 So in this last video for lecture 40 in our lecture series here, we're gonna do the same stuff that we've done in the last couple of lectures. Like we did with the integral test and the comparison test, it's not just good enough to know the convergence of the series. At some point we might wanna know how, what the series is actually equal to, what's the limit of that, of those partial sums. And so the convergence test allows us to know that the series converges, so the limit exists, but what is that limit? Now, so like we've done before, let's assume we have a series to take the sum where n equals one to infinity of a sub n. And let's say that we know that the series is convergent so that it adds up to be some value s. Well, since it converges, we can use the partial sum sequence to estimate that. So s sub n will be the partial sum where you take k equals one to n of a sub k. There's a typo there, sorry. And so we're gonna use the partial sums to estimate the series, but how good is the error, right? We use the remainder rn, which is the difference of sn minus one. And like we've seen before, if you take sn, which is the whole series, minus the first n terms, you'll get a sum starting at n plus one going off towards infinity. So the remainder term itself is a series. That's the framework we're having like before. But let's suppose that we know that the series is convergent using the alternating series test. So we have an alternating sequence. And so what we also know is that if you take the absolute series, a n plus one absolute value is less than the absolute value of a n. It's decreasing the absolute series and it decreases towards zero. So using the assumptions, the notation we have before b n will be our absolute series. So what we're saying here is that b n is decreasing. So b n plus one is less than b n and the b n's converge towards zero. So these are the assumptions we get from the alternating series test. So this is actually one of the reasons why we wanted to prove the alternating series test. It's not just to prove its validity, but it turns out that the things we learned while we were trying to prove the alternating series test actually help us out right here. So what we saw here is that if you look at the remainder term itself as a series, it'll be an alternating series. Now it's gonna go plus, minus, plus, minus, plus, minus or it could actually go minus, plus, minus, plus. We don't actually know. It depends on where we stop. Do we stop at an even number in or odd number? But with the proof, it didn't matter whether you start positive or negative. So as this puppy right here is an alternating series in its own right, we saw some important inequalities here that this first term is the largest term in the sequence since we're decreasing towards zero. And as we start adding more terms, we actually will shrink from where we are at a n plus one. And so in fact, the sequence, the remainder, rn, will be bounded above by b in plus one. So that is, you have a n right here, that could be positive or negative, so take the absolute value. The remainder, rn, will be bounded above by b in plus one. So as long as the next term in the sequence of an alternating sequence is sufficiently small, that'll guarantee that our partial sums here are sufficiently accurate. This is gonna be really nice. And let's look at a specific example. Let's find the sum of the series where we take n equals zero to infinity of negative one to the n over n factorial. Now, some things to note here is that this series starts at zero as opposed to one. That's not a huge consequence for us. We do have to remember that a zero factorial is equal to one by definition. That's not zero. So this thing is defined at n equals zero. But the first term doesn't really matter if it converges at zero or converges at one or converges at 17, whatever, right? So be aware that we're doing one here. So if we wanna find the sum of the series, we actually have to first figure out is this thing even convergent? Is this thing convergent at all? Now, if I was looking at this series, I actually would use something called the ratio test, which we'll talk about in lecture 41 here to determine the convergence there. But as we don't know the ratio test yet, what we actually can do is we can prove it's convergent using the all-training series test. So look at the absolute sequence, one over n factorial. It is bigger than zero, it's positive, and it's gonna be less than n, right? And that's because n factorial is gonna be bigger than n. So the reciprocals go the other way around, one over n factorial is less than one over n. And so notice in this situation here that our series, the one on the right here one over n is gonna go toward zero as n goes to infinity, right? And so by the squeeze theorem, this also shows us that one over n factorial would go off toward zero. So that's a way of showing, that's a way of showing that the sequence one over n factorial will go to zero, it goes to zero and it happens in this decreasing manner as well. And so we get