 So, we were looking at series of non-negative numbers and we looked at various tests of convergence. For example, comparison test, root test and integral test. We start looking at series which are not necessarily non-negative. So, let us define series a n to be absolutely convergent. If you take the absolute value of each term and make a new series, that is convergent. The big type of a is absolutely convergent if this is convergent. So, this last word, divergent should be convergent. And we say the series is conditionally convergent if the series is convergent, but it is not absolutely convergent. So, sigma a n is convergent, but sigma mod a n is not convergent, then we say the series is conditionally convergent. And there is a particular type of series in which the terms become alternatively positive and negative. So, either first term is positive, second term is negative and so on or other way around. So, such series are called alternating series. So, for example, if you remember that series minus 1 to the power n divided by n, that was a series whose first term was equal to minus 1, second was minus 1 by 2, plus 1 by 2 and so on. So, that was called alternating series. So, what we are going to look at is some of the properties of this series. So, already we have seen one example that the series, alternating harmonic series, that is conditionally convergent. Obviously, we proved actually it is a convergent series and if you take the absolute values, that becomes 1 over n, sigma 1 over n, that is not convergent. So, this is a series which is conditionally convergent. And if you look at the series cos n by n square, you can compare it with 1 over n square because cos is bounded by 1. So, by comparison test, absolute value of cos n divided by n square is less than 1 over n square. So, a n less than b n and b n is convergent, so a n will be also convergent. So, this is a series which is absolutely convergent. And you can give more examples like that. The series can be absolutely convergent. So, basically for a series to be absolutely convergent, the tests are basically that of non-negative terms because absolute value of each term is a non-negative term. So, checking whether a series is absolutely convergent or not, you have to apply the tests for non-negative term series like root test, comparison test and so on. So, here is a theorem which says that if a series is absolutely convergent, then it is also convergent. So, there is only one way, obviously, that a series is absolutely convergent, then it is also convergent. So, what we are going to do is to prove that it is convergent, we will compare it with a series which is convergent. So, let us, very simple constructions, let us define b n to be a n plus mod of a n. So, if a n is less than 0, what is b n? If a n is less than 0, then mod of a n will be equal to minus, so this will be 0. And if a n is bigger than 0, then this will be mod of a n is mod a n, a n itself, so it is two times that. So, this is a simple observation that b n is equal to 0 or b n is equal to 2 of mod a n. Can you say that the series b n is convergent? Because a n is given to be absolutely convergent. So, b n is either 0 or 2 mod a n. So, we can compare it with mod a n, which is absolutely convergent. So, series b n is convergent. So, series b n is less than or equal to mod a n, which is absolutely convergent. So, series b n is convergent. And what is b n? b n, what is a n? a n is b n minus mod of a n. That is how we define. So, if you go back, a n is equal to b n minus mod of a n. And we had already proved algebra of series, if a n and b n are two series, which are convergent, then the difference sum, they are all convergent. So, using that fact, we get that a n is sigma b n minus sigma mod a n. So, this is also a convergent series. The simple thing that define b n to be equal to a n plus mod a n and compare it with mod b n. So, it says that if a series is absolutely convergent, it is also convergent. It is an necessary condition. Obviously, it is not sufficient, because a series may be absolutely convergent, may be convergent, but not absolutely convergent. That alternating series example, alternating series is convergent, but not absolutely convergent. This is only conditionally. So, very simple result, which relates the two. You can have more examples like this. It looks quite complicated. It says a n is equal to 1 if n is 1. It is minus 1 to the power n by 2 n if n is a prime and 1 over 2 to the power n otherwise. It looks like a geometric series. But only when n is a prime, it is not 1 over 2 to the power n. It is minus 1 to the power n. So, it will be a negative term there. But if you look at the absolute value of this series, that is convergent, because that is a geometric series. If you take the absolute value of this series, so mod a n, that will be 1 over 2 to the power n for all n. So, that is a convergent series, because that is a geometric series with