 So, let us look at just some examples of that. I think that we have now p bigger than or equal to 1. You can analyze that like this one. I said n square. So, eventually, if I numerator I call as a n, denominator I call as b n. Then, eventually it looks like n over n, n square 1 over n, the limit looks like to be equal to 0. So, I can try to apply limit comparison test here. So, find out the limit of, I think there is something wrong here. Oh, I should not compare n with n square. Let us, if you directly try to compare numerator and denominator, then you will end up into problem, because you do not know either of them a convergent or not. You understand what I am saying? You should not take a n to be equal to n plus 5. I do not know that series is divergent. So, I cannot help it. So, this is my series. It looks like 1 over n, n over n square. It looks like 1 over n. So, let us try to compare this with 1 over n ratio a n by b n. So, a n is equal to this. b n is 1 over n. What is a n by b n? So, a n by b n, that will be n square plus 5 n, 1 over n. b n was 1 over n, a n. So, that limit is 1 over n square, a n square, that will be limit will be equal to 1. So, limit of a n by b n is equal to 1, which is not 0. 1 over n is divergent. So, that is what I said. If I look at the general term, it looks like 1 over n. It looks like 1 over n and 1 over n is not convergent. So, this series should not be convergent and that we are formalizing by comparing it with a n by b n limit of that. Eventually, it looks like 1 over n. That is how your thinking should go. So, that is divergent. n square n to the power 4. What does it look like? It looks like 1 over n square. So, I should compare it with 1 over n square. Same technique. a n equal to this. b n equal to 1 over n square. When you divide a n by b n, the limit will be equal to 3 n to the power 4 and so on. So, what will be the limit? What is the limit of that? It will be 3 by 1. So, that will be equal to 3. Is it okay? It will be equal to 3 by 1 here. So, that will be equal to 3. So, limit is not equal to 0. So, convergence of 1 over n square will imply convergence of this series, which is 3 n square minus 2 n plus 4 divided by that. Eventually, it looks like 1 over n square. Yes, but how do you formalize that? That is how you should think that the sequence general term looks like 1 over n square. So, eventually it should look like 1 over n square and that is what is the limit comparison test says that. Justification for that is limit comparison test. So, look at a n divided by b n. a n is this. b n is equal to 1 over n square. Compute the ratio and take the limit. That tells you eventually how does this a n compare with b n. So, that is a rigorous way of saying the same thing. So, that is convergent. So, here is a caution saying that in all these things, you have to think of what you should compare. You have to make a guess that I should compare it with this by looking at power up and down and so on and then only compare. So, these tests are not intrinsic tests. Once again, I am bringing that word intrinsic. Remember, we use the word intrinsic when we looked at convergence of sequences. Finding a sequence of a limit exists. A sequence is convergent. You require the limit, which is not given to you. But, saying is a caution. I do not require anything outside. I only have to analyze whether the terms are coming closer to each other or not. So, caution is an intrinsic property and the beauty is it is equivalent to saying the limit exists. Now, here comparison test and all these things ask me to guess something outside the given knowledge that I have to find someone like 1 over n square, 1 over n and then compare them. But, can I have a test which does not require me to do that kind of a thing? By looking at the series itself, I can say something. So, there are tests possible like that. So, let us look at that is called the ratio test. So, we are given the sequence a n. Look at the series a n. So, look at the limit of a n plus 1 divided by a n. The next term, it is ratio with the previous one. Look at that ratio. Why limit? Eventually. So, take a limit of this. That is L. If L is less than 1, then the series is convergent. Bigger than 1, it is divergent and equal to 1 is the same kind of problem that it can converge, it can diverge. So, what could be a proof of this? Tell now, what are the techniques we know? We only know geometric series is convergent. Then we knew that 1 over n square is convergent and then the comparison is and test gave me something for bigger than 2 and less than 1 and 1 over n p. So, those are known facts already. And here, L is equal to this limit. So, let us analyze what does the meaning of limit means here. If L is the limit of this ratio, that means what? And L is less than 1. So, L is less than 1. So, what does it mean? Here is 0, here is 1 and here is somewhere L. One should point out, it is strictly less