 We are at lecture 45 in the middle of 8.4 in the textbook So I'm kind of excited actually about what I'm going to do after this class that I'm moving a lot of my stuff from Harrelson Hall, which is a tired old building that I think is literally making me sick Through the ventilation system to the brand new building over there where old Riddick Stadium used to be across from Poe Hall So at about 11 30 today more than half my stuff will be in the back of my son's pickup and Moving into the new offices if you get a chance to walk in over there. It's really really a nice building So hopefully you'll get a Class in that building before you graduate. It's very very nice. I Don't think it has a name yet. I have an idea what the name will be Based on one of the donors, but it doesn't have a name yet Officially, what do you think it'll be? Griggs Hall. No There were based on my donors that the building would fit on this desk But someone else who has big deep pockets gave a lot of money. I think it eventually will be named after That man or his company which would be deserving the plan for Harrelson We still need the classroom space because our new building doesn't have Nearly the classroom space you would guess when you look at it because it's a big building, but it's a lot There's a lot of offices Mathematics will be over there and statistics in that new building And so everybody's offices in that those entire departments will be there including their graduate students So there are some there's a big auditorium Classroom that seats about 250 and there are several that seat in the 80s Some in the 40s, but we can't give up Harrelson yet. So I think the plan is to keep Harrelson active and used over the next five years or so and then Gradually like I think next year possibly the third floor of Harrelson will be closed off And then eventually when we've got enough classroom space the second floor And then the first floor and then the ground floor Things down there. So they'll phase it out And then eventually probably level it Yeah, that's built in 1963 It's unique, but it's just strange for a math classroom Okay, so let's um Start with what we left off with hopefully not the coughing fit But the content that was surrounding the math that led to the coughing fit We looked at this problem And I don't want to redo this one. I want to come up with another one That might actually get us to a reasonable error So we wanted the sum of this one accurate to three decimal places and we were kind of searching for that term That would not affect the third decimal place and then that's basically The first term that is not included So we would add all the terms together until we get to that point. So That one took quite a while to get there And there's another one that I want us to look at because I want to get factorial in a problem Before we use factorials in the ratio test So let's say instead of n cubed in the bottom Let's say we have n factorial and I think we'll See things getting there a little bit faster than we saw them getting there here as far as The level of accuracy that we want So it is alternating Uh, let's go ahead while we're kind of reviewing some things Let's see if this converges. I think probably not a doubt that it does converge with one over n factorial But let's go through the two-part test just to review. What's the first piece of that two-part test? Alternating series test Okay, is it ultimately decreasing So it's the n plus first term smaller in magnitude So we're ignoring the alternating part so the n plus first term with the Kind of the definition of what the value of the term is disregarding the alternating signs Would be one over n plus one factorial I'll put a question mark above this Is that smaller than one over n factorial? It is larger denominator smaller fraction. So that's true Um part of the reason I wanted to get factorial in a problem So I want to write out in a second what n plus one factorial is If you had to write out what n factorial is, how would you write that out? Okay, and let's throw that in reverse. So that's so the largest one is in right and we multiply n by n minus one n minus two Back to one right What's one factorial? Would be one times every integer that precede that Comes after it all the way back to one while it's already there What's zero factorial? It's one by definition. Okay, if it were not one by definition, we'd have some problems down the road So not that we get the zero factorial very often, but occasionally it comes up all right, so there's n factorial n plus one factorial should be n plus one times n times n minus one All the way back to one So could we say that n plus one factorial is really n plus one What is this? Times n factorial So we'll need that in the problem before we're finished today But clearly this denominator is bigger Then this denominator therefore the value of the fraction is smaller. So there's part one Uh, what else do we need the limit of the Value of the term other than the alternating signs Way out there to the right Is that zero That fraction one over n factorial as n gets larger and larger get closer and closer to zero Yes, so it is convergent by the alternating series test Suppose we wanted to let's go back to that same level of action. We went to the sum Accurate to three decimal places So it is convergent by the alternating series test So the question is can we find the sum Accurate to three decimal places