 So, here is a, keep in mind one thing. The sum of convergent series is always unique. If a series is convergent, its sum is unique. The limit of a sequence is unique. What is the definition of sum? It is a limit of partial sums. And partial sums can converge only to one limit. So, limit of a convergent sequence is unique. That implies that the sum of a convergent series is unique. Suppose you drop some of the terms from a a 1, a 2, a n is a sequence given. I want to know that whether it is summable from 1000 terms onwards or not. That is equivalent to saying whether it is the series itself is convergent or not. Because what is left is the only sum of finite terms only. So, you can say that the convergence of a series does not depend upon first few terms of the series. It is same as the fact that we did for sequences. Convergence of a sequence depends only on what is the tail of the sequence. So, convergence or divergence of a series does not depend on the first few terms, whether those are added in the series or not. But the sum may change. If you sum it from 1 to onwards or 1000 onwards, the sum may change. But convergence or divergence will not depend upon. It depends only on the tail of the sequence a n, which is given. Cauchy's criteria, which is a consequence of a sequence is convergent if and only if it is Cauchy. A series is convergent when this partial sums converge and partial sums will converge only when the partial sums is a Cauchy sequence. So, coupled with that fact Cauchy criteria, that sequence a n is convergent if and only if partial sums is a Cauchy. And that is same as saying given epsilon, we can write epsilon delta, epsilon n naught definition. Given epsilon, the difference between the nth and the nth term should be small. And that is that sum from mod of x n plus 1 to x n should be small for m bigger than n, s n minus s n. That should be small. So, there is nothing, it is a simple consequence of the fact that every sequence is a sequence is convergent if and only if it is Cauchy. So, apply it to the partial sums. Now, here is another simple fact that suppose a series is convergent, then what is the nth term? If a series is convergent, what is the nth term? This is a series, whether convergent or not, let us not bother. s n is the partial sum, then what is s n plus 1 minus s n? That is just a n plus 1 simple arithmetic. Now, if sigma a n is convergent, then in this left hand side, if I take the limit n going to infinity, then star implies what? What is the left hand side limit of the left hand side? Series is given to be convergent. Then what is the left hand side? That is 0 is equal to limit n going to infinity of a n plus 1. And if you like, this is same as limit n going to infinity of a n, does not matter. Or I could have just written here s n minus s n minus 1. I could have written that is equal to a n either way. So, I get a consequence, very simple observation that if a series is convergent, then the nth term in that sequence, the sequence of nth term must go to 0. So, a n must go to 0. So, this gives me a necessary condition for a series to be convergent. A series is convergent, then it should necessarily happen that the nth term should become smaller and smaller and go to 0. So, that is a necessary condition and very useful one, because if the nth term does not go to 0, then the series cannot converge. So, not convergence is useful proving. So, that is same as saying, if a n is not equal to 0, it is divergent. But if it is only a necessary condition, it is not sufficient. That means, if the nth term goes to 0, that does not imply that the series will always converge. For example, just now we said 1 over n, the series 1 over n is not convergent. nth term is 1 over n, that goes to 0. But the alternating series, again the nth term goes to 0, but that is convergent. So, this is not a sufficient condition that the nth term should go to 0. That is only necessary. So, may either converge or diverge. So, you can give examples, we just now given. You can apply, if you like, you can apply it to geometric series. We proved for mod r less than 1, for r bigger than 1, nth term will not go to 0, if r is bigger than 1, it goes to infinity, so that does not converge. Let us look at this kind of a series. It looks like n square plus 3 n plus 1 divided by 2 n square plus 1. It looks like the numerator and denominator both are increasing at the same rate, essentially, power is n square. So, when n goes to infinity, it will stabilize somewhere, because both are increasing at the same rate. So, how do you analyze that? So, look at the nth term. nth term is n square plus n plus 3 divided by 2 n square plus 1. I want to analyze what happens as n goes to infinity. So, the simplest thing is divide numerator and denominator by n square, because I know 1 over power of n goes to 0. So, that gives you 1 plus, numerator will give you 1 plus, 1 over n plus, 1 over n plus, 3 over n square. And the denominator will give you 2 plus 1 over n square, as n goes to infinity, the limit will be equal to, numerator will go to 1, denominator goes to 2. So, by the theorems on sequences, if a n is convergent, b n is convergent and b n is not convergent to 0, then a n divided by b n is convergent to limit of a n