 In this video, we'll work through a few problems to demonstrate the application of the ratio test. Let's first recall the ratio test. Consider the series, the infinite series, from k equals one to infinity of a sub k with only positive terms, and let rho equal the limit as k approaches infinity of a sub k plus one over a sub k. If rho is less than one, then the series converges. If rho is greater than one or rho is infinite, the series diverges, and if rho is equal to one, the series may converge or diverge, but another test is therefore needed. Each of the following series has only positive terms, so we can apply the ratio test on them to test for convergence. Let's apply the test to describe what it tells us. So consider this first example from k equals one to infinity of k over two to the k. Rho, as we know, is the limit as k approaches infinity of a sub k plus one over a sub k. Well, this here is a sub k, so I'm going to write as k approaches infinity of k plus one divided by two to the k plus one, all divided by k divided by two to the k. Rewriting this as a product limit as k approaches infinity of k plus one over two to the k plus one times two to the k divided by k. What we notice is that two to the k over two to the k plus one leaves me with a two in the denominator, and we know that the limit as k approaches infinity of k plus one over k approaches one. If we wanted to show this explicitly, we could use L'Hopital's rule, but we know that in the long term k plus one over k approaches one, so this limit is one half, which is less than one. So according to the ratio test, this series converges. Now consider the series k equals one to infinity of k to the k over k factorial. Now row equals limit as k approaches infinity of, as we said before, the quotient of the successive terms a sub k plus one over a sub k. In this case, that gives us limit as k approaches infinity of k plus one to the k plus one divided by k plus one factorial, all divided by k to the k over k factorial, rewriting this as a product of quotients. We get the limit as k approaches infinity of k plus one to the k plus one over k plus one factorial times k factorial over k to the k. We notice that between these factorials, k factorial divided by k plus one factorial were left with k plus one in the denominator. So we find that this is equal to the limit as k approaches infinity of k plus one to the k plus one divided by k plus one times one over k to the k. And this reduces to limit as k approaches infinity of k plus one to the k because this numerator and denominator shared a factor of k plus one divided by k to the k. I can rewrite this as limit as k approaches infinity of k plus one over k to the k. We're slowly recognizing what this looks like. Limit as k approaches infinity of one plus one over k to the k, which is actually equal to e. Now since this is greater than one, according to the ratio test, we know that the series diverges. For our last example, consider the infinite series from k equals one to infinity of one over two to two k minus one. Rho, which equals limit as k approaches infinity of a sub k plus one over a sub k, is equal to limit as k approaches infinity of one over two times k plus one minus one all over one over two k minus one. Rewriting this, we have one over two times k plus one minus one times two k minus one over one, which we find is equal to the limit as k approaches infinity of two k minus one over two k plus one. And we know using different methods, one could be L'Hopital's rule, another is just looking at long term behavior of two k minus one over two k plus one is actually equal to one. So with this series, we actually did not get a conclusion using the ratio test. We would have to use another test. And in this case, probably the integral test would be a good choice.