So we don't have to search quite so far with this one Because factorials get there a lot quicker Then something cubed which is what we had yesterday So the first term is positive then we'll alternate signs from there first term is what one next term minus One over two factorial, which is one over two probably should have done this example yesterday instead of the one we did The search is a little more confined back to positive one over three factorial, which is One over six back to minus one over four factorial Four factorials four times three times two times one 24 Five factorial five four three two one in a product 120 Six factorial six times 120 720 They're getting the getting larger the denominators getting larger faster than the cubed values yesterday um Five factorial six factorial seven factorial seven times 720 50 40 And we could find more I think we're good So we want accurate to three decimal places Uh, is there a possibility that we are there? Let's see one two three four five six Let's check this one out What's the value of the sixth term? I don't care if it's positive or negative What's one over 720? Not good enough yet, right? Because we want accurate to three decimal places I think that value as we decided yesterday would in fact affect the third decimal place Sure, we're going to get it here What's one over 50 40? one nine eight so If we use this term the seventh term as our error term it looks like we're going to get the level of accuracy We want so if this becomes our error term Then we want the sum of the first six right the sum of the first six. Let me see if I jotted that down Point three six eight so we would want Those six terms added together somebody can check that out, but I that's what I have written down And that is accurate to the third decimal place because the next value if we would add it in Is this value which would not affect the third decimal place? So the sum is point three six eight accurate to three decimal places So we might have some changes, but they're going to occur in the Decimal places beyond the third That's a little cleaner More searchable example than we had yesterday. Plus I wanted to get factorial in there Uh another thing that we're going to need in a problem before we finish today questions on this before we Leave this are you checking that out katie to see if point three six eight is right? Didn't get it. Okay Yeah, that's not right point three six eight can't be right. So let's I'm checking it out because one is too large right Then we subtract a half which we always end up subtracting too much So we then we get the sum of the first two is a half So the sum is actually between one and a half. So that can't be right. Would you get six? Six three two I don't know where that came from. Let's get rid of it Got that too. Okay. Thank you. Thanks for checking that out Um another problem we're going to encounter besides n plus one factorial We're going to have a problem that has n To the n in it And then we're going to be talking about the next term Well, the next term would be n plus one Raise to the n plus one So it's to our advantage to simplify A little bit n plus one raised to the n plus one Isn't that and it might not Be exactly clear where we're headed with this, but is this a legitimate equation? Is n plus one raised to the n plus one The same thing as n plus one times n plus one to the n same thing Right because this would be to the first we'd have the same base In a product two things In a product situation with the same base we could add the exponents So we'd add the one in the end and we're right back to where we started So that's something else we're going to encounter in a problem That we need to solve before the class is over today All right next topic in 8.4 after the alternating series test and how can we approximate a certain number of terms in an infinite Alternating series and what how do we guess at the error or remainder? Is absolute Convergence And it is what the name says it is is that it's Not only is the series under consideration, which might be alternating Convergent so is the version that is not alternating. So it's absolute value. So we're really kind of Considering in a sense. We're considering two series at the same time the one that alternates and the one that does not alternate So here's an example of an Absolute convergence problem If it converges and it's absolute value converges. We say it is absolutely convergent So we're really considering this one In this one In a sense simultaneously If the non alternating version converges Wouldn't it make sense that the alternating version? also converges We can take a series that it's Non alternating version does not converge But yet it's alternating version does converge What's an example of that one over in we've already done that one over in By itself not alternating is the harmonic series which diverges The alternating harmonic in fact converges so This is kind of the one we have to check if this one is convergent Clearly the alternating series will also be convergent But there is a possibility that that they could differ in solution So what about one over into the fourth the non alternating version? converges Not because I say so or you say so It's p series p is four So the absolute value of p is Greater than one or greater than or equal to one Greater than and if it were one then if we'd have one over into the one which is harmonic So we want it