divided by limit of b n. So, that theorem says that the limit of this nth term of this series is converging to 1 by 2, which is not equal to 0. So, this series cannot converge, because the nth term does not go to 0. So, nth term, that is how it is used to analyze not convergent of a series. So, this does not converge. Now, here are the theorems about algebra of limits, giving you algebra of convergent series. If a n is a series, which is convergent, b n is a series, which is convergent, you can add nth term of both, get a new series whose nth term is a n plus b n. Then, what will be the partial sums of a n plus b n? Partial sum of a n plus partial sums of b n, and if a n is convergent, then partial sums converges. So, partial sum of the sum will converge to sum of the partial sums by limit theorems on sequences. So, if a n is convergent, b n is convergent, then a n plus b n is convergent, and sum is equal to sum of a n plus sum of b n because of the limit theorems on sequences. Same logic applies to the other. You can have difference, you can have the scalar multiplication. One can wonder what happens if you multiply two series. Can you multiply two series? You can just look at a n, b n if you want. That is one way of multiplication. But in the partial sums, there will not be a multiplication of partial sums. So, that will not work out. So, it is only for the additions. One can think of what could be a way of multiplying series so that the corresponding result for a series is valid. You understand what I mean? I am throwing a question. Given a series a n, given a series b n, what could be the multiplication of these two series so that the limit of that product, whatever we define, is product of the sums. Think of it. It is a good thing to think. It is possible to do such things, but let us not do into that. So, this is the algebra of sums for the convergent series. This is for convergent only. If a n is convergent, if b n is convergent, then a n plus b n, that series is also convergent using the sequencing theorems or sequences. How is that useful? You can always come make examples. Look at this series, 2 by 3 to the power n plus 3 by 4 whole to the power n. Is it convergent? So, in sum of 2, 2 by 3 to the power n and you try to show that both of them are convergent and then sum will be the sum of that. So, I am leaving for you to check why both are convergent. More examples one can give. I think this is not convergent why? Because if it were, I know 5 by this is convergent when I subtract, it should give me 2 by n should be convergent, which is not true. So, one can play with this kind of things. The usefulness of saying that sum of convergent series is convergent. Here is something, this is okay. So, let us what I am going to do is, I am going to specialize for some time on series with non-negative terms only. a n's are all non-negative. When a n's are non-negative, partial sums are going to be increasing, partial sums are going to be increasing because we will be adding something on the time non-negative. So, either the series will converge if the partial sums are bounded above or what will happen? The partial sum will keep on increasing and go to plus infinity. So, in some in such case, when sequences of non-negative terms are given series of non-negative term, if convergent we write what is the sum or sigma a n less than infinity. Other only other possibility is it is divergent and in that case, partial sums converge to plus infinity. So, one writes sigma a n 1 to infinity equal to plus infinity. Just a notation, nothing more than that. So, just a notation saying that they are... So, here is the one of the simplest tests which can help us to analyze convergence or divergence of a series. Given two sequences of non-negative terms a n and b n, say that a n is less than b n. Ultimately, what does ultimately mean from same stage onwards? Ultimately means for some stage onwards because it is a tail which is going to matter for convergence. So, you can write ultimately that is for some n bigger than capital N, a n is less than or equal to b n. So, that means what? Each term of the sequence b n is dominating the term a n from some stage onwards. So, what will happen to the partial sums? Partial sum of a n will be dominated by the partial sums of the series b n because a n is less than b n. So, if the partial sums of b n converge, partial sums of a n are dominated by partial sums of b n. So, that will converge because it is less than or equal to b n. And if a n do not converge, then b n cannot converge because a n are less than b n. So, you get two ways of writing it, just writing the partial sums of the corresponding things that if b n is convergent, then the series a n is also convergent. Is it okay? Because the partial sums of a n will be less than or equal to partial sums of b n and that converges. So, no problem. And because a n is less than b n, if a n is divergent that means what? The partial sums go to infinity where partial sums of b n are bigger than partial sum of a n. So, b n also will partial sums of b n also will go to infinity. So, it implies that if a n is divergent, then b n is divergent. If b n is convergent, then a n is also convergent. Simple comparison of two series. From some stage