 In this video, we'll work through a couple of problems to demonstrate the application of the alternating series test. Let's recall the test, which states that if we consider the alternating series k equals 1 to infinity of negative 1 to the k, a sub k, or k equals 1 to infinity of negative 1 to the k plus 1, a sub k, such series converge if both of the following conditions are satisfied. Every successive term is less than the previous one, less than or equal to the previous one, or the absolute value of the terms are eventually decreasing, and that the limit of the general term, a sub k, as k approaches infinity, equals zero. So consider the series k equals 1 to infinity of negative 1 to the k plus 1, 1 over k. This series is known as the alternating harmonic series, and recall that the harmonic series diverges, which we proved using the integral test. Now, a sub k is equal to 1 over k. Now, 1 over k is greater than 1 over k plus 1, which tells us that a sub k is greater than a sub k plus 1 for any k. So the terms are decreasing, which satisfies the first condition of the test. We also know that the limit as k approaches infinity of 1 over k is equal to zero, which satisfies the second condition of the test. So we found that the alternating harmonic series converges by the alternating series test. Now consider the series from k equals 1 to infinity of negative 1 to the k plus 1 of k plus 3, divided by k times k plus 1. In order to test whether the first condition is satisfied, we're going to take a little different approach than we did in the previous example. If we can show that the ratio of terms a sub k plus 1 over a sub k is less than 1, this tells us that a sub k plus 1 is less than a sub k. So using this approach, we know that a sub k plus 1, divided by a sub k is equal to k plus 4 over k plus 1 times k plus 2, divided by k plus 3 over k times k plus 1. This equals k plus 4 divided by k plus 1 times k plus 2. Be careful to watch your algebra. Times k times k plus 1 divided by k plus 3. We notice that there's a factor of k plus 1 common to both the numerator and denominator. So we find that this is equal to k squared plus 4k. And expanding the denominator, we get k squared plus 5k plus 6. I'm going to rewrite that denominator in a way that allows us to see its relationship to the numerator. So what I've done is I've split this 5k into 4k and k so that we can see the relationship between k squared plus 4k, both in the numerator and the denominator. And we notice then that the denominator is greater than the numerator, which means that the ratio of a sub k plus 1 and a sub k is less than 1. Therefore, a sub k plus 1 is less than a sub k. So our first condition of the test is satisfied. For the second condition, limit is k approaches infinity of the general term a sub k. We find that that's equal to limit is k approaches infinity of k plus 3 divided by k squared plus k. We have a few approaches we could take with this. We could use L'Hopital's rule because the limit of the numerator and the denominator each approach infinity. And we know that this limit is zero. So our second condition of the test is satisfied. So the series we're investigating converges by the alternating series test since both of the conditions are satisfied.