greater than one So if the p series p is four the absolute value of p is greater than one then we know it converges If this one converges It's alternating version will automatically converge. So this is a series that is absolutely convergent Now we could go through the test Is the n plus first term Less than or equal to the nth term. Yes, larger denominator smaller fraction And is the limit as in goes to infinity is at zero Well, it doesn't it have to be zero because this one would not have converged if the limit were not zero In fact, if the limit were not zero, we'd say this one diverges, right? Which is one of the tests for divergent. So you can see if this one is convergent Clearly this one has got to be convergent. How could it go to zero if it were not ultimately decreasing? That'd be kind of tough to do And there's no way the limit could be zero if in fact this one did not converge. So once you get this one You're automatic on this one So that one as a first example as a second example one that we just talked about The alternating harmonic Compared to the harmonic itself So let's look at this one harmonic therefore Divergent we've validated that for three Really good reasons, but that's enough But Because this one was not convergent. We kind of have to test this one separately. We've already tested this one One over n plus one less than the predecessor. Yes Is the limit of the nth term zero as in works its way out there to the right? Yes So by the alternating series test this one is convergent This one we determined is divergent. So is this series Absolutely convergent It is not So if the positive term series is convergent Automatically the alternating one is convergent But if this one is not we can check the other one separately, but clearly already it's Not absolutely convergent All right, the bulk of what we need to spend our time on today is the ratio test Which is another test for convergence and we will use this Not only As a test to determine if a given series that doesn't have any variables in it Is convergent or divergent But we're also going to use it when we have variables in it as in a maclaurin series or a taylor series that have x's in there Under what conditions would one like that be convergent or when would it be divergent? So we'll use the ratio test in In the rest of this chapter quite a bit So here's how it's going to go. We're going to compare the values of the n plus first term In quotient form, so we want to divide these two the n plus first term divided by its predecessor We don't care if they were alternating or not because we're going to dispense with that We're going to take the absolute value And we want to see what happens to this quotient Way out to the right. What is the n plus first term? Divided by its predecessor So you can see why it's called the ratio test Ratio of the n plus first term divided by the nth term and we're going to get an answer l Now i bet you you can tell me What's going to happen based on what we get for l So let's say l Is smaller than one What does that say about the value of the n plus first term? Compared to the value of the nth term as we work our way way out to the right That this one Is bigger than this one right so if the top is smaller Then the bottom that means that the terms in a sense are Getting smaller as we go way out to the right so if they're getting smaller in this Passion now granted this is not a catch-all kind of test. We're not going to If that happens to the point where it's distinguishable way out to the right as n approaches infinity That's going to be good enough to determine convergence You say well, what about the harmonic? What if we did the harmonic? Well, we'll do that one and we'll show that this test isn't good enough to determine If it converges or diverges So you've got to at least trust the fact that if the limit is less than one It's convergent If the limit is greater than one, what's that say about the n plus first term? Compared to its predecessor Top's bigger than the bottom And that's not a good sign for convergence if the terms are getting a little bit larger as we go And you can tell which value we've left out Greater than one less than one What about if the limit is one? That means that the n plus first term Is for all practical purposes exactly the same as the nth term We can't we don't know we can't make a decision In this case the ratio test fails to give us a conclusion So they're basically they're indistinguishable one they might be getting slightly smaller But it's not enough to say that there's a distinguishable difference between the n plus first term and the nth term Hopefully that stuff Really makes a lot of sense that they're getting smaller very it's distinguishable Clearly that one the n plus first term is smaller than its predecessor That's got to be a convergent series If the n plus first term is larger, that's a problem So they're going to diverge and if they're indistinguishable We're not going to be able to make a decision So it's not a End-all kind of test. We're not going to be able to answer everyone Let's go the first example to the Harmonic we know already for three very good reasons that this is divergent What is the ratio test going to look like? What's the n plus first term? There's the n plus first term The nth term is one over n. I could take the absolute value, but It doesn't have any bearing on these because in the n values are all positive anyway So we'll multiply by the reciprocal What about n over n plus one as n approaches infinity? one right One in over one in plus one this plus one as we work our way way out to the right is going to be pretty insignificant, right? 