 In our previous video, we've learned about the comparison test and we saw how we could use the comparison test to determine the convergence or divergence of what we'll call a messy series, right? A series who has a sequence with a bunch of extra stuff floating around that we just kind of wanna ignore and simplify. And we were able to compare that to a much simpler sequence, a simpler series, maybe like a P series or geometric series for which convergence is much easier to ascertain. And then the comparison test allows us to infer the convergence or divergence of the messy series by comparing it to that simple series. But there are limitations to when we can apply the comparison test. Case in point, consider the series where we add together from n equals one to infinity, the sequence one over two to the n minus one. Now we saw an example very similar to this where we had two to the n plus one in the denominator. And what we did is like, hey, we're like, hey, if you get rid of the plus one, removing a positive quantity from the denominator actually makes the denominator get smaller. And as a consequence, that fraction is going to get bigger. And then the sum of all those fractions is in a bigger series. And so then we can compare it to a geometric series which would be convergent since it has a small ratio. Things are a little bit different though when we come to the question at hand. If we have a minus one there, if we remove a minus one as this is a negative quantity, removing a negative quantity from the denominator makes the denominator get bigger. And therefore the fraction becomes smaller and then the series becomes smaller. And so our geometric series we would like to compare it to is in fact going to be smaller than the one we started off with. And this is a problem. While the associated geometric series is gonna be convergent, it's smaller than the series we care about. And that tells us nothing about the series we have. If you have a smaller convergent series, the comparison test doesn't say anything about that because the larger series could be convergent or the larger series could be divergent. But it feels like we should be able to use the comparison test here because after all, if you look at this geometric sequence, one over two to the N, it's approximately the same thing as one over two to the N plus one. And it's approximately the same thing as one over two to the N minus one. This feels just like blatant discrimination going on here. Why is the positive considered so great and the negative is not? Why does the comparison test work in this direction but we can't work in the other direction? That's not gonna work there. It kind of feels somewhat arbitrary. Now it comes from inequalities because the comparison test requires limit inequalities and that's the limitation of the comparison test. But it turns out there's a good way to remove the limits of the comparison test and that's by taking a limit. I know it's a horrible pun, but the key to a limitless comparison test is to take limits. And so enter the so-called limit comparison test. The limit comparison test is able to work around the inequality limitation we saw before and we actually use a limit calculation to determine the asymptotic behavior of these sequences. And so then we can determine the convergence of the associated series and play. So we have two sequences like we did before, A and B. And the only requirement we have about A and B this time is that they're both positive sequences. They never obtain negative values. And so this is gonna be true for all N but honestly, this just has to be eventually true in order for this to work. We've seen this type of argument before. So suppose we have two non-negative sequences so that if we take the limit of their ratios, if we take the limit of the ratios this is equal to some number C. So this limit exists, this sequence is convergent. The ratio sequence is convergent. And in fact, the number C has two requirements. It must be a positive number and it must be finite. What does positive here mean? That means that C is strictly greater than zero. Zero is not allowed as an option. What does it mean to be finite? We're saying that C must be strictly less than infinity. So we're saying that C cannot equal zero, C cannot equal positive infinity. It has to be a finite positive number. Now if that's the case, if the limit of these ratios is some finite positive number, then it means that the convergence of the series will be the exact same. Either both series will be convergent or both series will be divergent. And so this is what we're doing right here. We're looking at the asymptotic behavior. How are these sequences growing? If they're growing at the same rate, at least up to a constant multiple, then this limit will go to that constant multiple C, right? If this limit is a positive finite number this tells us that the growth of sequence A and the growth of sequence B is proportional to each other. There's a proportional growth, which essentially for a calculus point of view means it's the same growth. Clearly the perfect choice would be to take C equals one, but honestly any positive finite number will be sufficient here because that just means up to some proportional constant. These two things are growing at the same rate. And if the sequences are growing at the same rate that means their series of the sequence partial sums will be also comparable. So let's actually see a quick argument on why this limit comparison test work. It's really kind of nice here. So take any what we say is epsilon greater than zero. How we want to interpret this is specify our margin of error. How much error will we tolerate for a problem? Do we need to be accurate to point one decimal places, point zero zero zero zero zero one decimal places, point zero zero zero zero zero zero zero zero zero zero zero zero, one decimal point, or something like that. I just realized what I said didn't make any sense. Like, you know, we want to be accurate to 0.00001. That is, we want to be accurate to 10 decimal places, 12 decimal places, a trillion decimal places. We specify our allowance there. So that's this epsilon, epsilon for error. And so then let's compute our margin of error, right? If you take the limit C, you subtract from it, epsilon, that gives you a little m. Take C plus epsilon, that gives you a capital M. And so because our sequence A over N, An over BN, is convergent to C and C is some positive number, we know that there's some number N, so that all later numbers, little n, that are bigger than big n, there's some sufficiently large n, it guarantees that this inequality will hold right here. And the basic idea is this. This just comes from the definition of the limit, that for all numbers larger than capital N, we have that the distance between An over BN minus the C will be less than epsilon. If you unravel this inequality of the absolute values, you get this one right here, this one's a little bit more practical to use. And then if you clear the denominators, you get the statement that we actually want, which is this one right here. M times BN will be less than An, which should be less than M times BN. That's a capital M and a lowercase M in presence there. And so now using this limit statement, A over B converges to C, that convergence statement of a limit translates into an inequality. And this inequality is where we can use the comparison test. Because consider the series of BNs, let's say that it's convergent. Well, if you times that series by a constant multiple, it'll still be convergent. So the sum of capital M times