common ratio less than 1. So, that is convergent. So, this is a series which is absolutely convergent and hence it must also be convergent. So, this is one way of analyzing series that sometimes the absolute convergence is easier to prove than the series being convergent directly. So, you prove it is absolutely convergent. As a consequence, the series becomes convergent. So, that is applicable here. So, geometric series which is convergent. So, here is a test specially applicable for alternating series when the terms are alternatively positive and negative. Only for those series is a useful theorem. It says that supposing a n is an alternating series. So, alternately either first term is positive, second is negative, third is positive or other way round. The first one is negative and positive. Since the two conditions must be satisfied, the terms of absolute values of the terms of the series should be a decreasing sequence. Mod a 1 is bigger than mod a 2 and bigger than mod a 3 and so on. So, it is a decreasing sequence. Not only it is decreasing, we should also have that limit of mod a n is equal to 0. It is decreasing to 0. Then the series is convergent. So, this is a test for alternating series. We will not go into the proof of this. We will assume it. So, for example, let us look at, remember we proved that alternating harmonic series is convergent. If you look at the terms, what are the terms of minus 1 to the power n plus 1 divided by n? So, you take absolute values that is equal to 1 over n. So, that is a decreasing and goes to 1 over n. Mod a n goes to 0. So, this theorem is applicable for that alternating series. So, as a consequence you can say, consequence of alternating series test that alternating harmonic series is convergent. We proved it by definition itself. But this is a consequence of this theorem also, how this theorem is useful. So, this is alternating series test. You may come across these things. So, proof is slightly long. So, we will not go through the proof. So, for example, this alternating series, the absolute values are decreasing and goes to 0. So, that is convergent. So, we can have more examples like this. Let us look at this series. We want to know whether this is convergent or this is alternating series. Minus and plus terms are coming. Let us look at mod a n, absolute value. So, this will go up. So, it is 2 to the power n divided by n square. What is the limit of that? That is a sequence. So, sequence is 2 to the power n divided by n square. What do you think will be the limit of that? Let us try to make a guess. That is what important is. Then you can prove it or disprove it, whichever way you want it. 2 to the power n divided by n square. Let us look at first few terms and try to make a guess. So, n equal to 4, for example. So, 2 to the power 4. So, what will it be? 2 to the power 4 is 16. And 4 to the power 2 also is 16. So, that is 1. So, let us look at 5. So, as n becomes larger and larger, 2 to the power n starts becoming much bigger than n square. So, that is how you should try to think of the numerator is becoming larger and larger, at a rate much faster than the denominator. So, this will not converge. So, it should converge to plus infinity. So, that is a guess, because numerator is becoming larger and larger compared to the denominator. So, to prove that, one way is you apply the allopithal rule here. So, when you apply the allopithal rule, you get this. And that clearly says the 2 n cancels at, it is same as a limit of 2 to the power n minus 1. So, that goes to infinity. So, what is the consequence of it? So, this alternating series cannot converge. Because mod of a n goes to plus infinity. Even you can apply the nth term test. If a series is convergent, then the nth term a n must go to 0. And mod a n is going to plus infinity. So, anyway a n cannot go to 0, because if a n goes to 0, then mod of a n also will go to 0 anyway. So, that cannot happen. So, either way you can say this series is not convergent. You can apply, no, no, previous theorem is only one way necessary. Sorry, these are sufficient conditions. The alternating series tests, all tests are giving you sufficient conditions. They are not necessary conditions. Test anywhere, whether in calculus or in series or anywhere, those tests always give you sufficient conditions, but not necessary. For example, maximum minima. If something happens, if a secondary derivative is bigger than 0, then it is a local maximum, not the other way around. So, that may not be true. So, alternating series test also is a test, which is giving sufficient conditions for something to happen, namely, alternating series. If mod a n is decreasing and decreases to 0, then it is convergent. That does not mean other way around also is true. So, always be careful. So, there are more examples that you can study later on. Let us not discuss. There is something in the series, which in some one way probably some