 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 Misseldine. In the last several lectures in this series, we've talked about various convergence test. We started off with very simple ones like geometric series and telescoping series. We moved on to the integral test with the p-test as a special case of that, very important special case. In lecture 39, we introduced the comparison test and the limit comparison test. In lecture 40, we're going to introduce one more convergence test for series. This is known as the alternating series test. Some of our examples here will have been derived from examples found from section 11.5 in James Stewart's Calculus textbook. So what do we mean by an alternating series? So an alternating series is a series whose terms alternate between positive or negative values. So what I mean is you're looking for a series that looks something like the following, n equals one to infinity. You can have this negative one to the n times some other sequence. Let's take one over n, for example. What you see right here is that these alternating series is going to have this power of negative one that sits in front of it. Because as you take different powers of negative one, you can alternate between plus, minus, plus, minus, plus, minus. You might see negative one to the n. Probably more likely you're going to see negative one to the n plus one. Or sometimes you might see negative one to the n minus one. The distinction here is do we want to start with a positive or a negative? If n starts at one and we take the nth power, this will start off as a negative. And so if you take a plus one, then one plus one is two. You start with an even power of negative one. That's going to give you a positive one. So if you take this series, for example, the sum equals n equals one to infinity of negative one to the n plus one times one over n. If you take away the negative sign, this is just the harmonic series. But now it has this alternating factor, plus, minus, plus, minus, plus, minus. And so this right here turns out to be the alternating harmonic series. And so while we've seen the harmonic series before, it's a p-series that is divergent. It turns out that the alternating harmonic series is actually a convergent series. And because you have some plus, minus, plus, minus, plus, minus, there's some cancellation. Some of the negative gobbles up the plus. And it does it at such a rate that actually the series will be convergent. And we're going to talk about that in a general idea. And this is the idea behind the alternating series test. So suppose we have an alternating series where n equals one to infinity the sum of the sequence a sub n. And we want the following conditions here. If we look at the absolute sequence, so this is a little bit of vocabulary I want to point out here. So if you have a sequence, a sequence a sub n, then its absolute sequence, the absolute sequence is when we take the absolute value of said sequence. And so what this does is the absolute sequence will guarantee there are no negative numbers in this sequence. Everything we pause for negative. And why that's sort of important is that when we do things like the comparison test or the integral test that we talked about before, those required as part of the assumptions that all the terms in the sequence have to be non-negative. It has to be an absolute sequence. And so sometimes we want things to be absolute here so that we don't have to worry about some issues that come up with negatives. And so that's also sort of something that comes up in this discussion with the alternating series test. So let's consider the absolute series the absolute sequence involved here. So we want that the absolute value of a n plus one will be less than or equal to the absolute value of a sub n. So we're requiring that the absolute sequence be decreasing. It needs to be decreasing. And we also want that the limit as in approaches infinity of the absolute sequence a sub n is equal to zero. So we want that the absolute sequence converges it converges to zero. Now, if these two conditions are satisfied, if the absolute sequence that is if you ignore this alternating factor of negative one, in that situation, the absolute sequence would be this guy right here would be the harmonic sequence. In that situation, ignoring the negative signs does that sequence decrease towards zero? And if so, that tells us that the alternating sequence will be convergent. And so those are some very nice conditions to look out for. And so let's see some arguments. Let's see some evidence. Why does this alternate series work? Why does the alternate series test work? So just for the sake of simplicity, let's start off with the assumption that a one is positive. We're going to assume that the first term of the sequence is positive. Now, if that's the case, if the first term is positive, that means our sequence has to look something like this. That is a sub n has to equal negative one to the n plus one times bn, where here bn is the absolute sequence, right? Now, because it's all channels switch negative or plus negative, plus negative, plus negative, the first term is positive and needs to start off positive. So we have an n plus one right here. This would happen if a one is positive. Now, the other option is maybe a n actually starts off as negative. We allow that as a possibility. A one could be a negative number. Now, if it is, that would actually mean that a n is going to look like negative one to the n times bn, where again, bn is just the absolute sequence right there. So the only difference between the starting term and being positive or negative really comes down to, should we have an n plus one or an n as our power of the negative one there? Now, the good news is, clean this up a little bit, the good news is that these two options, right? If you have negative one to the n plus one bn versus negative one to the n times bn, these only differ by a factor of negative one. They only differ by a factor of negative one, right? And as such, when it comes to the series in play, you have the series a sub n versus the series negative a sub n. And times in a series by negative one doesn't change the convergence or divergence of that series. So if one converges like this one, then so does that one. And if this one converges, then so does that one. So we are actually safe to assume that the first term is positive. And therefore, our sequence has the form a sub n equals negative one to the n plus one times bn, where b is this positive sequence. So we're gonna use that assumption for the rest of this proof right here. So the series, if the series is convergent, it's the limit of the partial sums. So the partial sums are gonna come into consideration here. Let s sub n equal the sum, the partial sum k equals one to n of the sequence a sub, that should be a k right there, typo, sorry about that. So let's consider this partial sum. And let's actually look at the first couple terms here. You're gonna like this. So looking at the first even partial sum, we're gonna take s sub two. This is gonna look like b one minus b two. Now, because of the original assumption that absolute sequence is decreasing, if this sequence is decreasing, that