 One of the first important tests are the comparison and the limit comparison tests. We'll make a couple of simplifications. Since the convergence or divergence of an alternating series can be determined by the alternating series test, we'll focus on series whose terms are all positive or zero. Suppose that I have two series of positive terms with AN, eventually less than or equal to BN. Is the sum of the AN's diverges than the sum of the BN's also diverges since every term is greater? On the other hand, is the sum of the BN's converges than the sum of the AN's also converges since every term is less? The easy way to remember this is that when using the comparison test, you're bigger than a divergent series, you diverge. If you're smaller than a convergent series, you converge. Now there's a bonus item here. We don't need to know if our series are eventually decreasing. It's enough to know that we're eventually less than a convergent series or eventually more than a divergent series. And this is a wonderful test with one minor flaw. When using the comparison test, you need a series whose convergence or divergence is known to you. So in the assignments, you'll actually look at the convergence or divergence of three important types of series. But let's go ahead and summarize the results that you'll eventually prove for yourself. First, if we have a geometric series with common ratio R, then the series will converge if the absolute value of R is strictly less than one and the series will diverge otherwise. Next, the series whose terms are one over N, the harmonic series, diverges. And the divergence of the harmonic series is a special case of a more general type of series known as a p-series, and we have a theorem resulting those p-series of the form one over N to power p. The series converge if p is greater than one and diverge for p less than or equal to one. So let's take a look at an example. Let's see if we can determine the convergence of the series whose terms are one over N times two to power N. And here we see that our series looks like one over N. It also looks like a geometric series. So let's see if we can compare it to one of these two series. We might begin by noticing that our series terms are less than the terms of the harmonic series. Unfortunately, that's not helpful because when using the comparison test you have to be bigger than a divergent series. Or you could be smaller than a convergent series. So let's compare to the geometric series. Our series terms are less than one over two to the N. And since the sum one over two to power N is a geometric series with common ratio less than one, this series converges, then the series of smaller terms, one over N times two to power N, must also converge. The problem with the comparison test is finding a suitable series to compare to. And so a more powerful version of the comparison test is known as the limit comparison test. Suppose I have two series where the limit as N goes to infinity of the ratio of the terms of the series, N over BN, is equal to L, where L is not zero and not infinity or negative infinity. And those restrictions are very important. If that ratio is not zero or plus or minus infinity, then the two series have the same convergence property, which is to say they either both converge or they both diverge. On the other hand, if the limit is undefined, or the limit is zero or plus or minus infinity, then no conclusions may be reached regarding the convergence of the series. Again, there's a bonus item here. As with the limit test, we don't need to know if our series is eventually decreasing. It's enough to know its ratio to a series whose convergence or divergence has already been found. And again, to use this test, we need to have a series whose convergence or divergence is already known, and those would be the geometric series, the harmonic series, or the P series. A useful way to think about the comparison test is that the terms of our series A and N are eventually L times the terms of BN. So the sum of the series A and N will be about L times as much as the sum of the series B and N. If the sum of the series BN is finite, then L times it will also be finite, while if the sum of the series BN is infinite, then L times it will still be infinite. So let's take a look at the problem, given that we know that the series 1 of our N squared convergence, what can we say about the convergence of this series N squared over N squared minus 4N minus 7? And here's an important idea. Problems in the real world don't come with section numbers. You won't know what to apply, so here's a key idea. Always check the easy things first. And when you're dealing with series, the very first thing you want to check is to see what happens to the terms as N goes to infinity. In other words, we want to apply the Nth term test. So you want to find the limit as N goes to infinity of the terms of our series. And while this term does have the form infinity over infinity, and we could use L'Hôpital's rule on it, remember we should avoid any unnecessary chips to the L'Hôpital. And this particular problem, we can multiply numerator and denominator by the highest power of N in the denominator, do a little bit of algebra, and take the limit as N goes to infinity. And that limit is 1. And so by the Nth term test, since the terms of the series don't go to zero, we know that the series diverges, and we don't have to do anything else with it. So let's try to find the convergence or divergence of the series whose terms are N squared over N to the 4th minus 4 and cubed minus 7. So first of all, we might want to check and make sure that the terms of this series do actually go to zero, which means that our Nth term test is inconclusive. Since we know the series whose terms are 1 over N squared converges, we might try to compare the two series. Now, since this is a convergent series, in order to be useful, we'll need to be less than a convergent series. So we'll compare the terms of our series, N squared over N to the 4th minus 4 and cubed minus 7, to the terms of our known convergent series, 1 over N squared. And since we don't know whether these are greater than or less than, we'll leave the inequality to be determined and follow the algebra. We'll cross-multiply, do a little bit of algebraic cleanup, and we see that 4N cubed plus 7 must be greater than zero, and so this will cascade back and our series terms will be greater than 1 over N squared. And this is a problem because we need to be less than a convergent series. Here, we see that we're greater than a convergent series, and our comparison test will be inconclusive. Rather than trying to find a different infinite series, let's try out this limit comparison test. So we'll look at the limit as N goes to infinity of the ratio of the terms of the two series. 1 over N squared to N squared over N to the 4th minus 4 and cubed minus 7. So we'll do a little algebra. And now we need to find this limit as N goes to infinity. Well, our insurance covers L'Hopital visits, so we'll try it even though we can avoid it. Numerator and denominator both go to infinity, so I'll find derivative of numerator and denominator. We'll simplify a little bit. And they still both go to infinity, so I'll repeat. And we'll apply L'Hopital's rule again. And remember, L'Hopital's rule does not actually give a limit. We still have to find a limit as N goes to infinity, which will be 1. And since the limit is 1, it's not 0, it's not plus or minus infinity, it's not undefined. Since the limit is 1, both series do the same thing. And since we know 1 over N squared converges, we know that both series converge. Thank you very much.