 The basic idea is given a sequence a n. So, given a sequence a n of numbers, how to add all its elements. So, that is saying that same as saying, what can be say a 1 plus a 2 plus, what this quantity can be? So, to define this properly, let us make a definition of a n. So, given a sequence a n of real numbers, let us define as n to be equal to a 1 plus up to a n, n because the 1. So, that is the sum of the first n terms of the sequence. So, this s n, it is called the nth partial sum of the sequence a n. So, this is called the nth partial sum and let us give it a name. So, the pair, so here is a sequence and here is a partial sum s n is called, we will call it a series. So, a sequence together with its partial sum is called a series of numbers and we say this series, this is a convergent series, if s n limit n going to infinity, s n exists. If you look at the partial sums and take the limit of that, if that limit exists, then we say that the series is convergent and if this limit is equal to s, s n, we write sigma a n, n equal to 1 to infinity equal to s. Now, let us observe something so that we do not have to write this cumbersome notation of series being this way that given a sequence a n, s n's are defined, partial sums are defined. So, we know s n's and supposing we say that as we give you a sequence s n of numbers which is the partial sums of some sequence, then the sequence also is known. So, how is that? So, given a n, s n which was defined as a 1 plus a n is known and conversely given s n's, what is a n? That is s n plus 1 minus s n for every n. So, giving the sequence or its partial sums, the data both are equivalent to each other. You give one data, you get the other data. So, for that reason, we do not write all the time a series to be like this. We just given a sequence a n. So, notation for a series is sigma a n. Given a sequence a n, we write the corresponding series as this. This does not mean you should not take it as if it says that the sum exists. It is just a notation for that series. If convergent, if the partial sums s n's which we defined as sigma i equal to 1 to n, a i converges to s, then we write sigma a n equal to 1 to infinity is equal to s and say a n is convergent. If it is not convergent, we say it is divergent. Then we say it is divergent. Let us look at some examples. They are relatively simple examples. Let us look at some examples. Let us look at the series sigma a n, where the nth term is minus 1 to the power n plus 1 for n bigger than or equal to 1. So, this is a sequence a n. What are the terms of sequence 1 plus n equal to 1 or minus 1, minus 1 plus 1, minus 1 plus 1 and so on. We know that as a sequence, this sequence is not convergent. It fluctuates. Let us try to form the partial sum s n. So, what will be the partial sum? It depends on whether the n is even or odd. So, partial sum will be equal to 0 if n is even. The terms will cancel out. Otherwise, it will be minus 1 or plus 1. So, partial sums do not converge. So, we can say that this series minus 1 to the power n plus 1 is not a convergent series by the definition itself. So, this is not convergent series. The simplest example of a convergent series is the one which we start looking at in our schools, normally called the geometric series. So, what is a geometric series? So, it is a series where the nth term a n is r to the power n, where r is a fixed real number. So, take a real number. First term is r, second is r square and so on. This number r is called the common ratio, because it is the ratio of a n plus 1 and a n. So, when is it convergent? We all know that it is convergent when mod of r is strictly less than 1. And why? How is that? What is the proof of that? One can find what is s n? 1 plus r plus this s n is equal to 1 minus r to the power n plus 1 over 1 minus r. That is easy to find. If you write s n something multiplied by a, it shifts the powers then subtract and you can easily compute what is s n. So, this formula that s n is equal to 1 minus r to the power n plus 1 over 1 minus r. We do it in our schools, but it is not difficult to find. So, the question is whether r to the power n plus 1 converges to something or not as n goes to infinity. We know, we have done it in sequence is that x to the power n converges to 0 if and only if mod x is strictly less than 1. So, using that fact, this is convergent if and only if mod x is less than 1 and in that case the sum is equal to r to the power n minus 1 will go to 0. So, it is 1 over 1 minus r. So, the simplest example of a series which is convergent and this will be sort of used again and again. A geometric series whose common ratio is less than 1 is convergent. You will see how this is one of the building blocks for analyzing series. Let us look at one more example. Let us look at s n equal to 1 over n. The first one was minus 1 plus 1 and now it is 1 over n. So, what is s n? s n is 1 plus 1 by 2 plus up to 1 by n. And the terms are non-negative. So, it seems s n is going to increase. You are going adding more and more non-negative numbers. So, s 1 is 1 and s 2 is equal to 1 plus half and so on. So, something is increasing. But the question is how much does it increase? Because if the partial sums s n are increasing, we know it is increasing, but if it is bounded then they will converge by the property of real numbers. So, is it bounded or not bounded? If it is not bounded above then it will not converge. So, to analyze that one has to make some estimates. So, let us look at for n equal to 2 to the power k, let us compute this quantity. So, n to the power k, so this is the 1 plus 1 by 2 and so on over 1 over 2 to the power k. And now you pair up, see 1 over 4 is 1 over 2 square. 