means b one is bigger than b two. And if b one is bigger than b two, that means their difference is gonna be positive. So this right here is a positive term. Well, I guess it could be zero because b one could equal b two, that's perfectly fine. So s two is gonna be positive. Now let's look at s four. S four is gonna involve s two, which is positive. Plus it's gonna take b three minus b four. But again, the sequence of b's is decreasing. So b three is bigger than b four, which means that the difference of these b three minus b four will likewise be positive. So you have a non-negative number, s two plus a non-negative number, b three minus b four. This tells you that collectively, s four is gonna be a non-negative number as well. And this keeps on happening. If we do s six, you're gonna get s four, which is positive, plus b five minus b six. Now, since b five is bigger than b six, their difference is positive. And a positive plus a positive gives you a positive, or I should say non-negative be more precise. And this is gonna happen over and over and over again. If we apply induction here, we could eventually get up to the term s two sub n. This will be s two times n minus one plus, we're gonna get b two n minus one minus b two n. So why do I keep on getting plus minus, plus minus, plus minus? That's because of the alternating factors, right? So this first term has a positive sign in front, then a negative sign, right? And then we're gonna have a positive sign, then a negative sign. A positive sign and negative sign, a positive sign and negative sign. That's the alternating bit right there. We're always getting plus minus, plus minus, plus minus. And so b two n minus one is bigger than b two n. So their difference will be a positive number by our inductive hypothesis. We know that s two n minus one is positive. So this thing will add up to be something non-negative. You get right there. And so what we see here is that s two n by induction will always be positive. It's always gonna be positive. But we actually say something a little bit better than that, right? We actually see that the sequence is in fact increasing, right? Because notice that s four is s two plus something. So it's bigger than s two. And s six is s four plus something positive. So it's bigger than s four. And s two n is bigger than s two n minus one because it's that partial sum plus a little extra. I mean, you might be adding zeros that might have not got bigger at all, but that's okay. Our inequalities have less than or equal to. So this tells us that our sequence is increasing. We have an increasing sequence. I mean, it's a monotonic sequence. It's an increasing sequence, but it's also bounded, right? Because it's increasing, it's gonna be bounded below by wherever it started from. All of these things are positive. So we're gonna be greater than zero. Who's an upper bound? Who's an upper bound for this thing? It turns out with a slightly change of perspective, if you take any number in this sequence, we're gonna start off with s two n again. Well, actually you can see it right here. If you take s two n, if you list all the numbers b one, b two, b three, b four, b five, all the way up to b two n. In that case, you're gonna get plus minus plus, sorry, plus minus plus minus plus all the way up, right? Now, if you factor away the negative sign right here, you're gonna see that you have b one, and then you're gonna get minus b two minus b three. Notice if you distribute this negative sign through, you're gonna end up with negative b two and plus b three. The next negative sign, if you distribute it onto b four, that'll be negative. If you distribute it onto b five, that'll be positive. So this does have the alternating plus minus plus minus plus pattern that an all-terrain series is supposed to have. Now, what do you see here? You have b one minus something. Now, b two is bigger than b three, so b two minus b three is non-negative. So you have a positive number minus something. Again, b four is bigger than b five, so their difference is positive. You're gonna take b one minus something. Then you're gonna have b one minus something, b one minus something. Those things are always positive numbers. So you have b one minus something, minus something, minus something, minus something. These things are getting smaller, smaller, smaller, smaller. And so what we see here is in fact, that S2n is bounded above by B1. And so we see that all terms here are bounded below by zero and bounded above by B1. So in summary here, S2n, it's bounded above by B1, it's bounded below by zero. And so this tells us that our sequence is bounded. We have a bounded sequence and we also have an increasing sequence. So the monotone convergence theorems applies and tells us that the sequence of even partial sums converges and thus it has to converge to some value which we will call S. That gets us halfway there, right? We know that the sum of the even partial sums will converge towards S. But what about all of the partial sums? What about the odd ones? Now, if you look at the odd ones, where I take the limit as n approaches infinity of two Sn plus one. Well, if you have a partial sum of S2n plus one, well, that can be broken up as you take the partial sum of Sn, that should be an S2n. Sorry, there's a couple of typos in this lecture here today. So if you break up S2n plus one, this breaks up as S2n plus the next term in the sequence which will be B2n plus one, which you can see right here, right? Now, separate these things. This then becomes a sequence as in approaches, the sequence of evens, S2n as in approaches infinity right here, that sequence will go off towards S as we determined over here with the monotone convergence theorem. And then the other sequence is the limit as n goes to infinity of B sub n. By assumption, since B sub n is the absolute sequence, this will actually converge towards zero. So we end up with S plus zero, which is equal to S right here. So this tells us that the odd partial sums, they converge towards S. We've already determined that the even partial sums converge towards S. So when we merge these two odd and even partial sums together, we see that the entire sequence of partial sums converges towards S. And this, and as the series is the limit of the partial sum sequence, we see that our series is convergent. And so this gives us, this shows us that the alternate series test does guarantee convergence here. And what were the assumptions that we needed? Remember, we needed that the absolute, the absolute sequence is decreasing. We needed those decreasing to make all of the inequality statements we use throughout the thing. We needed this to be decreasing to employ the monotone convergence theorem. But we also needed that this limit goes to zero so that we could merge together the odd sequence with the even sequence of partial sums and get that it's all convergent right there. It's a pretty nice argument. And it turns out that this proof doesn't just prove the validity of this test here. Also, when we talk about remainder estimates for alternating series, these inequalities we derived and the proof will be useful there as well. So put a pin in those ones. We'll come back to it in just a minute.