of you might have already come across a series of the form sigma a n x minus series to power n. So, what is this? We are trying to add up terms a 1 x minus series to power 1 and so that is a 0. You can also have a 0 if you like. You can take n equal to 0 also here. This is n equal to 0. You can put a constant term also. So, what are the terms look like? They look like some constant, nth term looks like a constant a n x minus a scalar c, which is fixed raised to power n. So, it looks something like geometric series. You can think of, where the coefficient a n's are also varying, powers are increasing. This x, what is x? x is a variable, which can take any real value. So, such a series is called a power series. Because it is x minus c, we say this power series is centered around the point c. So, this is a power series with variable x centered around the point c. So, we want to say that this series is convergent or not. We want to analyze the convergence or divergence when x is varying. But we only know when x is fixed. When x is fixed as a real number, then there is a series of real numbers. So, we can analyze convergence or divergence of this. So, we say that the series converges at the point x is equal to x 0. If you put the value of x is equal to x 0, that is a series of numbers now, that is convergent. So, basic problem is to analyze for what value, given a series, power series like this, for what values of x it will converge. How to analyze that? To analyze that, the simplest thing is, let us look at x to the power n. This series x to the power n is centered around x minus 0 to the power n. So, it is centered around x equal to 0 or c equal to 0. x to the power n, so it is a geometric series with common ratio x. So, we can analyze for what values of x this will converge that we have already seen. So, it converges for mod x strictly less than 1. So, this series is convergent for mod x less than 1, strictly less than 1 and we know what is the sum. Sum of geometric series 1 over 1 minus common ratio, so 1 over 1 minus x. So, this is a geometric series which is convergent. So, this power series is convergent for all the values x between minus 1 and 1 and the sum is equal to 1 over 1 minus x. So, now, this looks like a function equation. For the domain x is equal to minus 1 to 1, the function f of x is equal to 1 over 1 minus x is equal to x to the power n for x between minus 1 and 1. So, this looks like a function being defined by a series and is an important question in mathematics. You will see it comes at various places. When can a function be represented in the form of a series? So, one particular case will come across today, but others are quite important. There is something called Fourier series. There is something called in statistics probability will come across characteristic functions of distributions and so on, trying to express a function f in terms of a series. So, here is the simplest case of power series. So, this is an example x to the power n is convergent power series with whenever mod x is between minus 1 to 1. So, look at this, for example, in the power series centered at x is equal to 3, because x minus 3 is to power n. We are giving very simple examples. So, how do you analyze this convergence of this? It is again a geometric series. It is again a geometric series. What is the common ratio? Minus 1 by 3 into x minus 3. So, we know apply the geometric series test. So, this will be convergent if and only if the common ratio is between minus 1 and 1. So, minus 1 by 3 multiplied by x minus 3, that should be less than, mod of that should be less than 1, common ratio. So, that gives you the range for which values of x. So, if you analyze that, so it says x is between 0 and 6, mod of 1 by 3, x minus 3 less than 1. So, simplify that. So, that means this series, the given series is convergent when x lies between 0 and 6. Simple geometric series we are looking at. So, that is the advantage of power series that you can apply in one way or the other geometric series and get some results. And the sum is 1 over 1 minus x. So, you can find out the sum. So, there are more examples you can look at. Now, I think probably, yeah. So, it makes sense to define what is called the domain of convergence of a power series. So, given a power series a n, x minus c raise to power n, one of the things will happen. So, what will happen? The series converges only when x is equal to c. Or there is a number r, such that it converges for all values x minus c less than r and diverges for mod of x minus c, strictly bigger than r, strictly less than r and strictly bigger than r. Or the series converges absolutely. That is also a possibility. The series converges absolutely. So, one can prove that only one of these possibilities can hold and one will hold. Again, the argument why one of them only will hold will not go through that. So, for a power series, the possibilities are the series converges for only