1 over 3 is 1 over 2 square minus 1. So, make this pairing and 2 to the power k is even. So, you can pair up. Once you pair up, now this quantity 1 over 2 square minus 1, it is bigger than, if I increase the denominator, it is bigger than 1 over 2 square. So, I get bigger than, I do it everywhere and this is 2 by 2 square and k by 2 to the power k. So, this is bigger than 1 plus k by 2. So, this kind of estimates one has to do to analyze a series. So, what we are saying is for n equal to 2 to the power k, the sum s 2 to the power k is 1 plus k by 2. So, what happens to this partial sums for n equal to 2 to the power k, as k goes to infinity, it is bigger than k by 2. So, it goes to infinity. So, at least for the partial sums, we have got a subsequence. When n is equal to 2 to the power k, the partial sums has a subsequence which goes to infinity. It is non-negative. So, it is not bounded above. So, there is a subsequence which is not bounded above of partial sums. So, the sequence itself cannot be bounded above. So, the sequence of partial sums is not convergent because it is non-negative. It is not bounded above. Now, given any n, you can always find 2 to the power k such that 2 to the power k is bigger than n. Given any natural number n, you can always find a k such that 2 to the power k is bigger than n. That increase is faster than n. You can easily prove that. So, s to the power 2, s partial sum up to 2 to the power k will be bigger than the partial sum up to n. So, that also will go to infinity. So, that shows it is not bounded. So, it is not convergent. So, what does it imply? It implies that the series 1 over n is not convergent. So, this is how by definition itself alone, we are trying to analyze because it is non-negative. We can make estimates. Let us do one more estimation like this. This is an interesting, we had 1 over n. Now, let us look at, so the series 1 over n is called harmonic series because of a different reason. Let us consider the harmonic series, but now the terms are coming plus and minus. So, minus 1 to the power n plus 1 divided by n. So, it starts with n equal to 1. It is 1 minus 1 by 2 plus 1 by 3 and so on. Let us try to find out whether the partial sums for this converge or not somewhere. So, let us make some estimates. Once again, as before, let us try to look at n equal to 2 k. So, n equal to 2 k, look at the partial sums up to the terms 2 k. I think there is something wrong here. This is not 2 to the power 2 k, it is just 2 k. So, there is a typo here. Now, how many terms are there? Even number of terms, 2 k. So, I am taking the sums of first 2 k terms. So, I can pair them. So, the first one, 1. So, what I am doing is I am pairing up so that it is sum of non-negative terms, 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 and so on. So, I have grouped them in 2 to a pairing and 1 by 3 minus 1 by 4, that is non-negative. So, each bracket is a non-negative number. So, what does it tell you about this partial sums? That for n equal to 2 k, the partial sums are increasing because each bracket is non-negative. When it k plus 1, one more bracket will come. Some non-negative thing will be added up. So, s 2 k is a sequence of non-negative real numbers. Let us see what happens when it is odd. So, let us compute the same thing for 2 k plus 1. So, s 2 k plus 1, one more term will be added there. So, what is the relation between them? s 2 k plus 1 over 2 k plus 1, that is equal to s 2 k plus 1. One more term is added there. And it is plus here because 2 k plus 1 is odd. So, minus 1 to the power n plus 1 it was. So, this is a relation between 2 k and 2 k plus 1. So, if I look at this sequence of 2 k and 2 k plus 1, what is the difference between these two? This one is increasing s 2 k. s 2 k plus 1, when I pair them up, what can I say about the sequence s 2 k plus 1? 