you know formed from existing sequence by summing up elements of this particular sequence. First of all, we introduce n-th sum which is from n let me use different index here. Let's say i for i equals 1 to n. So, this is n-th sum so the first n elements of our sequence now all s n's all s n's form a sequence in its own right. So, if this converges then we say that the series converges to whatever the limit actually, the limit of this is. So, our theory is if all members of this particular sequence are positive and all partial sums are bounded from above then the series converges. Which means that all partial sums will go to certain limit. Now, how can I prove it using the previous theory? Well, that's actually very simple. Because considering that all members of our sequence are positive, it actually means that s n plus first is always greater or equal than s n's. Actually, greater even. If this is greater or equal then this is greater or equal. Because we are always adding something, right? What is s n plus first? It's the sum of the first n plus first elements of our sequence. Which is first n elements plus n plus first and this is greater or equal than zero. So, this is greater than or equal to s n. So, our partial sums represent the monotonic sequence. It's monotonously increasing in this particular case. And it's bounded from above as we said. Actually, that's the condition of our theory. Which means from the previous theory it has a limit. That's the end of the proof. Very simple. So, basically what we have done here, we concluded that from the positiveness of elements follows the monotonousness of the partial sums. And that's what was very important actually in this case. Okay, next. Next is not related to supremums or infinums. It's just about limits in general. If you have two sequences which are converges to the same limit then the sequence which constructed from their differences, the corresponding differences. So, I subtract members with the same number. Converges to zero. Intuitively it's obvious. So, if you have one sequence goes to some limit and another goes to the same limit the difference between them should go to zero. How can they prove it? Well, it's very easy but you have to remember one very important things which I actually used before. Now, there is such an inequality. Actually, it's kind of a triangle inequality. But it's obvious with numbers as well. If x and y are positive then all the absolute where you can be dropped and it would be exactly equal. If x and y are both negative then again this would be just an opposite and this would be just an opposite and this would be just an opposite. It would be exactly equality. Like, for instance, minus five plus minus three it's minus eight. Absolute value is eight, right? And this is absolute value of minus five and absolute value of minus three. This is five and this is three so it's equal, right? However, if the signs are different then these guys will cancel partially cancel each other and these will always be added together. So, this is an obvious inequality and I will use it to prove that the a n minus b nth are going to zero. Very simply. Look at this. It's less than or equal to a n minus l plus l minus b nth. Right? I can add and subtract the same number which is the limit which is, well, actually in this case it's equal. I don't have to put less than that. Now, in this case now I will consider this as my x and this as my y so I will put less than or equal sign now a n minus l plus l minus b nth. Now, l minus a n or a l minus l is the same thing because there is an absolute value here and now let's think about this. This is as n goes to infinity is very, very small and it goes to zero and this one as well. So, intuitively we feel that that smallness of these two guarantees smallness of this one because this is greater than this and all this everything is positive. We are talking about absolute later. Now, how to rigorously prove it? Very simply. Again, let's just choose any kind of a g which is greater than zero and find the number n such that this particular thing a n minus b n is less than g. Alright? How can I do? Now, since l is the limit for a n's I can find number n1 such that if n greater than n1 it follows that a n minus l less than d over 2. Now, I can also find n2 because this is also limit for b n. Right? Such that if n greater than n2 what follows is that b n minus l is less than d over 2. Right? d is any number right? So, I can always find d over 2 and find n1 from the convergence of this and n2 from the convergence of this. Now, if I will choose n equals maximum of n1 and n2. Now, then both of these inequalities will be true and that's exactly is a sufficient condition for their sum. This is less than d over 2 and this is over d over 2 so the sum would be d and that's exactly what's necessary to prove here. Alright? So, obviously as intuitively felt the difference between two sequences converging to the same limit converges to zero. And another which is very much like this that's the next problem. If you have two sequences which are converging to the same limit and you have a third sequence which is in between these two. Now, what do you think happens with xn? Well, obviously if a n is going to this limit and bn going to the same limit and xn is in between there is no other way but also go to the same limit. So, what I'm saying is that xn should go to the same limit. Now, how can I prove it? Again, choose d greater than zero now we have to find number n such as the difference between l minus xn less than g for n greater than capital n. That's what we have to find we have to find number n if we choose d. Well, let's just do exactly the same thing. l minus xn less than an minus xn plus l minus and again, this is an equality so far, right? I added and subtracted an. Less than or equal to now I will use the same I will group it in two so it will be an minus xn plus l minus an. Now, the difference between an and xn between an and xn is less than the difference between an and bn, right? So, I can put bn minus an plus l minus an. Now, let's use the fact from the previous problem that bn minus an converges to zero. l minus an is also converges to zero since l is the limit, right? And these are two very small variables as n increasing to infinity. And that's what we're going to use in exactly the same fashion as before. Now, since bn minus an tends to zero, then I can always find for a d over 2 I can find number n1 such that bn minus an is less than d over 2 if n greater than n1. Similarly, for the same d over 2 I can find number n2 such that my difference between l and an less than d2 if n greater than n2. Exactly the same as in the previous problem. Now, I will choose the number n which is maximum between n1 and n2 and for any lowercase n which exceeds this maximum, both will be true and their sum will be less than d d over 2 plus d over 2. So, that concludes the proof. Okay, next one is I've done this, I've done this, aha, all right, a new concept very simple though, the concept of subsequence. Well, it's very similar to the concept of a subset. You have a set and then you pick certain elements from that set and they constitute subset. Same thing with the sequence. If you have some kind of a sequence a1, a2, a3, an, etc. Subsequence is you just pick certain numbers instead of 1, 2, 3, 4 when you pick 10, 20, 25, etc. Some numbers and 1 and 2 and 3 preserving the order. So, here 1, 2, 3 goes in increasing order and so n1 should precede n2, should precede n3, etc. And this is your new sequence. This is b1, this is b2, this is b3, etc. Now, let's think about if this sequence is converges to some sequence. Now, the problem, the next problem is to prove that the subsequence, any subsequence from this converging sequence converges to the same limit. Well, it's kind of obvious. Let's just think about it. If this converges it means that from any distance from the limit, so limit is L, right? Limit of this sequence. Okay, so for any G, I can find some number n such that for all subsequent numbers, difference is less than G. That's the definition of L being a limit for ams, right? Now, what does it mean? It means that any number in this sequence and 1 and 2 and 3, etc. So, from certain point, all these mk and ak plus 1, etc. would also be greater than m. So, basically, I have to find out of these numbers and 1 and 2 and 3, etc., the first one which is greater than this number n. And it's always possible because the sequence is infinite and this is the finite number. And that particular mk's would be exactly the boundary after which all members of our subsequence would be within the distance from the limit. Why? Because all corresponding members of the sequence are there. Including picked up numbers from that infinite tail which is included into our sequence. So, that's the last one which I wanted to talk about today. I do recommend you to go through notes for this lecture at unizor.com and try to create all these proofs which I have just provided or any other, whatever you decide. But anyway, it's very important. Especially the very first theorem about if the monotonic sequence is bounded then it has a limit. If it's monotonically increasing and it's bound from above it has a limit. Or if it's monotonically decreasing and is bound from below it also has a limit. This logic is very important and I do recommend you to go through it again and again. And obviously as usual I recommend to register as a student, have somebody else or yourself if you're very disciplined as your parent or supervisor registered and go through exams very, very useful for you. Alright, that's it for today. Thank you very much.