one point, x is equal to c. Or there is a interval around x, x minus r to x plus r, so that this is convergent. And if the value is strictly bigger like geometric series mod x less than 1, strictly less than 1, it was convergent. Bigger than or equal to 1 was divergent. So, it says bigger than r, it is divergent. Equal to r, we do not know what can happen. So, that may depend upon the series and converges absolutely for all values. That is another possibility. So, three possibilities. So, given these three possibilities, let us skip the proof. We define what is called the radius of convergence of a power series. So what is the radius of convergence? First possibility is all, it was converges only at one point. So, in that case, we say that the radius of convergence is 0. Because only at x is equal to c, it is convergent. Around x equal to c, there is no interval only that point. So, there is no length of the interval, it is a only one point in that interval. So, and if it converges for all, then you say r is equal to plus infinity. It is a whole real line. So, that is plus infinity. And if there is a positive number r, such that for x minus c bigger than r, it diverges and converges absolutely for x minus c less than r, then you say that this r is called the radius of convergence and x minus r to x plus r is called the interval of convergence, no open interval. It is called x minus c. So, that means this is the interval around c, c minus r to c plus r, x lies between that. So, this says that. So, that is called the interval of convergence. So, either it will be 0, r is equal to 0, only one point, r is the whole of real line, or it is an interval, open interval in which surely it will converge. But we do not know what happens at the end points of that interval. For example, in geometric series, when x is equal to plus 1 or minus 1, the series diverges. So, at end points, anything is possible. So, let us look at one example. Look at the power series x minus 2 raised to the power n divided by n square. So, how do you analyze convergence of this series? Power n, n square. So, which test do you think will be more suitable? I cannot apply comparison test here, because our numerator and denominator both are increasing and becoming larger. It is a power n. So, it looks more suitable to take the ratio test here. When I take the ratio, powers will tend to cancel out. So, let us apply the ratio test to this and see what happens. So, take the ratio and limit comes out to be mod of x minus 2. So, that is simple. Powers cancel out, limit mod x minus 2 n over n plus 1. And that gives this, remember that series goes to, that sequence goes to 1, limit of that. If you want to see why n plus 1 minus 1. So, it is 1 minus 1 over n plus 1. So, that goes to 1. So, this is mod x minus 2. So, it says that this series will converge absolutely when mod x minus 2 is less than ratio test, less than 1. So, ratio test less than 1. So, it will converge absolutely and that means between these limits. And it will be divergent outside. When it is equal to 1, so what is the interval? Minus 1 to 3. That is the interval of convergence. When x is equal to 1, what is that series? n point x is equal to 1. So, minus 1 to the power n divided by n square. Is that convergent? That is the alternative series. You can apply the alternative series test. Mod of, that is 1 over n square, that goes to 0. So, that is convergent for at the n point. So, it is absolutely convergent for that also. When x is equal to 3, what happens? x is equal to 3. That is the other n point. So, x is equal to 3, 3 minus 2, that is 1 over n square, that also is convergent. So, not only the radius of convergence is equal to 1 to 3, that open interval, but even the n points it is converging for this series, particular series. So, n points you have to analyze separately. So, that is convergent. So, interval of convergence is 1 to 3. That may not happen always. So, let us, why we are bothered about this interval of convergence? Because in the interval of convergence, the power series converges absolutely. And when it converges absolutely, the limit you can call it as f of x. So, if the series converges absolutely in the interval of convergence, you can define the function f of x to be equal to this. Now, you can ask a question. So, what is the question? You would like to ask. Each term on the right hand side seems to be differentiable. x minus x 0 raised to power n, that is a differentiable function. So, we are adding up differentiable powers of a differentiable function. Is the sum differentiable? Is the sum integrable? Because they are all integrable also, x to be minus a to the power n is integrable. So, the question comes that in a power series is the function defined by the sum in the domain of convergence differentiable or integrable? So, these are two theorems. We will not again go to the proof, because so it says, consider the power series this and as a