than 1. So, it is not equal to 1. It is something in between. That means what? That means after some stage, all the terms of the sequence must be close to L. So, let us say they are here. So, let us L minus epsilon and L plus epsilon. So, for every epsilon bigger than 0, such that L minus epsilon less than L less than L plus epsilon, still less than 1. Let us keep it less than 1. Is it possible? L is between 0 and 1. So, for there exist some n naught such that a n plus 1 by a n is less than L plus epsilon and bigger than L minus epsilon for every n bigger than n naught. Is it okay? Definition of a limit. Now, see what is happening is it says, look at this part. It says a n plus 1 by a n is less than L plus epsilon. That means a n plus 1 is less than L plus epsilon times for every n bigger than n naught. So, this is telling me how much a n plus 1 is in comparison with a n. And this thing, it says a n plus 1 by a n is less than 1. This is less than 1. So, now, let us try to use it inductively. This is for any n, n naught onwards. So, what will happen? a n. So, let us take equal to a n, n equal to n naught. So, I got a n naught plus 1 is less than L plus epsilon of a n naught. Is it bigger than n naught? Let us go to the next step. a n naught plus 2 will be less than L plus epsilon a n naught plus 1. But a n that is already less than. So, it is L plus epsilon to the power 2 a n naught. So, what is happening? So, what will happen to the n plus k? If we look at a n naught plus k, it will be less than L plus epsilon to the power k of a n naught inductively. So, what we are saying is the terms of the given sequence a n naught from some stage onwards are bounded by this times a constant that is fixing a n naught. And what kind of? So, if I call this as b, b to the power k, what is b? It is less than 1. So, that is geometric series. So, that is convergent. So, I am saying a n's are less than b n's and b n's are geometric series from some stage onwards. So, that is convergent. So, that implies a n will be convergent. So, that is how this is useful. So, what it says, if L is less than 1, the series is convergent. If L is bigger than 1, what will happen? a n by b n will stay away from 1 on the right hand side. So, a n by b n will be bigger than L minus epsilon, which is bigger than 1. So, inductively again powers. So, it will be a geometric series with common ratio of bigger than 1. Again, comparison test will give me that this should be divergent. So, that will say the series is divergent. When it is equal to 1, either thing is possible. One can give examples. So, definition of the limit of the ratio a n plus 1 and a n is falling back upon comparison test and geometric series. The proof involves writing the limit and using the fact that the geometric series is convergent if the common ratio is less than 1. So, that gives us the result. So, we have already seen that. So, you can write it out. The proof a n plus 1 is less than that. That is less than 1 and so on. L is bigger than 1. So, we will keep on increasing comparison. So, that is bigger than 1. So, let us look at some applications of simple applications. a n is 1 over 2 n minus 1. It looks like 1 over n anyway. So, I do not have to really do anything just compared with that. But, let us take the ratio. The ratio is equal to with 1 over n. So, this is or ratio of this itself, that limit is equal to 1, so divergent. So, as we guessed, you can compare it with that if you like, either way. So, limit is equal to 1, so divergent. a n plus 1 divided by a n, if the limit is equal to 1, then the series is divergent. That is what we do. Less than 1, it was convergent. Pardon? L is strictly less than 1 convergent, because in the geometric series, common ratio strictly less than 1 only will give you convergence. And common ratio is equal to 1, that gives you divergence. 1. So, this is not strictly less than 1. Which one? Yes. So, here it is divergent. That is all. Limit is 1, but it is divergent compared to that. Which theorem? We are giving examples that when it is equal to 1, anything can happen. For the third case, I said l equal to 1, anything can happen. Strictly less than 1, it is convergent. Strictly bigger than 1, it is divergent. So, in this case, it is strictly equal to 1. But, if you apply the comparison test, it is divergent. So, equal to 1, but divergent. You can look at n square. Similarly, the ratio will be equal to 1, but it is convergent. If you look at n square, the ratio will be equal to 1. It is convergent. So, either is possible in that case. We are not applying the theorem, but we are saying that counter examples for the third case. So, there is something called the root test. It is something similar to the ratio test. It says, a n's are non-negative. Look at the nth root. Look at the limit. If that exists, less than 1 is convergent, bigger than 1. Or infinity, it is divergent equal to 1. Anything can