1 minus something, minus again something, I am subtracting and each bracket is non-negative. So, more and more things are being subtracted. So, s 2 k plus 1 is decreasing, s k increases, but s 2 k is increasing. And this is a relation between them. s 2 k is less than 2 k plus 1 and all are bounded in between 0 and 1. So, what does it imply? s 2 k is increasing and bounded. So, that will converge. s 2 k minus 1, that also is decreasing is a monotone sequence bounded below. So, that also will converge. So, both of them converge. And what is the difference between the two? Between this s 2 k and s 2 k plus 1, the difference is 1 over k plus 1. So, the difference can be made as small as you want. So, the sequence of partial sums, the even partial sums and the odd partial sums both converge to the same value. We have got a sequence where the odd and the even both converge. The subsequence of odd terms and the subsequence of even terms both converge to the same value. So, here is an exercise show that the sequence itself is convergent. So, take it as an exercise in sequences. You have got a sequence a n of numbers. Say that if I take the subsequence of even. So, a 2, a 4, a 6 that subsequence and look at the subsequence a 1, a 3 and so on both converge to the same value. Then claim that the sequence itself should converge to that value. The sequence itself is convergent. It is a very small exercise. It is a good exercise to go back and revise your notion of sequences. Just definition. So, that will mean what? What will that mean? Odd and even both converge. So, the sequences themselves converge. So, that means s n is convergent. s n itself is convergent. That is the exercise we are saying. And as a result, this series is convergent. So, the interesting thing is the series 1 over n is not convergent. But minus 1 to the power n plus 1 divided by n, that is a convergent series. Alternate plus and minus terms if you make it. That is called the alternating harmonic series. So, this series is called alternating harmonic series. That is convergent. So, I am just giving you some examples to illustrate that how definition can be used to prove something is convergent or not. And it becomes slightly cumbersome every time estimating the partial sum and trying to see whether it is convergent or not. So, these are the examples which illustrate that. So, naturally, let us look at one more probably. 1 over n square. So, the series is nth term is 1 over n square. There is a series of non-negative terms. 1 over n square is non-negative. So, partial sums will be a monotonically increasing sequence of numbers. The question is whether it is bounded above or not. If it is bounded above the partial sums, then the series will converge. If not, then it will diverge. Again, let us try to estimate. The claim is that if I look at n k, which is 2 to the power k minus 1, then S n k is less than or equal to sum of the geometric series, 1 by 2 plus 1, 1 by 2, 1 by 2 cube and so on. So, we are trying to bring in somewhere, again, something known kind of a thing. And this proof, for every k, we want to prove something. So, what is the technique of proving something for every natural number? The only thing we know is by induction. So, induction, so, apply induction. S n 1 is equal to 1. So, it is true. n k plus 1, what will be the, that is, S n k plus something plus the remaining terms, which are being added. And that squares 2 to the power k minus 2 k square. So, you can make it less than 1 over 2 to the power k is less than 1 over 2 and so on. So, this becomes less than the geometric series. So, estimates basically. So, once you do that, once you know this is true, so, what happens to the series S n k? It is less than this. And this is a convergent series. We know that. So, what happens to the limit as n k goes to infinity? The geometric series. Is it okay? It is bounded by the geometric series. So, we know the sum of nth terms and goes to infinity. So, sorry. So, what we are saying is there is a subsequence n k. There is a subsequence n k of partial sums, S n's, which are bounded between 0 and 2. Can I say that that implies S n itself is bounded? What is n k? What was n k? n k was 2 to the power k minus 1. We are saying that if I take n to be, n k to be this, then S n k is bounded. Can I claim that S n itself is bounded? Keep in mind, they are non-negative. Once again, given any n, given any n, you can find a k, say that n is less than n k. Is that okay? Given any natural number n, you can find a power of 2 to the power k, say that n is less than 2 to the power k minus 1. Is that okay? Yes or no? Yes. Natural numbers, 2 to the power, they are going to increase faster than n anyway, much faster. So, that means what? They are non-negative terms. So, given any n, there is a k, say that n is less than n k. Can I say S n is less than S of n k? Yes, because they are non-negative terms. S n is increasing and that is bounded by 2. So each S n is bounded by 2. Each partial sum is bounded by 2 because the sequence of partial sums is monotonically increasing and for n equal to 2 to the power k minus 1, it is bounded by 2. And couple this with the fact that given any n, you can find a natural number k, say that n is less than 2 to the power k minus 1. So, S n will be less than the partial sum up to 2 to the power k minus 1, which is less than 2. So, each S n is bounded by 2. It is monotonically increasing. So, they will be convergent because they are non-negative. So, 1 over n square is a convergent series. So, this is convergent, mononically increasing and bounded. So, it is convergent. Here, what was helping us is because this is a series of non-negative terms, partial sums are monotonically increasing. We have to only analyze whether they are bounded above or not.