 In this video, we'll work through identifying the interval and radius of convergence of another power series. Let's first briefly recall the ratio test. We consider a series with only positive terms and define rho as the limit as k approaches infinity of the k plus first term divided by the k-th term in that series. If rho is less than 1, the series converges. If rho is 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. Consider the series, the sum from k equals 1 to infinity of x minus 5 to the k over k squared. What are the radius and interval of convergence of this power series? Now applying the ratio test for absolute convergence to this series, we obtain the following for rho. The limit as k approaches infinity of the absolute value of a sub k plus 1 over a sub k is equal to the limit as k approaches infinity of the absolute value of x minus 5 to the k plus 1 divided by k plus 1 squared all divided by x minus 5 to the k divided by k squared. Using algebra, we can rewrite this as limit as k approaches infinity of the absolute value of x minus 5 to the k plus 1 divided by k plus 1 squared times k squared divided by x minus 5 to the k. This is equal to the limit as k approaches infinity of k squared divided by k plus 1 squared times the absolute value of x minus 5 because we have a factor of x minus 5 left in the numerator after we perform this division. Now what we know about the limit as k approaches infinity of k squared over k plus 1 squared, that's actually equal to 1. We can verify this using multiple methods, one of which is L'Hopital's rule. But ultimately we know that this limit is equal to the absolute value of x minus 5. Now the ratio test tells us that the series converges if rho is less than 1. So the series converges, the ratio test tells us that the series converges if rho is less than 1. So the series converges for absolute value of x minus 5 being less than 1. Another way to state this is that x minus 5 is between negative 1 and 1. Even further, this means that x is between 4 and 6. Now unfortunately again we find that rho is equal to 1 when x is 4 and 6. So I need to use another test to check about the convergence at those endpoints. When rho equals 1, that means x equals 4 is 1 value for x. Our series becomes the sum from 1 to infinity of 4 minus 5 to the k divided by k squared, which is equal to the sum from 1 to infinity of negative 1 to the k over k squared. And we recognize this as an alternating series. So applying those two conditions to see if they're satisfied, we find that 1 over k squared is greater than 1 over k plus 1 squared because k plus 1 squared is greater than k squared and the limit as k approaches infinity of 1 over k squared is equal to 0. So since both conditions are satisfied in the alternating series test, we know that the series, the power series converges when x equals 4. So we want to include this in our interval of convergence. Now also, rho is 1 when x is 6. We recognize that our series is the sum from 1 to infinity of 1 over k squared and this is a convergent p-series with p being 2, which is greater than 1. So our series converges here as well. So in summary, our radius of convergence was 1 because the absolute value of x minus 5 being less than 1, we're looking around the value of 5. Our interval of convergence is 4 to 6 inclusive.