Welcome back to our lecture series, Math 1220, Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. This video is the first video of Lecture 39. We're getting to the tail end of our lecture series for Calculus II. In the previous lecture, we had introduced the notion of the integral test and used that to show that series were convergent or divergent or such. In this lecture 39, we're going to introduce two more convergence tests. The first one, which we'll talk about in this video, is known as the comparison test. Now, if you've been part of this series for a while, then the word, the name of the comparison test might actually resonate with you. We did something strikingly similar with integrals, with improper integrals, in which, given the discussion we had with the integral test last time, we see that series and improper integrals are actually deeply connected with each other. The same reasoning behind the comparison tests for integrals actually gives us the comparison tests for series. We have two sequences in question here. We have a sub n and b sub n. Now, what we know about these sequences is that we're going to assume that a is bigger than b. They both need to be positive. They're both above the x-axis. The sequence is always greater than or equal to zero. We have something like the following if we have our x-axis, like so. If you take the first sequence, it's above. In order for this thing to converge, it's going to have to go towards zero. Then we have a second one, which maybe in each instance is smaller than the above one. In this instance, the first yellow is bigger than blue. In the second instance, yellow is bigger than blue. In the third instance, yellow is bigger than blue. We have this idea going on right here. What we see is the following. If the larger sequence, if the corresponding series is convergent, that is, if we add up together all of the terms in this sequence, that's going to be a bigger number than this one. Let's take one step back. These statements right here tell us the very following thing. We see that the series involving n equals 1 to infinity of a sub n, the sum of all the terms of the sequence, this has to be greater than or equal to the sum where n equals 1 to infinity of the b ns. That is, the sum of the larger sequence must be larger than the sum of the smaller sequence. Both of these series have to be greater or equal to zero. With that in mind here, this inequality comes immediately from the assumptions we have right here. If we assume that the bigger series is convergent, well, since it's positive, the only way that this series can be convergent is if it's strictly less than infinity. It's going to be some positive number. Well, if the bigger series is less than infinity, then the smaller series must also be less than infinity. That's going to imply convergence of this second series here because it can't go off towards negative infinity or something like that because it's positive. Therefore, the larger series being convergent being less than infinity implies convergence of the smaller one. We saw this exact same thing with the integral comparison test. The larger series being convergent implies the smaller one is convergent. But what if we go the other way around? What if we take the smaller series to be divergent? Well, the only way that a positive series can be divergent here is going to basically because it actually goes off towards infinity. It adds up to be an infinite value like so because since the sequence is positive, the sum, each partial sum, is going to be increasing, increasing, increasing. That's the thing we're using here. Since the sequence is positive, the sequence of partial sums must be an increasing sequence. We're basically using the monotone convergence theorem in this regard. They're bounded below by zero. In the convergent case, they were bounded above by some finite number. In this case now, the smaller one's divergent, which would imply it has to go off towards infinity, that will force the bigger one to be greater than or equal to infinity. Well, that only leaves one option. It's infinite as well. Thus, it's a divergent series. Remember, if a series equals infinity, that makes it divergent. So when the smaller series diverges, the bigger series must diverge. And when the larger series is convergent, the smaller series must be convergent as well, identical to what we saw with integrals previously. And you have to make sure that you go in the right direction. If the smaller series was convergent, that says nothing about the bigger series. And if the bigger series was convergent, sorry, if the bigger series was divergent, that says nothing for the smaller series. And also I want to mention that the numbers n equals one, n equals one has really no significance on this situation. If these inequalities are eventually true, if it's eventually true. So after 5,280, if a sub n is greater than equal to b sub n, which is greater than or equal to zero, then we can make these statements. So don't fix it on the starting value. We just need that this thing will be eventually true in order to apply it. So let's look at some examples of series in the situation. Let's determine the convergence of the following series. Take the sum where n equals one to infinity of one over two to the n plus one. Now, one thing you want to remember when you're working with these comparisons here is that if you have a fraction, right, so you have some fraction a plus b, and you make the denominator, sorry, if you let's start with this easier one, if you make the numerator bigger, like you added something to it a plus c over b, this is going to make the whole fraction get bigger. Again, this is assuming that c is some positive number. If you make the numerator get bigger, the fraction gets bigger. But on the flip side, if you take a over b, and you add something to the denominator, if the denominator gets bigger, that actually makes the fraction get smaller. So that's the important thing here to remember is that smaller denominators, this actually means bigger fraction. And the other direction is true as well. A bigger denominator actually means a smaller fraction. So that often comes into play when you do these type of comparisons. Because when you look at this sequence one over two to the n plus one, notice that if I were to get rid of the plus one right here, this would look like one over two to the n. That's a geometric sequence. I could determine the convergence of that pretty quickly. And so I kind of want to get rid of this plus one. Now, if I went about doing that, we take the series where n equals one to infinity, like so, if I got rid of the one, one over to the n, notice I'm actually making the denominator get smaller because I'm removing a positive value. And a smaller denominator makes a bigger fraction just like we talked about. So this second series where we take the plus one out of the denominator is a bigger series. And now we want to mention that this thing is geometric. It's geometric with a constant ratio of one half. We get one half is equal to one. And so that's, that's, that actually implies that this series right here is convergent. The second series is convergent. And this comes from the geometric series test we've learned about before. Remember, in that situation, geometric series will be convergent if and only if its ratio is small, that is, its absolute value is strictly less than one. So this geometric series right here is convergent. And since it's a bigger convergent series, this plot implies convergence of the smaller series. And this follows from the comparison test. And you should make mention of this. When you write these things up, you should be writing up that, oh, the comparison test says that the original series is convergent