 In this video, we're going to talk about what's known as the test for divergence. In previous videos, we talked about how and test whether geometric series is convergent or divergent, and it all comes down to the constant ratio. If the constant ratio is small, the geometric series is convergent. If the constant ratio is big, then the geometric series is divergent. For other sequences, their series won't be as easy to determine, but it turns out there's going to be one very nice test that one could use here. It turns out that if a series is convergent, so you take the sum where n equals 1 to infinity of the sequence a sub n, if the series is convergent, then it means that the sequence converges to zero. Now, one common mistake that students make here is that we have sequences in play and we have series, right? The series is the infinite sum of a sequence. The sequence could be convergent or divergent. The series could be convergent or divergent. The two are related, but they're not the same thing. What this theorem is telling us is that if the series converges, what it converges to, we don't know. But if the series converges, the infinite sum converges, then we know that the sequence must have converged and it must converge necessarily to zero. So let's take a look at why that is. So just an argument here. The proof is not that complicated, it turns out. If we take the sequence of partial sums, s sub n, so this is going to be the sum where k ranges from 1 to n and we add together the sequence a sub k. So you take the sequence of partial sums here, then notice that if we look at s n minus s n minus 1, this is just going to look like a 1 plus a 2 up to a n. And then if you subtract from that s n minus 1, which is going to look like a, sorry, this is going to look like a 1 plus a 2. And this will go all the way up until you get to a n minus 1. When you subtract these things, the a 1s will cancel, the a 2s will cancel, all the way up to the a ns will cancel. And the only one who doesn't cancel will be a n. So s n minus s n minus 1, this is just equal to the term a n. That's an important observation to notice here. And so now by assumption, we're assuming that the series is convergent. And since it's convergent, we get that the limit as n approaches infinity of s n. This equals something we're going to call an s as well. So this limit exists, so the limit exists, that's the important part and it's equal to s right here. Well in that case, then consider the limit as n goes to infinity of a sub n. Well like we observed earlier, a sub n is just equal to s n minus s n minus 1 as n goes to infinity right here. And if we break these things up because s n is a convergent sequence, we're going to get the limit of s n minus the limit of s n minus 1 as n goes to infinity. Now as the sequence s n is convergent, it converges to the number s that we saw before. But this sequence right here, this is the sequence that's just one step behind this one. As n goes to infinity, that difference of one step won't make any bit of a difference as well. If s n is going to go to s, then s n minus 1 will likewise go to s. And so you end up with s minus s which is equal to 0. And this then verifies the statement we were looking for here that if the series is convergent, then the sequence must also be convergent and it must converge to 0, the sequence. If we take the logical contrapositive of that statement, we get the very following useful observation. If a limit does not converge to 0, then the associated series must be divergent. Now we're not saying that the sequence is divergent, we're just saying that if the sequence doesn't converge to 0, then the previous theorem guarantees that the series would be divergent. Because if the series was convergent, then the limit would have to go to 0 and we would get a contradiction in that situation. This is commonly referred to as the test for divergence. We can tell whether a series is divergent if the sequence doesn't go to 0. And so let's illustrate this with an example. Let's show that the series n equals 1 to infinity of n squared over 5 n squared plus 4. Let's show that this thing diverges. So consider the sequence for the moment, not the sum, the sequence, the things that we're adding together right here. Take the sequence of n squared over 5 n squared plus 4. Now if you take the limit as n goes to infinity, since this is a balanced rational function you see n squared on top and bottom, you're going to end up with one-fifth as the limit here. But one-fifth is not 0, right? This sequence does not converge to 0 and so what this tells us is that the series, so notice the series n equals 1 to infinity of n squared over 5 n squared plus 4, this series is divergent by our so-called divergence test, which we talked about on the previous slide. So one thing you're going to see here is we're showing that series are convergent or divergent. It's not just good enough to say whether the series is convergent or divergent. We have to also describe why. There's going to be these convergence tests that tell us whether it converges or not. And so by the divergence test we see that this series diverges because the limit of the sequence doesn't go to 0. And sort of the idea of this is the following. If we have the x and y axes here, so the x-axis, the y-axis, what we're seeing is that we take the line, the line of one-fifth, y equals one-fifth. Let's say it's right here. I'm going to make it a little bit bigger so we can see this. So this is the line y equals one-fifth, one-fifth. So based upon our sequence we saw before that as n gets bigger and bigger and bigger, our sequence starts off when n equals one. You're going to get one over nine, which is smaller than one-fifth. So you get a point right here. Then the next one you're going to get when n equals two, you're going to get four over. Well, let's figure out this one here. You're going to get four times five, which is 20 plus four. So you get four over 24, which equals to one-sixth. That's still less than one-fifth, but you're getting a little bit closer. And this is what you're going to see closer and closer, that as n gets bigger, bigger, bigger, this number is going to get closer and closer to one-fifth. And so after a while, you won't be able to tell the difference between the sequence a sub n is this equal to one-fifth? It'll be so hard to tell because it'll be so close to each other. But think about the infinite sum here. What we're basically getting is we're going to be getting an infinite sum that looks like something plus something plus something that's approximately one-fifth plus something that's approximately one-fifth, plus something that's approximately one-fifth, et cetera, right? You're going to be getting things that look like an infinite sum of one-fifths. And so that's going to converge towards infinity. And that's the basic idea behind the divergence test right here. If the sequence is not getting smaller, right? If you're not getting closer and closer, closer to zero, that means the terms and the sequence are not shrinking. And so the infinite sum is going to explode. It cannot converge. The series cannot converge unless the terms and the sequence gets smaller and smaller and smaller. So the sequence needs to get smaller, it needs to converge towards zero. But be warned, the converse of the divergence test is in fact false. That is, there do exist sequences. There do exist sequences which converge to zero, but the associated series diverges. So as an example of this, consider the so-called harmonic series. The harmonic series. This is the series where you take the sum, where n equals one to infinity. You take one over n. This harmonic series diverges towards infinity. This infinite sum turns out to be infinity. But if you look at the sequence one over n, it actually converges towards zero. So if the sequence converges to zero, that does not mean that the series is convergent. But if the sequence does, if it doesn't converge to zero, we can infer from that the series will be divergent there. So one thing to be aware of with the divergence test is that the divergence test can only show divergence. The divergence test can never be used to show that a series is convergent. That's a very important distinction. As we learn about more convergence tests in the future, we might need them to show convergence. But the divergence test is a very nice quick test to show when a series is divergent. Now if you're wondering how do we know that the harmonic series is divergent, that's an argument we could make right now. But in lieu of the integral test, which we'll talk about in the not too distant future, we can actually very quickly show that the harmonic series is divergent there. And so we will actually postpone that conversation until we talk about the integral test in a future lecture.