We're going to try to determine whether this alternating series converges or diverges using the alternating series test. The first thing we have to figure out is actually a formula for b-n. If you recall, b-n is basically the absolute value of each of the terms. And if we're looking at this, we just ignore the part that causes the sign to alternate, Hopefully you would agree that b-n is just 1 over n. The first term is 1, the second term we're subtracting is one half [1/2], and then one third [1/3], then we subtract one fourth [1/4], and we add one fifth [1/5]. That would be our expression for b-n. There are two conditions that need to be met for this test. So the first condition is that the n plus first term is supposed to be less than or equal to the nth term for all values of n beyond a certain point. And to test that, we just have to figure out what the n plus first term would be; of course, that would be 1 over n plus 1. If we compare that to the nth term, which is 1 over n, clearly, 1 over n plus 1 is less than 1 over n, so that satisfies the first condition. The second condition is that the limit as n goes to infinity for b-n needs to equal 0, so that's the next thing to test. And in some examples, we'll actually do this first because if this is not true, then the whole test is going to fail. But in this case, if we look at the limit as n goes to infinity for 1 over n, hopefully everybody would agree that that definitely is 0. Since this is an alternating series and these two conditions have been met, that implies that this series right here, just like we drew out the diagram of in the first video, converges. This series, n goes from 1 to infinity, negative 1 to the n minus 1 divided by n converges, and it converges by the alternating series test. We've got another alternating series here. This one starts with a negative term, but the formula that we have is a little bit different. You can see I've listed out the first few terms. I've chosen not to reduce all the fractions just so that we can see the pattern that we've got going on here, and we're going to use the alternating series test to try to determine whether this series converges or not. To begin with, let's figure out what b-n would be. That's the absolute value of each term. Basically, the only thing that affects the sign here is this part. That means the b-n would just be 3n divided by 4n plus 1. Now, it's not immediately obvious if these terms are actually shrinking as n goes to infinity, we could look at the first few and try to figure out whether those fractions are getting smaller or not, but I would actually suggest ignoring step number 1 for the time being (just because that's a tougher question to answer), and let's look at step 2. Let's try to figure out if the limit as n goes to infinity for b-n is equal to 0. So if we actually write in the formula for b-n, we're going to wind up with 3n, divided by 4n plus 1. And to do this limit, we can just divide everything by the highest power of n, which is actually just n to the 1st. Now, of course, the n’s are going to cancel in the first two terms, but this last term is going to wind up approaching 0 as n goes to infinity, so we're going to be left with 3 over 4 plus 0, or three fourths [3/4], and that is clearly not equal to 0. Since we failed this second condition, that actually means that the alternating series test doesn't apply, so we may as well not even try to figure out whether that first condition is met or not. But how do we determine whether the series converges or not? Well, fortunately, back in Section 11.2, we found out about something called the test for divergence, and what that says is, if these terms right here of the original series do not approach 0, then that means that the series would be divergent. If we were to look at the limit as n goes to infinity of the original terms, (negative 1 to the n, times 3n, divided by 4 plus 1), what we would wind up finding out is that the absolute value of the terms approach 3/4. But because of this alternating portion here, that means that for large values of n, we're going to be approaching numbers that are close to positive 3/4 and then negative 3/4, and then positive 3/4, and then negative 3/4. And since that means the terms are not actually coalescing around a single value, what does that tell us about this limit? Well, what that tells us is that this limit does not exist. And if we look back at the test for divergence, it says that if the terms approach any limit other than 0 or if the limit of the terms does not exist, that means that the series is going to be divergent. Therefore, by the test for divergence, it's actually not the alternating series test that tells us this result; it's actually the test for divergence. Because of that, we can say that this series has to diverge because if the individual terms don't approach 0, then the series automatically diverges.