by comparing it to a bigger convergent series. If we look at the next one here, take the series n equals one to infinity of five over two n squared plus four n plus three. So what we can first do is kind of like we did a moment ago, we can get rid of that plus three, right? That makes the denominator smaller, which makes the fraction bigger. So we get n equals one to infinity, we would get five over two n squared plus four n. So that made it get smaller. Well, that's not, so the smaller denominator makes the fraction get bigger, makes the series get bigger. That's not quite good enough, right? Because it's still the, the convergence of this series is a little bit difficult to describe. What if we could get rid of the plus three and the four n? Well, we have to be a little bit careful, right? Getting rid of the plus three, that's definitely going to shrink it. But if we get rid of the plus four n, does that make the denominator get smaller or bigger? Well, it depends on the number n, right? If n was a positive number, then removing it from the denominator would make the denominator get smaller, which makes the fraction get bigger. Great. But if n was a negative number, removing it actually makes the denominator get bigger, which makes the fraction get smaller. So we have to be careful about the direction of our inequalities. Now, the good news is, we're going from n equals one to infinity. So in that range, notice we're saying n is positive here. And so in particular, that also means that four n is positive. If we remove it from the denominator, like from here, that makes the denominator get smaller because we removed a positive quantity and then now makes the whole fraction get bigger. Therefore, the series will get bigger by removing the four n. So we do this one more time. You're going to get somewhere where n equals one to infinity of five over two n squared. If we factor out the five over two, we see that this series n equals one to infinity, this will actually look like a p-series one over n squared, where in this situation, we have a p-series. We have a p-series where our p-value equals two. Remember that a p-series is convergent. It's convergent if the p-value is greater than one, which it is in this situation. So that implies convergence of this series. Now, this series is greater than or larger than another series. So by the comparison test, we get that this series will be convergent as well. So the first series over here, we got that was convergent by the p-test. This smaller one, we're going to get that it's convergent by the comparison test. And since this one, this will likewise, since it's bigger than that one, that implies convergence, likewise by the comparison test. So the original series is likewise convergent, because it's smaller than it's smaller than convergence series. Now, frankly speaking, when I look at this one right here, I actually cut out the middleman entirely. We don't need the middle series. Honestly, if I was to approach this one, I would just start with this right here. Right? So I actually basically would compare, I would get rid of the four and plus three from the get go. And then you compare your series to a p-series. This series over here is convergent by the p-test. And then by the comparison test, this series is likewise convergent. What is interesting about working with series in chapter 11 of Stuart's textbook here is that the frame of mind is very different than what we did before. And when we talked about these continuous integrals, we focus so much on computation, compute the antiderivative, compute the antiderivative, compute the antiderivative. With series, we're not as interested in computations anymore. It's more of a conceptual understanding that we're seeking. The answer is always going to be convergent or divergent, convergent or divergent. And heck, if you were to guess, you're going to be right half the time on average. But the answer is not actually what we're seeking here. What I want is for students to gain this conceptual understanding. It's not on how did you get the answer. It's about why is that the correct answer? Why is it convergent? And can you provide an argument on why it's convergent? That is what we seek when we talk about these series and such. Let's look at one more example in this video. Let's take the series where n equals one, the sum n equals one to infinity of natural log of k over k. What are we going to do here? Which admittedly, in this series, one actually could use the integral test, right? You could try integrating from one to infinity of the natural log of x over x dx. You could do a u substitution and go from there. That's perfectly acceptable. But be aware that although we could use the integral test if we're only interested in finding the convergence of the series, if we're only interested in whether the series is convergent or divergent, we could find that answer using something maybe easier than integral test such as the comparison test. So one thing I want to mention is the following is note that if n is greater than or equal to three, then we see that the natural log of n are, in this case, k. I'll stick with that. Oh, I used n over here a moment ago. I guess, oh, there's some mismatch I see here. It should be natural log of n over the natural log of n. That's a typo in my part. Sorry, everyone. Note that as n is greater than equal to three, then the natural log of n will be greater than one. The issue is that the natural log of one is zero. The natural log of two is some decimal. It's less than one. But eventually, remember, eventually all that matters when it comes to the comparison test, is it eventually true? Eventually, the natural log of n will be greater than one. Therefore, eventually, this series, n equals one to infinity, will be greater than the series n equals one to infinity of one over n. And again, we're kind of fudging this right now. We really should be saying when n equals three. So we should be saying, okay, the series when n equals three to infinity of natural log of n over n will be greater than the series n equals three to infinity of one over n. And now the second series is actually divergent. This is the so-called harmonic series. This is a series we're going to see all the time. So we just kind of remember it. But we can also see this by the p-test because your p-value is equal to one. It's the harmonic series. So this one is divergent. And now notice we found a series which is larger than a divergent series. So the comparison test says that this original series is likewise divergent. Now, this shows that the series starting at n equals three is divergent. But like we said before, the initial value is not of much significance here. So if these qualities are eventually true, that's all that matters. So since it'll be diverging when you start at three, it'll be diverging when you start as one as well. And so we compare this to harmonic series. And with the harmonic series, which is divergent, and it's smaller than the series in question, we get that it's divergent as well. And so this gives you some examples of how one can use the comparison test to compute convergence or divergence of a series. Now, sort of the general rule of thumb is the comparison test is really good test to use if you can easily compare it to a p-series or a geometric series. Because those two series are often simplifications of the series we have in question. And they have very simple convergence test to use. If you compare it to a series which has a difficult convergence test to use, you might as well not have used comparison test and just try that difficult test on the function to begin with. So those are some tips I want to mention about the comparison test.


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