onwards, a n is less than b n. Very simple proof. So, no problem about that. So, let us skip the proof. Try to write a proof yourself. What will be involved? Let me just write probably once so that you understand why some minor modifications are required. So, we have got a n less than equal to b n for every n. For every n or from some stage onwards, that does not matter. So, what is s n equal to a 1 plus a 2 plus a n will be less than or equal to b 1 plus b 2 plus b n and that is s n of this. What shall I write for s n of that? Some notation. So, let me write s n dash. We are calling that as s n dash. We are calling it as s n. So, this implies this. So, s n dash convergent implies s n convergent. Is it okay? Why? How should I justify this statement? These are partial sums only. s n is partial sum. This is a partial sum. Now, we can write this. This is okay. But I am saying this implies. So, this statement implies convergent is or hence s n dash convergent implies s n convergent. Why is that? b n is not convergent. Yeah, series b n is convergent is same as saying this is convergent. So, this is because series b n is convergent. Why does it imply s n is convergent? Yes, you have to say something more, bounded by what? This partial sum is less than or equal to partial sum of that. So, the claim is s n is convergent. Why? What is the reason s n is a sequence of numbers? Why is it convergent? s n prime. See, both are non-negative terms. So, s n dash is increasing. So, limit of s n dash will dominate all s n's. Limit of s n dash will dominate all s n's. Is it okay? And s n itself is also increasing. It is non-negative terms. So, partial sums are again increasing. So, this is also an increasing sequence of non-negative terms which is bounded above and hence it must converge. So, we are using both things. If a sequence is monotonically increasing and convergent, what is the limit? It is upper bound, least upper bound. So, all s n's dash for every n is less than or equal to the upper bound which is the limit which exists. So, s n's are bounded, monotonically increasing. So, they converge. So, that is the argument that we have to supply in between. Similarly, if we want to say that a n's are, sigma a n is divergent, that means what? That the partial sums s n are converging to infinity and s n is less than s n dash. So, s n dash also will converge to infinity because they are bigger than s n and s n is going to infinity. So, if a n is divergent, then b n also is divergent. Simple observations about sequences only. So, that is the comparison test. So, let us skip the proof of that. So, let us look at examples of comparison test. So, we are comparing two. So, look at the series. Remember, we did 1 over n that was divergent. 1 over n square was convergent. So, now we are looking at for any p between bigger than 0 less than infinity. What can we say about that? So, let us assume, so it depends on p of course, because n equal to 1, we know it is a divergent. So, let us look at when p is between 0 and 1, then n to the power p is less than or equal to n because p is between 0 and 1. So, what happens to 1 over n p? That is bigger than 1 over that will be bigger than 1 over n and 1 over n is divergent. So, 1 over n p will be divergent. The sigma 1 over n p is divergent for p between 0 and 1 by comparison test, comparing it with 1 over the series 1 over n. Simple observation. So, that is divergent. So, this is divergent between 0 and 1. Let us look at when p is bigger than 2, we know 1 over n square was convergent. p was equal to 2. So, let us take p bigger than or equal to 2. What happens to n p and n square? If p is bigger than 2, compare n to the power p and n to the power 2 n square. n to the power p will be bigger than n square. So, convergence of 1 over n square will give you convergence of 1 over n to the power p for p bigger than or equal to 2. Already proved convergence of 1 over n square, already proved divergence of 1 over n, comparison gives you for p between 1 and 0, n p is divergent. For p bigger than or equal to 2, 1 over n p, sigma n p is convergent. And 1 and 2, we have not done anything yet. Because we have just known things, we have to compare it with something known, a n b n, a n less than or equal to b n. If you know something about b n convergent, then you can say a n convergent, sigma a n convergent. We will do that also a bit later. For example, let us look at this kind of a thing. So, how these things help us analyzing 1 over 2 n square plus n plus 1. It looks like 1 over n square, it looks like 1 over n square. So, can we compare 1 over 2 n square plus n plus 1 with 1 over n square nth term? Already 2 n squares, we are increasing the denominator, so making it smaller anyway. So, 1 over 2 n square plus n plus 1, that you call as a n is less than or 1 over n square, which is b n, that is convergent. So, this is convergent. So, how you think and what you have to compare with, that you have to sort of keep in mind. So, this is convergent. I think there are more examples, you can study later on. Let us look at something which requires a bit of thought, but not much. Suppose, is again kind of comparison, but we are going to look at, see we looked at 1 over 2 n square and all that and 1 over n square. So, 