 Hi, I'm Zor. Welcome to Unisor Education. I continue talking about different problems. In this case, it will be related to series. The last lecture was about sequences, now it's series. So, first of all, we need a couple of definitions. We will talk about some limits for series. So, question is, what actually does it mean, limits for a series? Well, to define it properly, let's first define a partial sum of a series. Now, partial sum of a series is, first you have a base sequence, and then the series is basically the sum of the elements of the sequence. So, I introduce the partial sum, which I call SN, which is a sum of all elements of a sequence from, let's use index i for this index changing from one to n. Now, I assume that this sequence is infinite. Well, obviously, when we are talking about limits, we aren't talking about infinite series, so n is actually goes to infinity. But partial sum is the sum of all elements from the first one to the nth, and this is a finite sum. Now, let's consider a sequence of partial sums. So, basically, what the definition is, the limit of a series is a limit of a sequence of its partial sums. So, right now, for instance, you have the base sequence, one, one-half, one-quarter, one-eighths, et cetera, one over two to the n minus one, et cetera. For instance, this is a base sequence, infinite. Now, the s first would be one, s second would be one plus one-half, s third would be one plus one-half plus one-quarter, et cetera. So, that's how partial sums are created, and they, in turn, can be considered as sequence of certain values. This is one, this is one-and-a-half, this is one in whatever, trig waters, et cetera, right? So, the limit of this sequence of partial sums is called the limit for the series. So, that's the definition. Now, once we have defined what is a limit for a series, now we can investigate different cases. When we have the series limit, when we don't have the series limit, et cetera. So, I have a few problems related to this. So, the problem number one is, I have this little plan for myself, let's consider a series which has a limit. Intuitively, you feel that the elements of the sequence, which is the base sequence for that series, so, you have a one, a two, a n, et cetera. This is the base sequence, and the partial sums are these. So, now you have a sequence of partial sums, and this sequence has a limit. That's exactly what the limit for the series is. So, you feel that since we are actually increasing the number of elements in this sum, for this one it's only one element, for this it's the first and the second two elements, for this is the first and elements of the base sequence. Now, you feel that if this thing has a limit, then it should be closer and closer to this limit, which means that additional element which we are adding every time, which is basically the elements of the sequence, should be smaller and smaller. So, the proposition is, if the series has a limit, then the base sequence should actually converge to zero. The elements should be smaller and smaller. And I would like to prove it. Okay, how to prove it? But let's think about this. If Sn is the sum of all elements from one to n, then I can say that the element An is equal to Sn minus Sn minus first. Am I right? Because this is the sum of all elements from A1 to An. This is sum of all elements from A1 to An minus first. So, the difference is only the element An. Right? Now, how can I prove that An is converging to zero? Well, very simply. We have to basically examine this. Choose G greater than zero. And we have to find out if there is some kind of a number n, after which if n greater than n, An is less than zero. Now, how can I find this number n? Does it exist? And if it exists, how can I find it? Well, elementary. I will use this thing. And here is that. Instead of An, I will use this difference. And I will change it to this, where L is limit of Sn when n goes to infinity. We can basically put it as a condition of this theorem that partial sums Sn's, they have limit. So, let's say L is that limit. So, if L is that limit, then this difference, which is actually An, absolute value, is equal to this. Now, since as you know, so you know this property of absolute values, right? So, we have this is less than or equal to Sn minus L plus absolute value of L minus Sn minus 1. Okay? So, I divide, split basically the whole sum into two halves. Now, I know that L is a limit of Sn, which means I can find such a number n1 after which the absolute value of this distance is less than d over 2. And I can find the number n2 such that Sn minus 1 minus L greater than d over 2. Actually, if I will find just this one, the number after which this difference is less than d over 2 since it is true for all n which are greater than n2. This will also be true. So, I don't really have to have two numbers. I can have only one number n when this thing is true, because then this also will be true since it's for all n greater than n, all right? And n is greater than n minus 1 of this, which means that if this condition n greater than n, or actually n minus 1, sorry, n minus 1, it doesn't really matter since it's all greater than n. So, if this condition is true, then this and this are true, which means this would be less than d, right? d over 2 plus d over 2. So, I have found exactly the number n from the convergence of partial sums Sn. I can found for any g, based on the convergence of Sn, I can find the number n after which this would be less than this, which means a n is converging to zero. And the proof. Geometric series with quotient less than 1 has limit, all right? Now, obviously, geometric series with the quotient greater than 1 would not produce a series which has a