[MUSIC] Check what we have introduced so far through the example. First, example one. Okay. I would like to determine the interval of convergence of, okay. I'm interested in IOC of the policy is some 0 to infinity -2X+1 to the end over 12-1. Okay, right. We can re write it as some of those little to infinitive -2 to the end X -1/2 to the end over 2x-1. Okay, I think now we can recognize the center and the coefficients. Right, what is the center then? Centre is 1.5, right? Correspondent co efficiencies minus to the end over 12 minus 1, right, okay. So syntax 0 is equal to 1.5, right? And the system of N is equal to -2 to the n? Over 12-1, right, okay. To find this the the interval of convergence forced to find the radius of convergence, right? So then what is r? Okay, by the formula you can compute it as a limit of intensity, infinity of absolute value of c sub n. C sub n is minus 2 to the n over 12 -1, right. Okay, divided by 67 plus 1, 67 plus 1 is minus 2 to the N + 1/2N +1, right. Upside down you will get 2N plus 1 over minus 22 D, N+1. Okay, so that is a cord 2, okay. This is 1.5 of the limit of N tends to infinity of okay, 2n +1 over 2n-1 that is equal to 1.5, right. So that's the radius of convergence, right, okay. What does that mean? So I said there's a convergence, right, okay. To give them policy series away. Give them policies converges absolutely four. Absolutely value of x-1.5 is less than 1.5, right, okay. Absolute value of X -1.5 in less than 1.5. In other words for axis between 0 and the 1 right, okay, converges absolutely. It diverges okay, outside of that interval so diverges 4 X less than 0 or X greater than 1 y. Okay, at 2 boundary points say X is equal to 0 in the one we do not know, okay. We must check its convergence or divergence individually, right? So when X is equal to 0, okay, plugging in X is equal to 0 here, right. Then you will get some of the 0 to infinity of trivially one to the end is one. So 1 over 2 minus 1. Okay, can you add those two numbers? Those numbers 1/10-1. Okay, I think you can easily see that this is a plus infinity. Okay, easy way to conform. It is by the so called the integral test. Okay, that's right, okay. On the other hand there X is equal to 1. The c becomes fx is equal to one then started to become summation. 0 to infinity of -1 to the n over 2n-1, okay. No to that, every time has a changing signs, right, okay. So this is the so called the alternating series and we know that it converges right, converges by. Okay, alternating see it as a test, okay. Combining what we have so far, converges absolutely one actually is strictly between 0 and 1. Okay, divergence at X is equal to 0, convergence at X is equal to 1. So finally you can say the interval of convergence is half open interval0 to 1, not including 0 but including 1. Okay, that's the interval of convergence, right? Let's look at one more example. Okay, now interval of convergence of the following cities. Okay, following Power Cities. Power City is the sum of from 0 to infinity of to the 2n+1 and x+1 and 2n+1. Okay, we are interested in this interval of convergence. Right, forced to know to that. Okay, this number 2 to the 3n+1. This is the coefficient okay, 2n+1 the term of the power siries, right? In symbol this is A over 2n+1, okay. Only the other powers of express one survive and we do not have any even powers of X+1, right, okay. So that you cannot compute r by intensity infinity of a 7 over 7+1. Right, this limit does not exist because if anything even then this is 0. If n is older then that is even so the denominator will be 0, so that okay, this rate is undetermined, right? So no way to computer are by this approach, okay. Here's a very simple trick. Okay, so this method does not work, right. I will approach this problem in the following way. Okay, convergence of some of those little to infinity of 2 to the 2n+1 and X+1 to the 12+1. Okay, let me say it this way, okay. It converges if I don't leave. Okay, converges okay, sum of 0 to infinity up to 2n +1 and X+1 and to the 2N converges 2n. Think about it, right. Do you agree? Right, I'm just neglecting one power of X+1 from this expression, okay. It has no effect times X+1, right? Because you can write it as a X+1 times this way convergence of this. It's the same as the convergence of only this part. Right, trivially okay. Okay then I'm changing okay, Lexus said these equal to X+1 scare and consider the G power series right. 2 to the 12+1 to the end. It becomes okay as a policy Z, okay, coefficient is 2 to 2n+1. Right, okay, so from this expression okay, sure. It's radius convergence, right from this one. Right of which radius of convergence is, okay. Now let me apply the second formula for the r, okay, it's a limited and tends to infinity right? And through the tub, absolutely 2 to the 2n +1, okay. And it's in bus, right? That's the second formula for the radius convergence, right, okay. How much is it? And this is equal to this is a limited and tends to infinity right 2+1 over N and its inverse, right, okay. So when the intestine infinity one over and tends to 0, so that this is a 4 and 4 to the -1, that is equal to 1 quarters, right? That's the radius convergence of this deep our series, okay. So what does that mean then? Okay, summarize the conclusions. Okay, series 2 to the 2n+1 and Z to the n, okay. In fact, this is equal to summation 0 to infinity of 2 to 2n+1, X+1 and 2N right, okay, it converges absolutely. Okay or diverges. Okay, if absolutely value of Z. Ok, less than 1 quarter. Right, okay, that is equal to X+1 skill, right, okay. If this is less than one quarter, right, okay diverges respectively, okay, if absolutely value of Z in other words x+1 squared if this is greater than 1quarter than it diverges, right, okay. Read this information in terms of this exit polls citizens. Okay then what this policy is a converges, absolutely. If Absolutely Valley Express one scare is less than one quarter or diverges if this is greater than one quarter. Okay, so now we have a 0 to infinity however, 2 to the 12plus 1 X+1 to the 2 N. Ok, converge, absolutely. If absolutely, value X plus one can solve this. It less than 1.5 diverges if absolutely value x+1 is greater than 1.5. Right, Okay. Yes, simply find this inequality. Okay, say that we can say that. Okay, summation from 0 to infinity of 2 to the 2n+1 X+0 to the 2n. Now we have a convergence if right, absolutely. Value of x+1 is less than one half, okay. If it's already then this is a -3.5 and between access between this and -1.5 and diverges right? If actually X less than -3.5 or if X is greater than -1.5, right. And still undetermined. If x is equal to either -3.5 or minus 1.5, right? That's the conclusion, right, okay. So let's look at those two points, right. Convergence of those two points separated, okay. If you're plugging those two numbers into the power cities, right then okay. Yeah, I'll leave it to you, but my claim is at both these points, okay, diverges. Okay, easy to see that. So finally, the what is IOC, okay? IZ equal to in the second example, that'll be open interval from -3/2ves to the -1 half, right? That's the the corresponding interval of convergence, okay.