600 trillion over 600 trillion plus one Not a whole lot of difference between the numerator and denominator So the limit is one We already know this diverges, but according to the ratio test We're right here. It's Distinguishable the n plus first term and the nth term Way out to the right. So this test wouldn't tell us That it did one or the other, but we already know it diverges. So I guess the reason to do that as an example is You've got to be able to trust the fact that if the if it is distinguishable and the limit is less than one That would be a convergent series Okay, first real example That's kind of a pseudo example alternating In cubed Over three to the n So when when would be the situation where we would need to use the ratio test If it's an infinite geometric series, we don't need to use it. We have a better way In other words, if we can use a more straightforward way We would use that I don't think we have a more straightforward way for an alternating series That has n cubed in the numerator and three to the n in the denominator I don't think we've dealt with anything quite like that before As a certain type of series other than alternating So the ratio test Is going to look like this I'm going to write this the first time, but you won't need to Write this every time negative one To the n plus one n plus one cubed Three to the n plus one does that look like the n plus first term? Everywhere there was an n there better be an n plus one if i've done this correctly So there's the n plus first term We're going to divide it by the nth term which is normally the way it's handed to us So what does the absolute value do to the fact that these might be alternating one positive one negative Won't that always be the case between two consecutive terms one will be positive one will be negative But doesn't the absolute value more or less take care of that? So we can kind of dispense with those you don't even need to write those into your problem See anything numerator and denominator that can reduce three to the n And three to the n plus one Leaves a three where In the bottom right this has three to one higher power than this one And we've got something cubed and something else cubed So i'm going to bring that three out in the denominator. I'm going to bring that out as a one third I don't know. I don't know if I like that any better than the other This one Or this one not going to change the answer We look at it this way n plus one over n as n approaches infinity I don't like that and I'll I'll come back to why I don't like that as well as this one I'm going to abandon this one. It's true, but As long as we have a finite exponent We're in business with that if our exponent is something that is itself increasing and getting larger That may not be the best way If n plus one were cubed what would be the first term? I'm going to scrap that we could use that but i'm moving over here What's the first term of n plus one cubed? Incubed is everything else of lesser degree? Yes, so we've got some other stuff up here Not needed. Why is it not needed? Because they won't really matter. This would be the dominant term as n approaches infinity, right? So we've got n cubed plus some stuff of lesser degree Down here. We've got n cubed So what happens to n cubed? Plus some stuff of lesser degree Divided by n cubed as n approaches infinity One right that approaches one For the same reason that we had earlier would we have an n plus one over n? That was one n over one n here. We've got one n cubed and one n cubed So our answer is one third. So do we get a conclusion on this series? right so the limit Which was equal to one third is less than one So this one converges by the ratio test This one would work. I just think this might be a little better line of thinking Since we've seen things like that in the past Get the first term everything else is of lesser degree n cubed over n cubed no matter what other stragglers they have with them that approaches one All right, last example. This is kind of cool actually It's got some factorials in it. It's got n to the n in it Not that we do non-cool examples They're just not always as cool as this one is n to the n over n factorial I don't know if you have a intelligent guess as to whether that converges or diverges That one's kind of tricky to make a make a stab at But let's see what happens with the ratio test So wherever there's an n we now want an n plus one So there's the n plus first term over the nth term Don't tell Jacob about that what I said for not to tell him Just kidding Yeah, that's not just n. It actually has mathematical significance So we have n plus one to the n plus one It is always interesting to get that question. You know, what does that mean? So we'll multiply by the reciprocal of the one in the bottom This is why it was hesitant in the previous problem to Do n plus one Over n put them together when it was cubed because we do not want to do that here Because it's an unknown exponent In fact, that becomes the crux of this problem All right, let's do some simplification in the top and also the bottom And we've done both of these already So we've got an n factorial in the numerator And we've got an n to the