 This lesson is on radius and interval of convergence. Well, remember the ratio test. We use the ratio test in determining these radius and intervals of convergence. And the ratio test read the limit as n approaches infinity of the absolute value of a of n plus 1 over a of n. And again, three conditions here. If the limit as n approaches infinity of a of n plus 1 over a of n is infinite, then there is only one point of convergence, and that is at 0. And the radius, of course, is 0, 2. So the interval of convergence is at the point 0, and the radius is 0. If the limit as n approaches infinity of the absolute value of a of n plus 1 over a of n is equal to 0, then the interval of convergence is over all real numbers, and it is infinite. And the radius of convergence also is infinite. If the limit as n approaches infinity of the absolute value of a of n plus 1 over a is between two values, then the interval is between these two values, and we must check the endpoints. And the radius is halfway between these values. So let's go on and do some examples. The first example, we have the sum of x to the n. What is the radius and interval of convergence of this? So we'll use our ratio test limit as n approaches infinity of a of n plus 1, which is x to the n plus 1 over x to the n. And we'll take the absolute value of this. And this is equal to absolute value of x. Now, this has to be less than 1 if it is to converge. So we can say that x is between negative 1 and 1. This is our interval of convergence, but we have to take it one more step, because we have to check the endpoints. So let's go back into our series and check each endpoint. If we put in a negative 1 for the endpoint, we have the series negative 1 to the nth power. And of course, that becomes negative 1, positive 1, negative 1. Of course, we remember from before that this diverges. Let's check the other side, 1. So we'll have the series of 1 to the n. And this also diverges. So our interval actually is x is between negative 1 to 1. If either of these converge, then we would put an equal to at the endpoint. What is the radius of convergence? It is just 1. Let's go on and look at another example. Determine the interval and radius of convergence of the series n factorial times x to the nth power. So again, we'll do the ratio test. We'll have n plus 1 factorial x to the n plus 1 over n factorial x to the n. We will reduce this and see what we've got. n plus 1 factorial over n factorial is n plus 1. And then x to the n plus 1 over x to the n is x. Now, if we put n approaching infinity, we get infinity out of this, which means there is only one point where this converges. And that's when x is equal to 0, which means also that the radius of convergence is 0. Let's go on. What about the series x minus 1 to the nth power over n? Let's try this one. Limit as n approaches infinity. Absolute value of x minus 1 to the n plus 1 over n plus 1 times n over x minus 1 to the nth power. And we'll reduce this one too. Limit as n approaches infinity. Absolute value x minus 1 to the n plus 1. x minus 1 to the n is x minus 1. And over n plus 1 as n approaches infinity is 1. So now we'll have absolute value of x minus 1. And when we do that one, we can say, with the ratio test that this has to be less than 1. So now we look at it as x minus 1 is between 1 and negative 1. Solve that. We get x is between 0 and 2. Let's go on and test the end points. Put a 0 into our limit, and then put a 2 into our limit. We'll have the series. And once we put a 0 into it, we'll have negative 1 to the nth power over n. And that is our harmonic series. But it is an alternating harmonic series. So it converges conditionally. So that means at 0, we'll put the equal to on the inequality. And let's try the other side. 1 to the n over n. And that one diverges. So we do not put an equal to on the 2. So our interval is between 0 and 2, including 0. If we want the radius of convergence, it is 1 because we're going from 0 to 2. Halfway is 1 unit long. Let's go on. Determine the interval and radius of convergence of negative 1 to the n times x to the n over 2 to the n times n factorial. Again, we have to use our ratio test. Take out the alternating piece. So we have x to the n plus 1 over 2 to the n plus 1 times n plus 1 factorial times 2 to the n n factorial over x to the n. Clean this up. x to the n plus 1. x to the n gives us just an x in that numerator. 2 to the n over 2 to the n plus 1 gives us a 2 in the denominator. n factorial over n plus 1 factorial gives us n plus 1. As we take a limit as n approaches infinity, we see all of this becomes 0. This means our interval of convergence is negative infinity to infinity. And our radius of convergence is infinite. So we've checked all kinds so far. We've had one between two numbers. We've had one that went out to infinity. And our interval and radius of convergence was 0. And then we have this one, which is infinite. Let's just try another one. 5 to the n over n cubed times x to the n. Again, we'll use that ratio test. 5 to the n plus 1, x to the n plus 1 over n plus 1 cubed times n cubed over 5 to the n, x to the n. Reduce this one. Limit as n approaches infinity. 5 to the n plus 1 over 5 to the n is 5. x to the n plus 1 over x to the n is x. n cubed over n plus 1 cubed. Limit as n approaches infinity makes that 1. So we'll have 5x. That needs to be less than 1. So we'll have 5x is between 1. And of course, on the other side, it's negative 1. Determining x, we get 1 fifth on the right-hand side. And we get negative 1 fifth on the left-hand side. So now we have this interval, which means we must check the endpoints. So let's first check the negative 1 fifth endpoint. And we have our series. And if we put x as negative 1 fifth, we'll have 5 to the n negative 1 fifth to the nth power all over n cubed. Well, 5 to the n times negative 1 fifth to the n is a negative 1 over n cubed. So this converges. So that puts an equal to sign on that side. Try the other one. I think you'll see that both of them will converge. 5 to the n times 1 fifth to the nth power over n cubed. This means we'll have just 1 to the nth power. 1 over n cubed does converge also because of our p-series idea. On this one, the interval of convergence is between negative 1 fifth and 1 fifth, including both ends. The radius is 1 fifth. Let's try one more. Determine the interval and radius of convergence of negative 1 to the nth power x to the 2n plus 1 over 2n plus 1 factorial. This is actually your sign as a power series. So we will only write in the non-alternating part. So this is x to the 2n plus 3 over 2n plus 3 factorial times 2n plus 1 factorial over x to the 2n plus 1. Reducing this, x to the 2n plus 3, x to the 2n plus 1 gives us an x squared. 2n plus 1 factorial over 2n plus 3 factorial actually reduces to 2n plus 3 times 2n plus 2 in that denominator. And as n approaches infinity, the whole absolute value approaches 0. So the interval of convergence is negative infinity to infinity. And the radius is infinite. You will be working with more and more of the sine and cosine and arc tangent, which are the important ones that we work on with Taylor polynomials. This concludes our lesson on intervals and radius of convergence.