B is a convergent series. Well, because that's bigger than the smaller series, and this is a positive series, this also tells us by the comparison test that the series of the A's will be convergent well. Again, that's by the comparison test. We are able to use an inequality there. But let's switch the direction here. What if BN was divergent? Well, then any multiple of BN would also be divergent as a series. So the sum of little MBs will be divergent as well. And then look at the inequality here. If our series is bigger than a divergent series, then it likewise has to be divergent again using the comparison test. So this limit comparison test piggybacks off the comparison test by turning a limit statement, an asymptotic statement into an inequality for which the comparison test applies. So this tells us that when B is divergent, A is divergent. And when B is convergent, that A is convergent. So the convergence of A implies the convergence of B. But then who's on top and who's on bottom in this ratio is somewhat irrelevant. You could take the reciprocals because if C is a positive finite number, its reciprocal will likewise be positive finite. And that'll be the limit of B over A. That'll just be one over C. So the same above argument works, you switch the appropriate parts. And so then we get this limit comparison test. Let's see some examples of how one can use the limit comparison test. Well, let's take the series that started this whole conversation. Take the sum where N ranges from one to infinity of one over two to the N. And so let's look at two sequences. There's the sequence we started off with, one over two to the N minus one. And then there's the sequence we wanna compare it to, one over two to the N. We just kinda wanna ignore this negative one right here. If we take the limit as N goes to infinity of A N over B N, this looks like the limit as N goes to infinity. You're gonna get one over two to the N minus one over one over two to the N. You have a fraction divided by a fraction. So if we multiply by the reciprocal, we'll get the limit as N goes to infinity of two to the N over two to the N minus one. Now at this moment, you could notice, and at some point you should, it does have this indeterminate form, infinity over infinity. L'Hopital's rule does apply, but honestly in this situation, we have these, look at the dominant terms, two N is the only term on top, and on the bottom you have two N, that's the fastest growing term. This is a balanced rational function here, a ratio here, they're balanced. The top and bottom are growing at the same rate. And so we can see that this limit's gonna go towards one. If you want a little bit more justification than just sort of like this, I just eyeballed it type argument, you can times top and bottom by two to the negative N, two to the negative N, whoops, two to the negative N there. And that's because there'll be some cancellation of things. You'll be having this two to the N right there. And so now as things go towards infinity, you end up with one, and then on the bottom you have one minus two to the negative N, which will go to zero, and you get a one over one, which is equal to one. So there's a couple of ways one could do this. Frankly speaking, I feel quite comfortable just the argument like, okay, look at the dominant terms, those are balanced. So it's gonna go to their coefficients. This is gonna go to one over one, which is just a one. In particular, this is our limit C, and C is a positive number that is less than infinity. It's a finite number. So this tells me that the convergence of the series involving one over two to the N will be the same as that involving one over two to the N minus one. And so that asymptotically speaking, the series that we start off with is essentially the same series as N equals one to infinity of one over two to the N. Now, I'm not saying the series are equal, but I'm saying that as N gets sufficiently large, the two terms will be so close together that we really shouldn't distinguish between them and that the sums of these two series will be really close to each other. So asymptotically, these two series are the same. Now, the second series, one over two to the N, this is a geometric series. It's a geometric series whose ratio equals one half. And so since our ratio is small, now if the value is less than one, this tells us that the geometric series is convergent. It is a convergent geometric series. Those are usually fairly easy to see. And so the geometric series test implies that the geometric series is convergent. But what about the original series? The original series is gonna be convergent by using this limit comparison test. Now, we couldn't use the comparison test directly, but the limit comparison test allowed us to go around the obstacle, the so-called inequalities that the comparison test requires. So this series will be convergent by the limit comparison test. Let's look at the example B right here. Same basic idea. Let's look at the dominant terms. We have a two to the N squared on top. We have a square root of N to the fifth on the bottom right there. And so we wanna compare our sequence. So look at the sequence. We're gonna get two N squared plus three N over the square root of five plus N to the fifth. We wanna compare this thing to B, two to the N squared over N to the five halves, taking the square root of the same thing as the one half power. Now, if you simplify these things, cause after all two on top is same thing as four over two, this thing would simplify to be two over N to the one half power. And so that's the comparison we wanna see here. And so what's gonna happen again, if you take the limit as N approaches infinity of A N over B N, this limit is gonna equal one. And that's because of the balance we had here. Notice the leading terms. We have an N squared on top. We have a square root of N to the fifth on the bottom. We have an N squared on top, a square root of N to the fifth on the bottom, plus also there's a two, a constant than the multiple of two in the numerator. These things will asymptotically approach the exact same, they're going at the same rate. Therefore, we get that this limits can equal one. Kind of skirting around the details here, but we could go through the details. We could do a L'Hopital's argument or like in the previous example, we see that it'll be a balanced rational function. And so we end up seeing that this limit equals one. The limit comparison test applies. Take the limit comparison test. And so what is it that we're comparing to? Well, we're comparing to the series that involves this sequence B, right? Our series will have the same asymptotic behavior as the sum when N goes from one to infinity of two over N to the one half. Now this new series in question, it's a P series. It's a P series whose P value, which we can see right here is one half. For a P series to be convergent, P has to be greater than one as our P is less than or equal to one. This implies that the series is divergent. Now the series involving two over N to the one half power, that's divergent by the P test. And then the limit comparison test shows that the original series is likewise divergent. And that's all we have to do for this limit comparison test. We have to take, look at the simplified sequence by looking at the dominant terms of growth. And if we're at all uncertain about the dominance going on here, use L'Hopital's rule. But we look at the dominant terms and then look at that simplified one. And that oftentimes will be a geometric or P series is much easier to determine the convergence of.