n sorry In the denominator. Let's leave that alone Let's rewrite this Isn't that the same thing as n plus one Times n plus one to the n. Is that correct? We did that a few minutes ago Question Nicole Yes So before we get there Everybody okay with that n plus one raised to the n plus one is the same thing as n plus one Times n plus one to the n There's a reason why that becomes beneficial And Nicole you were going to rewrite this one as what? Everybody all right with that? So what are the advantages to simplifying that term in the numerator and that term in the denominator? Yes, so n plus one over n plus one Clearly that's one no matter what n is approaching n factorial over n factorial Here's what I think is the cool part of this problem We've got n plus one to the n Over n to the n Since we have them both to the n We can do the division first and then raise the quotient to the n now The tendency with this which I didn't want to develop that mindset in the previous example is to look at what's inside And say n plus one over n as n approaches infinity The inside piece goes to one And then we'd have one to the n which would be one which would mean the we fail on the ratio test But when this is headed toward infinity One to the infinity may in fact not be one That's an indeterminate form One to the infinity So let's break this piece down right here a little bit further We've done this before but not in a problem just like this Let's take everything inside this quotient And the key issue to this when would you want to do this you would only want to do this when the power Is getting infinitely large If it were something cubed something to the fourth You're not going to experience the same thing that we're going to experience in this problem by letting it go all the way To some infinitely large power So the n gets divided by n The one gets divided by n and this n gets divided by n inside the parentheses So n over n is one One over n we're stuck with one over n and in the denominator We also have n over n So anything over one is just that now that's Maybe A whole lot like what we started with but it in in essence is very different. Does that look familiar? One plus one over n to the end I guess not something yeah, one of those yeah pi b One of those Let's put in some values If n is a hundred What's one plus one over a hundred raised to the one hundredth power somebody else put in ten thousand What happens to one plus one over n to the end as n gets infinitely large You look at that and you say well, that's just one to the end It isn't because This is probably what 2.70 and this is probably 2.718 I would imagine we're getting pretty darn close Pretty darn close to what? e So that's one of the definitions of e or one of those numbers right in this case. It's e So there's our conclusion Not really our conclusion, but it's going to help us with our conclusion If the limit is e What's our conclusion about this? It diverges Because what the limit was larger than one right? It's kind of a unique problem that we end up with n plus one over n raised to the end and Everything tells us that that's one But if we really analyze what happens to that it isn't one. It's the value e So e larger than one the limit is greater than one therefore the series diverges Uh, let me make sure that i'm not leaving anything out here. This is kind of some interesting wording I guess we should look at it before we leave it So back to the absolute convergence by definition a series is absolutely convergent. This is in the book If the series of absolute values Of that series is convergent So basically if the positively termed series is convergent So is its alternating series companion The series a sub n is absolutely convergent that that actually just kind of sounds funny If it's absolutely convergent then it is convergent, right? So why what why would that be stated? If a series is absolutely convergent Then it is convergent It referring to possibly the alternating version If it's convergent Absolutely In its positively termed series. We don't even need to examine the alternating series So if it's absolutely convergent, certainly it's convergent both of them are positive and negative versions Glad we looked at that Here's the ratio test. Let's make sure that we Came up with the same conclusion We take the limit of the absolute value of the n plus first term divided by its predecessor way out to the right We get a number hopefully L L is less than one then that series is absolutely convergent I guess we didn't throw that word in Probably should because what did we do? We did away with the alternating part and we looked at just the positively termed series because of that So if the positively termed series is convergent so is its alternating counterpart So we can throw that word in absolutely convergent If we get an answer that is greater than one certainly infinity is greater than one Then the series is divergent So they don't even mention what happens when l is one. So in other words, there's no conclusion Absolutely no conclusion. Let's let's add that in absolute in that decision Okay, we have five minutes left, but I'm not going to start power series Which is section 8.5 with five minutes remaining So let's start 8.5 tomorrow Web assigned when do we have web assigns do? Two days, okay So tomorrow would be a good day probably if you have web assigned questions Since we're meeting tomorrow and not meeting Thursday