2 n square and n square sort of compatible with each other kind of a thing, we could compare. So, here we are looking at a n and b n, two sequences of positive real numbers, only for the time being positive. And look at the limit of that, suppose that limit exists and is l. So, what does the meaning of saying this l exists, a n over b n. That means, eventually a n and b n are stabilizing. They are becoming sort of coming to a common kind of proportion kind of a thing. So, the claim is, if this limit exists and if this limit is not equal to 0, then you can have a inequality which says for some alpha and beta, alpha b n will be less than or equal to a n is less than or equal to beta times b n, if this limit is not equal to 0. In what way that is useful? It helps you to compare a n and b n. It says from some stage onwards, you can compare a n and b n. So, convergence of one can imply the convergence of other or divergence of one can imply the divergence of the other. So, this limit becomes important. Let us look at, first why is this thing happening? I want to look at a simple analysis of real line is coming back to a picture. So, a n divided by b n limit n going to infinity equal to l and that is not equal to 0. So, here is 0, here is l. Either on the positive side or on the negative side, either side does not matter. It is away from 0. So, what is the meaning of saying that the limit exists? Limit exists means the terms should come in a neighborhood of that point after some stage onwards. So, let us define a neighborhood like this, say l minus epsilon and l plus epsilon. So, let us choose epsilon, say that if l is bigger than 0, choose, say that l minus epsilon is also bigger than 0, then convergence there exists some n naught such that a n by b n belongs to l minus epsilon and l plus epsilon for every n bigger than n naught. That is simple convergence. It comes in a neighborhood. I want to stay away from 0. So, I have just taken l minus. So, what does that mean? That means, l minus epsilon is less than a n by b n is less than l plus epsilon. Call this number as alpha. Call this number as beta. So, then alpha times b n is less than a n is less than beta times b n. You get the inequality. Simple definition of limit. If the limit is not 0, a n over b n should stay away from 0. That is all. And if this limit is equal to 0, it will mean what? So, this is l bigger than 0. Similarly, l less than 0 does not matter. This is l bigger than 0. If it is less than 0, what will be your argument? So, if l is here, then you will have l plus epsilon n. Then you will choose epsilon such that l plus epsilon is bigger than 0. Here, we have taken this. In this case, we will choose epsilon such that l plus epsilon is bigger than 0. So, everything will be inside this. Again, this will be your alpha that will be your beta. They are positive. I am saying this. So, it is a valid point. I am just giving you general arguments for any sequence a n and b n, not necessarily positive. In our case, in other case, this does not matter. But in general, I am saying the limit of a sequence. If it is not 0, then everything should be in a neighborhood of away from 0. Now, if the limit is equal to 0, that is possible. So, let us say that limit of a n by b n is equal to 0. That means what? By the same argument, then there exists. So, for every epsilon bigger than 0, there is a n naught such that a n by b n between minus epsilon n plus epsilon in a neighborhood of 0. So, that means what? Minus epsilon b n is less than a n is less than a n plus epsilon b n times epsilon. But a n, in our case, all the a n's are non-negative. So, 0 less than a n less than epsilon b n, because all the a n's are non-negative. So, if the limit is 0, now you get only one inequality a n and less than or equal to b n. But still, it gives you a comparison between a n and b n. You can compare a n with b n. So, that was the lemma. Now, you can use this lemma in view of the lemma. So, what will that give me? If the limit is not equal to 0 of this thing and supposing b n is convergent, then look at this part of the inequality a n less than or equal to beta times b n. b n is convergent. So, that will imply a n is convergent. And if a n is convergent, then I can use this part to say b n is convergent. So, if the limit is not equal to 0 of a n by b n, then either both a n and b n converge or both diverge. If sigma a n is convergent, then sigma b n is convergent and other way around. If the limit is equal to 0, then only convergence of b n can apply convergence of a n or divergence of a n can apply divergence of b n. It will not give you if and only if. The idea is that comparison test. When you want to find the sum of a series, it is only some point onwards the things matter. How are we getting this? We are saying that l minus epsilon l plus epsilon a n by b n is in between for n bigger than or equal to n naught. So, this comparison is valid only for and that is good enough for convergence. So, that gives me that test which is the limit comparison test. So, either sigma a n is convergent if and only if sigma b n is convergent. If limit is 0, then b n is convergent implies a n is convergent. So, how this test helps in analyzing series of non-negative terms that is important.