limit. Why? Because we have already proven before that geometric sequence with quotient greater than 1 by absolute value. Is going to infinity, plus infinity, minus infinity, or whatever it is, it's not converging. To converge, we need every element to go down to zero. So, the sequence should converge to zero. This is a necessary condition, as I just proved before, for a corresponding series to have a limit. So, right now we are only considering q by absolute value less than 1. Only this, in theory, it might or might not, but in theory it might produce a series which has a limit. Now, question is, does this series, does the geometric series have a limit for q by absolute value less than 1? Answer is yes. And here is the proof. What is Sn? A plus aq plus, etcetera, plus aq n minus 1. Now, I don't remember the formula, which I actually proved in one theoretical lecture. So, to do this I will, that basically, but I do remember how I derived this formula. I just multiplied it by q, right? So, q times Sn is equal to aq plus, blah, blah, plus aq n minus 1 plus aq n, right? So, if I will subtract one from another, Sn minus Sn times q, this minus this, is equal to, this is all cancelling out, that's a minus aq to the n. And Sn is equal to a1 minus qn divided by y minus q. Okay, fine. So, I found this formula using the approach which I do remember, because I don't remember the formula. Now, the question is, does this have a limit? Well, a and 1 and q are all constants, right? Only n is variable, n goes to infinity. And I do remember that q by absolute value is less than zero. Now, obviously, having this formula, and if you remember, we have a few properties of the limits, like limit of the sum is equal to sum of the limits, limit of the product is equal to product of the limits, if all the limits exist, etc. So, basically, what I can say that limit of Sn, if n is equal to, if n goes to infinity, would be equal to, this is a constant, this is a constant, this is a constant, and the only thing which is remaining is this. So, obviously, limit of constant goes outside of the limit, if I multiply it, this is the difference, so it's difference of limits, one is a constant, so it's only a limit of q to the nth degree. Now, what is the limit of q to the nth degree if q is less than one? Well, obviously, zero, we have already proven that many times. So, I can say that this is equal to a over 1 minus q. So, as you see, there is a concrete, there is a concrete result of sum of all these geometrical values. This is a over 1 minus q. This is the sum of internet geometric series with q by absolute value going down to zero. All right, fine. So, basically, let's just talk about one little example, if you wish. Let's say I have this geometric series. So, a is equal to one-half and q is equal to minus one-half, right? So, one-half times minus one-half is minus four, then minus one-half is one-eighths, minus one-half minus one-sixteenths, etc. So, let's use this formula and find out what is exactly the result of this. So, we take one-half, we split it in half, this is one-quarter, we cut it out, and then half of this we add it back, half of that we add it back, half of this we add it back. So, that's basically the process. So, what will be the result? Well, let's just substitute to this formula. So, we have one-half over one minus minus one-half equal to one over two, one plus one-half is three-half, one-third. Interesting, right? This is one-third. It's a little bit unusual, let's put it this way. But, however, it's absolutely true. It's proven, so there is nothing to do about it. But what's interesting about this is the following. You probably know from geometry, if you didn't study it yet, you will, that there is a famous problem of trisection of the angle using only the straight edge ruler and the compass. If you have an angle, and then you would like to divide it in three equal parts, it's called trisection of angle. Now, it's proven that you cannot do this using straight edge and the compass. However, what you can do always, you can divide it in half without any problems. So, what I can do is the following. I cannot divide it by three, the angle, but I can do it with any kind of precision using this mechanism. So, instead of dividing it in three parts, which I don't know how, I mean, it's not possible to do, which is proven, I divide it in half. Then, out of this half, I will add half, and then I will subtract half, and then I will add another half, I will subtract another half, etc. So, that's the process which will allow me to trisect an angle, not precisely, but with any precision, because as we understand, the elements are going smaller and smaller. Okay, next. Now, question is, next question, very easy. Now, we have actually learned that if a series has a limit, then the elements of underlying base sequence should go to zero. That's obvious. Question is, is reversed through? So, for instance, I have a sequence, I have a sequence, and I know that elements are converging to zero. Is it sufficient condition? I mean, we know it's a necessary condition for the corresponding series to have a limit, but is it sufficient condition for the corresponding series to have a limit, which means if this is true, does it mean that sum of a n's s n? Does it mean that limit s n, if n goes to infinity, exists? Well, the answer is no. This condition is not sufficient for the corresponding series to converge. And here is an example, very simple example when it's not correct. Let's consider a sequence 1, 2, 1, 3, 1, 4, etc., 1, n, etc. Now, this sequence is converging to zero. Now, how about a series which is made out of the sequence? Does it have a limit? The answer is no. And here is why. Let's group one element, two