Definition of a power series. A power series about X = 0 is a series of the form. The summation from N goes to 0 to Infinity of C sub NX to the N power. So that really just equals C sub, not plus C sub One X + C sub two X squared, et cetera. Now if the power series is about a different location, let's say X equal A, then it's a series of the form, the summation n = 0 to Infinity of C sub NX minus A to the north equaling C sub naughty plus C sub One X -, a + C sub two X -, a ^2 plus... in which the center A and the coefficient C naughty C1C2... CN are constants. An example, the geometric series. If we take all coefficients equaling 1 and we let our R be our X. So we'd have the summation of n = 0 to Infinity of X to the north, or 1 + X + X ^2 + X ^3, etcetera. When will this converge? Well, we know in the geometric series, it converges when the ratio, the absolute value of the ratio ratio is less than one. So in this case, the absolute value of X is less than one. Or we could think of it as -1 less than X less than one for it to converge. So now we're going to talk about the radius. And the radius convergence is the distance here divided by two. So the distance of the convergence is two 1 - -1 and when we divide it by two we can see that the radius the series radius for the geometric series is 1. And we actually can figure out what it converges to because in the geometric series we know that it converges to a sub one or the first term over 1 minus rural route. So 1 / 1 -, X for -1 less than X less than one. Now, what's nice about this is we can actually start to look at what a graph's going to look like for a series. So if we know that the partial sum 1 / 1 -, X is going to be the same as the sum if it converges. So remember, it converges from -1 less than X less than one. So what we're going to do is we're actually going to start graphing these individual terms, and at some point we're going to have an error that we're not going to worry about the rest of the terms. So if we thought about Y not equaling just the plain one, that's going to be this horizontal line. And then if we thought about Y1 equaling the 1st 2 terms, 1 + X, well, that's a line with a slope of one. And then if we said Y 2 equal 1 + X + X ^2, it's getting a little closer. And then if we said Y8 somewhere, one plus, that's this pink line. And the actual graph is this blue line. So depending on how close we need it to be, how much precision we need can determine how far out we're going to go with our different terms if we were thinking about to graph it. So let's look at some other examples. We're going to determine where the series converges and we're going to use the ratio test. So this is a power series because it's something to a power. So we're going to use the ratio test. The limit is N goes to Infinity of X + 5 to the n + 1 / X + 5 to the N. When we simplify that down, we just get X + 5 because the NS will actually cancel when we subtract. Well, we know that using the ratio test, if the ratio is less than one, it converges. So we're going to take this ratio here and figure out when it is less than one. So absolute value of that less than one -1 less than X + 5, less than one. Subtract the five so we can see -6 less than X less than -4. So our radius of convergence is 1. The distance between the two is 1. Now we know that it's converging in between -6 and -4 and we actually know it's absolutely converging in there. So our next question should be what happens to the endpoints? What happens at -6? Does it converge? Does it absolutely converge? Does it conditionally converge, Or does it diverge at the endpoints? So when I put X equal -6 back into the original, I get. Simplifying it down, the summation of n = 0 to Infinity of -1 to the N Well that diverges because it's oscillating. And when X equal -4 N equals 0 to Infinity of -4 + 5 to the N or n = 0 to Infinity of one to the N. This also diverges by the NTH term test, so the interval of convergence is always absolutely convergent. The endpoints may absolutely converge, diverge, or conditionally converge. So in this example, the radius that converges was one. The interval of absolute convergence is -6 less than X less than -4, and there were no conditional convergence locations. Let's look at another example. If we have N equal 1 to Infinity of the quantity three X -, 2 to the N all over N we're going to use the ratio test, and we want that absolute value of P to be less than one for it to converge. So the limit is N goes to Infinity of three X -, 2 to the n + 1 / n + 1 all divided by three X - 2 to the n / n So we get it to simplify the three limit as N goes to Infinity of n / 3 X -2 / n + 1. Now when N goes to Infinity, we realize that these are just going to end up being one because of the leading coefficient test. But we could, if we wanted to, divide everything through by just an N. So n / n would give us a coefficient of one here, the leading coefficient idea. And then if we took this denominator and divided each term by N, we'd get 1 + 1 / n. And then when N goes to Infinity, we see this one goes to 0. So now we have the absolute value of three X -, 2, and we need it to be less than one for it to be convergent. So we're going to put it in between -1 and one. We're going to add 2 to each side. We're going to divide by 