 In this video, we'll discuss how and why we use the ratio test for absolute convergence when identifying the interval and radius of convergence of power series. Recall from our study of the ratio test for absolute convergence that for any series of the form, sum from k equals 1 to infinity of a sub k with only positive terms and rho is equal to the limit as k approaches infinity of the absolute value of a sub k plus 1 divided by a sub k. If rho is less than 1, the series converges. If it's greater than 1 or infinity, the series diverges. And if rho is equal to 1, the series may converge or diverge, but another test is needed. Well, since power series are of the form, the sum from k equals sum value to infinity of a sub k times x minus x naught to the kth power, the ratio test is a natural choice as a test of convergence since ratios of successive terms are usually quite reasonable to work with. Let's apply this test to an example. Consider the series shown here. What are the radius and interval of convergence of this power series? Let's take a look at rho, which is the limit as k approaches infinity of the absolute value a sub k plus 1 divided by a sub k. Specifically for this series, that's the limit as k approaches infinity of negative 1 to the k plus 1 times x to the k plus 1 divided by 3 to the k plus 1, all divided by negative 1 to the k times x to the k divided by 3 to the k. Now the nice thing about absolute convergence is we don't have to worry about our alternating parts. So I could leave those off, in fact. This would be the same as the limit as k approaches infinity of the absolute value of x to the k plus 1 over 3 to the k plus 1 times 3 to the k divided by x to the k. What I notice is that between x to the k plus 1 and x to the k, I differ by only a factor of x remaining in the numerator. Similarly, I have that happening here with 3, where I'm left with 3 in the denominator. So this limit is in fact absolute value of x divided by 3. Now the ratio test tells us that the series converges if rho is less than 1. So this series converges for the absolute value of x divided by 3 being less than 1. Now what's another way of writing that? That means that the absolute value of x is less than 3. This means that our radius of convergence is 3. Since we're looking at this series, which is generated about x equals 0, because absolute value of x is less than 3, that means that the radius around 0, at which I will converge, is 3. Now we know that unfortunately the ratio test does not give us a conclusion when rho is equal to 1. For our series, rho is equal to 1 when x is 3 and when x is negative 3. The ratio tells us that we'll likely need, we will need a different test for convergence. When x equals 3, our series becomes sum from k to infinity, is k equals 0 to infinity of negative 1 to the k, 3 to the k, divided by 3 to the k. This is equal to the sum from k equals 0 to infinity of negative 1 to the k. Now by the divergence test we know that since the limit as k approaches infinity of negative 1 to the k is not equal to 0, then this series diverges. When x equals negative 3, our series is the sum from k equals 0 to infinity of negative 1 to the k times negative 3 to the k divided by 3 to the k. I can rewrite this as the sum from k equals 0 to infinity of negative 1 to the k times negative 1 to the k times 3 to the k divided by 3 to the k. This gives me the sum from k equals 0 to infinity of negative 1 to the 2k times 3 to the k divided by 3 to the k, which then leaves me with the sum from k equals 0 to infinity of negative 1 to the 2k. Now since this is an even power all the time, this term is always equal to 1. And again, by the divergence test, since the limit of this general term is not 0, the series diverges. So in summary, our radius of convergence is 3 and our interval of convergence is negative 3 to 3, not including the end points.