non-zero radius of convergence r. So, this is the interval of convergence. Then it says, then the derivative f is a function defined by the series. So, what is the derivative mean? f of x, f of x plus h minus f of x divided by h, limit h going to 0. So, we are asking whether you can compute the derivative of that function. It says, yes, you can not only compute, you can take the derivative inside the series. So, it is again interchange. Derivative of the sum is equal to sum of the derivatives. So, it says that the function is differentiable and this series also has the same radius of convergence and the sum is equal to f dash. So, that is a very nice thing. If a function is defined by a power series in a radius of convergence, the function is differentiable and the derivative is the derivative of each term added up together. So, that is what it says. Because x is power n is any number of times differentiable. So, when you add up a series, power series, so a power series in the radius of convergence is infinitely differentiable. Every k, the derivative exists. Every kth derivative exists and that is given by this. So, this is our derivative. So, the function defined by power series in the interval of convergence is differentiable and derivative, first, second, third, anything is equal to the sum of those corresponding derivatives of x minus a raise to power n. What about integration? Something similar happens that f of x is the function defined by the series. Then, this function f has an antiderivative and how do you get the antiderivative of a function? Fundamental theorem of calculus by integrating a to x. So, it says the antiderivative is given by integrating each term one at a time to get the antiderivative of f. You have to just take the antiderivatives of x minus a raise to power n and what is the antiderivative of that? We know. So, that is the antiderivative. So, f of x. So, as a consequence of this, f of x is equal to sigma of the integrals. So, integration, which is again a limit process integral a to b f x dx is a limiting process, limit of partition sums and so on, upper sums, lower sums. It says you can take that limit also inside the summation sign. So, in power series is a very useful result that you can take differentiation inside. Derivative of the sum is equal to sum of the derivatives integral of the sum is equal to sum of the integrals. It is a very useful result. And if you will, proof is there in the slides, you can read if you like, but from examination point of view, we will not ask you the proofs. But those who are keen to know how the proof goes, you can have a look at it. And essentially, you will have to look at the upper sums and lower sums, because integrals are coming into picture. In the power series, there is a very special case of power series. So, let me just say that what is the special case? This is a special case of power series. Remember, power series is at an x minus a raise to power n, where an's are the coefficient of the nth term. If the coefficient of the nth term is the nth derivative at the point a, where it is centered divided by n factorial, then this series is called the Taylor series for the function f. f is a function given to you and suppose the function is any number of times differentiable. nth derivative exists. So, if you can write f of x, so this is a function with the sum as f. So, f of x is equal to this thing and this is called the Taylor series. So, you can ask the question for what functions have Taylor series expansions. This series is called Taylor series expansion, because it involves derivatives of the function f. But it may or may not converge. It is a power series. So, what is the domain of convergence of this series? In that domain of convergence, will it represent the function f itself or not? That is the question. So, most of you must have done this in your undergraduate course is Taylor series. When the a is 0, this is called Maclaurin series. All those series sin x, cos x, e raise to the power x, they have the series expansions. What are those series? Precisely, Maclaurin series x is equal to 0, a is equal to 0. So, you ask, I hope you have done these things. If not, anyway probably you will do it sometime or you should revise. So, you want to know that when will this converge to the function f? So, what are you interested in? You are interested in knowing f minus the sum, partial sum, when there is a series converge and the partial sum goes to 0. So, the sum from some stage k to infinity, that is the remainder term. In the series, that remaining thing is called the remainder term. So, you have many ways of representing remainder terms and analyzing when does the remainder term goes to 0, in what domain it goes to 0. So, for the trigonometric functions, exponential functions, you show that the remainder terms go to 0 for all values of x. So, that is a particular form of our series. So, with that, we end the course.