3. So the radius of convergence, we're going to take the distance here, which is 2/3, and we're going to divide it by two. So the radius of convergence is 1/3. You could think of this as a diameter, and the diameter to get a radius, we divide it by two, the distance. So now our next question is what happens at the endpoints? So we're going to stick the endpoints back into the original N equal 1 to Infinity of 3 * 1/3 -, 2 to the north over N Well, this is actually the alternating harmonic series, and it converges conditionally. So at this end point, it's a conditionally convergent. What happens at 1:00 when we stick one in? We end up simplifying it up to one to the north over N That's a harmonic series, and it diverges. So our interval of convergence, it is going to converge at 1/3. So it's going to be 1/3 less than or equal to X less than one, but it's going to be absolutely convergent just within one third and one. So 1/3 less than X less than one, and it's going to be conditionally convergent at X equal 1/3. If we look at another one, we're going to do the same steps every time. So we're going to do the ratio test. We're going to simplify this out. We're going to get this limit as N goes to Infinity to just be the absolute value of X + 2. If we stick that in between -1 and one to find where it converges, absolutely going to subtract our two. We're going to get -3 to -1. Then we're going to test the endpoints to figure out is it diverges at the endpoints, absolutely converges or conditionally converges. So when X is -3 we literally stick -3 in for the X of the original, and we can see that we get one to the north power because -1 * -1 is 1 to the north all over N. Well, that's diverges because that's a harmonic series. And at -1 we stick in and we see -1 to the north over N that conditionally converges by alternating harmonic series. So the radius of convergence this time is 1. The interval of convergence is -3 less than X less than or equal to -1 because now it's the right end point that was conditionally convergent. Absolute convergence within the interval conditionally convergent at -1. If we look at another one, n = 0 to Infinity 3 NX to the n / n factorial. We do our ratio test and this time we get it equaling 0. Everything cancelled out. So we know that zero is always less than one. Thus it's true for all values of X. So our radius of convergence is going to be Infinity. Everything works. Our series converges on negative Infinity to Infinity. It's absolutely convergent the entire time, and it's nowhere conditionally convergent. If we look at another example, if we have n = 0 to Infinity of N factorial X -, 4 to the north. Doing our ratio test X + 1 factorial X -, 4 N plus 1 / n factorial X -, 4 to the north. Simplifying it up, we get the limit as N goes to Infinity of the absolute value of n + 1 * X - 4. Well, this is going to equal Infinity other than if the X itself is the number 4, because if I have X being four, 4 - 4 is 0 and 0 times anything is going to be 0. But other than that, this Infinity is going to make this X -, 4 really really insignificant except at the location when X = 4. So the radius of convergence here is 0 because it's a single point, there isn't a distance the series converges at X equal 4, it absolutely converges at X equal 4, and it never conditionally converges. If you stuck 4 back in here, 4 - 4 be 00 to the north is 00 times north factorial. So at X equal four 0 + 0 + 0 + 0, so it converges to 0. So a summary the convergence theorem for power series. If the power series from n = 0 to Infinity of a sub NX sub n = a sub naughty plus a sub One X + a sub two X ^2 plus... It converges for X = C where C is not equal to 0. Then it converges absolutely for all X where the absolute value of X is less than the absolute value of C it diver. If the series diverges for X equal D, then it diverges for all X with the absolute value of X greater than the absolute value of D. So the convergence of the series n = 0 to Infinity of C sub N X -, A to the north is described by one of the following possibilities. There's a positive number R such that the series diverges for X with X the absolute value of X -, a being greater than R, but converges absolutely for X with the absolute value of X -, a less than R. The series may or may not converge at either of the end points X = a -, r and X = a + r That was our first several examples. The series converges absolutely for every X or the real numbers. Infinity. The radius is Infinity. So that was when everything went away. But it was always always true. We got 0 which was less than one or The series converges only at a single point X equal A and diverges everywhere else. In that case, the radius is 0. How to test a power series for convergence? We're going to use the ratio or the root test to determine where the series converges absolutely. If the interval of convergence is finite, we're going to test for convergence or divergent at each of the end points using comparison test, integral test, alternating series test are usually the ones that are the most common to use. If the interval of absolute convergence is a -, r less than X less than a + r, the series diverges for all the absolute value of X -, a greater than R because the NTH term test does not approach 0 for these values of X.