happen. Basically, eventually analyzing the limit and bring it back to something that is already known, that is the idea of the proof. So, if limit of a n raise to be 1 over n, that is, l is less than 1, then what happens? It should say in a neighborhood of 1. It should say in a neighborhood of limit is 1. If the limit l is less than 1, it should say on the left side of 1. That means what a n raise to power 1 over n will be less than something, which is less than 1. So, when you raise the power, a n will be less than that small quantity, which is less than 1 raise to power n. Again, geometric series will be convergent. So, basically, definition and geometric are giving these tests. Similarly, for l bigger than 1, again geometric series, the common ratio of bigger than 1 will be divergent. Equal to 1, we have to give examples to illustrate that. So, a n will be less than alpha to the power n, while alpha is less than 1. I said, that will be geometric series and convergent. Similarly, by the comparison test and comparison with geometric series. When bigger than 1 a n will be bigger than on the right hand side of 1. So, l minus epsilon will be bigger than 1. So, power a n will be bigger than l minus epsilon raise to power n. That is bigger than 1. So, again, comparison will give you, it is divergent. So, basically, you can give examples. So, I think these examples you should study and try to do it. I can explain examples and it will be okay. You will nod your head. Yes, it is okay and all that, but you should understand why. So, look at the examples when the limit is less than 1, bigger than 1 and so on. I do not know whether the limit of this quantity is equal to e. Did we prove that in the tutorial classes or something? Limit of 1 over 1 plus n or n plus 1, 1 plus 1 over n raise to power n, that limit is equal to the number e, Euler's number. So, that is being used here actually. That is an interesting thing. I do not know. Was it a part of a tutorial? 1 plus 1 over n raise to power n, that limit exists is equal to e. Anyway, so that is being used here because that is actually the definition of the number e. e is a number which is called Euler's number and the same which comes in the exponential function also, e raise to power 1, exponential of 1 is the same number as this. So, there are connections. I think let me not go into that. Those who are interested, read, see the slides and try to figure out why it is the same number as that, if you are interested in mathematics. As such, unity, air coming here, coming there, both are same or not. So, I think that is a root test. There is an integral test. I will not discuss much about this because this is something, it is not difficult, but anyway it is 1 to infinity, that is improper Riemann integral kind of a thing. So, let f be a continuous function from 1 to infinity. Evaluate the value of f at the point n and if that is a n, then the series and this integral either both converge or both diverge. Remember, we had defined what is called the convergence of an improper integral. So, this relates with improper integral. I think let me not go into the proof of this. It is easy, but let me not go into the proof and let me not go into the proof of this statement of this. This you can look at to apply with something like function being 1 over x to the power p and p between 1 and 0 and then you show it is improper integral is convergent and then, remember I said between 0 and 1, 1 over n to the power p, we did not analyze. We analyze only when equal to 1, 1 over n or bigger than that. So, this is the integral test gives you, but that requires effect that this is a convergent improper integral. So, one requires that. Let me probably sum it up. What we have done today? How do we analyze testing or convergence of a series? The general rule is, check first of all whether the nth term goes to 0 or not. If it does not go to 0, it is not convergent. So, proceed only when it goes to 0, analyze convergence. You can apply integral test. If it looks like a rational function, something divided by something that n square divided by something, then the limit comparison test may work. For the standard series, you can try to compare it with geometric series, p series and so on. The ratio test works when there are factorials and powers coming, because when you divide, powers will try to cancel it out. So, you should try that. The root test is useful when nth root of somewhere it is coming. So, the general rules kind of, not rules, general hints of how to analyze convergence of series. So, we have looked at only today for non-negative terms series, but we saw one series, alternate series was convergent. If a series is not convergent, you can always take the absolute values each term and see whether that is convergent or not. So, there is something called absolute convergence of a series and series for alternating terms. So, we will look at it in next time.