elements, four elements, one, two, four, eight elements, etc. Now, let me just explain for a couple of groups. Now, this is one third plus one fourth. It's greater than one fourth plus one fourth, which is equal to one half, right? Now, next four are one fifths, one sixths, one sevenths, and one eighths. It's greater than one eighth plus one eighths plus one eighths, plus one eighths, which is equal to one half again. So, what I'm saying is that each group is, if summed together, is greater than one half. And how many these one halves we have? Obviously, the infinite number. And one half plus one half plus one half plus one half is definitely not converging. It's a constant actually, which is summarized. So, it's basically some of the same number. It's one half times whatever the number is, and number isn't integer. So, this is greater than this, and this is not limited. So, basically, this particular example is an example of a sequence which does converge to zero, but the corresponding series has no limit. Now, the last thing which I wanted to show today is, I think I mentioned in one of my lectures, Zeno's paradox about Achilles, which is trying to, which is raising a tortoise. So, here is Achilles, and here is a tortoise. Now, Zeno, the Greek philosopher, was saying the following. Let's say Achilles is 100 meters behind the tortoise. By the time he moves to this point, the tortoise will be able to move to the next point. Let's say it's only one meter, but it's still further. Then, by the time Achilles reaches this point, tortoise will move a little further, maybe one millimeter or centimeter or whatever, but still a little further. So, it looks like it will take for Achilles an infinite number of steps, which means, as Zeno says, he will never catch the tortoise, which is obviously wrong, because we all know that Achilles is a very strong and fast runner, and he will definitely catch the tortoise. So, where exactly this paradox is? What's the contradiction in this particular logic? Well, let's consider from a mathematical standpoint. Just for definiteness, let's say that the speed of Achilles is 10 meters per second, and the speed of a tortoise is 0.1 meter per second, 100 times less. So, what happens? Well, Achilles covers this distance in how much? Well, in 10 seconds, right? So, he will go, Achilles, go 100 meters in 10 seconds. What does tortoise do during this time? In 10 seconds, it will cover one meter, right? So, tortoise will cover one meter and the same 10 seconds. Next, what happens? Now, we have one meter distance. One meter will be covered, one meter will be covered in 0.1 seconds by Achilles, right? One meter with a speed 10 meters per seconds, it's 0.1 second. Now, with a speed 0.1 meter per second and 0.1 second, so the same 0.1 second, tortoise will cover 0.01 meter. So, what happens? Every time, my distance is diminishing in 100 times and my time is correspondingly diminishing in 100 times. So, if I will summarize all these little segments, because next segments will be what? 100s of one meter, so it will be 0.01 meter and 0.001 seconds, right? And this is 0,0,0,0,0,0,0,0,0,0,0,0,0,1 second, it will cover 0.001 meter, etc. So, what we have here, we obviously have a geometric progression with, if we are talking about distances, the beginning is 100, this is a equals 100, and q is equal to 100. In case of time, a is equal to 10 and q again is equal to 100. So, in any case, it's a geometric progression which has a specific sum. So, whenever we are summing up all these intervals, what will we have? Now, let's just think about it. In case of a distance, if I will summarize to infinity, we are talking about infinity, right? So, if I will summarize to infinity, a over 1 over q would be, in this case, 100 over 99 hundreds, which is 10,000 divided by 99 whatever it is. So, that's how many meters, basically, and that's just a little bit more than 100 because 99 is almost 100, right? So, it will go 100, but slightly more, like 102 or whatever. So, it will be obviously more than 100, the initial distance, but not very much more than that. So, this is the sum of all these little segments, and it means that during these segments, these meters on this particular segment, which is the result of summation of the infinite series, q is indeed will not catch up with the tortoise, but as long as he crosses this particular limit in distance, he will. Same thing with the time. I can say that during the first, now it's 10, right? So, this is the distance, and the time is 10 over 1 minus 100, which is 10 over 99 hundreds, which is 10000 over 99, which is slightly more than 10. So, slightly more than 10 minutes would require a helus to catch up with the tortoise. So, these are two different approaches, obviously, from the distance standpoint and from the timing standpoint, when we definitely have a limit during which he will not catch up with the tortoise, but as long as this limit is crossed, more time than this particular sum or more distance than this particular sum, then he will catch the tortoise. Well, obviously, the same result can be achieved by just subtracting the difference between their speed and dividing initial distance 100 by the difference in speed, and you will get exactly the same answer, obviously. So, that's it. I don't think I have, yeah, that's it. That was my last problem. Try to go through these problems again. Do it yourself. It would be great if you can register and do some exams. Your parent supervisors or group leaders can enroll you in any course, and you will pass the exams. That's probably a better way to approach the Unisor.com system. But in any case, lectures are open for everybody. You are free, and that's it for today. Thank you very much.