by the all-training series test since one over n factorial will decrease toward zero, we get by the all-training series test that it's convergent. So the answer would be yes, yes by all-training series test. Like I said, we wouldn't have to do this sort of squeeze theorem argument right here when we have the ratio test in hand. Use a squeeze theorem to help us get that limit. But the all-training series test could be used here as well. So almost we can determine one over n factorial decreases toward zero and we do that by sandwiching it between two decreasing sequences toward zero. All right, so we got its convergent, great. So we can approximate the series using partial sums but how accurate is it gonna be? Well, the accuracy depends on our remainder term, right? How big is the remainder term? Well, since we want this to be accurate to three decimal places, our error needs to be no worse than 0.001 which is the same thing as one over 1,000 which is the same thing as one over 10 cubed. This last figure right here comes into play because we're trying to be accurate to three decimal places, right? But as this is an all-training series, we proved it was convergent by the all-training series test then the remainder will actually be bound above by this bn plus one. That is, for getting the negative signs, this needs to be bigger, our bn plus one term is one over n plus one factorial. So when is, those are the players in play here, one over n plus one factorial needs to be less than one over 1,000. So if we think about that for a moment, we have one over n plus one factorial is less than 1,000. Taking the reciprocals, we see that n plus one factorial needs to be greater than 1,000. Now, how do you get rid of the factorial symbol? It really just comes down to looking at the sequence, right? If we look at the sequence of factorial numbers n versus n factorial, zero factorial is one, one factorial is one, two factorial is two. It might sound like it starts off slowly but this thing actually grows quite rapidly. Three factorials, six, four factorials, 24. Go a little bit farther down. Five factorial is 120, six factorial is 720. So that actually is really close to 1,000. I anticipate the next one's gonna jump over. So seven factorial is, let's see, what is that one, 5,040? So winner, winner chicken dinner right there. We see that n plus one factorial will be greater than 1,000 only if n plus one is greater than or equal to seven, right? And so subtracting one n needs to be greater than equal to six. That's pretty nice, right? In order to be accurate to three decimal places, we only have to choose n to be greater than six. We only need six terms here. So our series, our series, right? It equals zero to infinity. Do be aware we started zero here. You have negative one to the n over n factorial. This will be approximately the same thing as s sub six which is just gonna look like one. Well, you're gonna get one over zero factorial minus one over one factorial, plus one over two factorial, minus one over three factorial, plus one over four factorial, minus one over five factorial, plus one over six factorial. So we take those numbers here and so we're gonna get the numbers we had here. We're gonna get one over one, one over one. Sorry, one, you're gonna get one over one, minus one over one. Notice the first two just cancel each other out entirely. They're donors. So the next thing we get here comparing our list, we're going to get one half minus one sixth plus one 24th minus one over one 20 and plus one over 720. That's gonna take a little bit more to do. I would put that in a fraction. If you do, sorry, put that into a calculator. The fraction you would get would be 53 over 144 or a decimal expansion is probably more appropriate here. This thing would be approximately 0.368056. So that's the sum of this number right here. And according to our estimate, our estimate will be no worse than B7, which is one over 5,040, which is approximately 0.0002. So while we required it to be accurate to three decimal places, we are almost accurate to four decimal places. And so this alternating series actually converges quite quickly. And we can actually find out its error bound much, much simpler. And so I actually like trying to compute the error of alternating series compared to some of the other techniques we did. Because for the integral test, it's a little bit more complicated because it involves counting some integrals. We'll do some other stuff later on with like Taylor's inequality, that's applicable. But whenever a series is alternating, you're gonna wanna use this error bound estimate to determine how accurate your estimate is. It's gonna turn out to be the best one possible. I mentioned already that and to determine the convergence of this series actually is a lot easier using something called the ratio test. And so in our next lecture 41, we will introduce both the ratio test and the root test as one, some final convergence test for series. And honestly, they're kind of my favorite, best for last in this chapter.