 Welcome back. We're at lecture 41. We are probably almost literally in the middle of 8.3. We will finish this section. I hope if my jaw and mouth and tongue cooperate. I went to the dentist this morning We will finish this prior to test to our next test, which is say I told you it may not work exactly like I wanted to Probably everybody said well, I think he went by the bar on the way to class this morning. No that's not the case but We will finish this section, which is different from what the syllabus looks like Actually, it's it's doing real well. I was really worried about 915 that I Couldn't make any you know, I could make some guttural sounds, but I couldn't form any words. So it's much better Just don't laugh too loud So we are at the comparison test Actually, let's go back a half a step to the p-series So we want to be able to shortcut this stuff We don't want to have to re-prove everything every time we come across a new Series so we have something in this form and we've already validated some things about p-series when Do series like this converge and when do they diverge? P greater than one converge that correct Anything that we've done in class that will remind you that this is the category for convergence We did one over n squared, right? We did that a couple of different ways Clearly p larger than one it converges P less than or equal to one well equal to one We know that's the harmonic series so that has to diverge I think we did one also that was less than one didn't we do the square root of n into the one half as an example and Validated that it diverges. So that's all you need to do. We've done the background work for this You identify the one in question as a p-series Identify what p is in the problem Validate that it's in one category or the other then you write down your conclusion You don't have to redo this every time. I think the sheets work their way around class today Let's take a brief look at that. Did they make it around? Extras okay, just pass them up Because you didn't get one this was put together by Marilyn the column it's it's excellent and it's a great summary sorry I didn't Copy I copied it myself. So didn't kind of you have to flip it and turn it It's kind of the way they're presented in the book, but it might be the way you want to approach this thing as far as if you look on the first page which is It has at the top summary of convergence and divergence tests But the nth term basically is the Test it the nth term and the nth term does not disappear go to zero as in a person's infinity Decision already made in divergence now. What if the limit is zero? Might converge, but we need some other tests that's going to tell us that But if it is not zero for the limit of the nth term, then you work your way down the page geometric series What do they look like? Under what conditions did they diverge and converge? Common things to watch for maybe a kind of a classic Example over here to the right P series we have that one now integral test Basically if you can Integrate it and the integral diverges. So does the associated series you can integrate it and it converges So does the associated series so we are Basically the bottom of this first side at this point now This is not something you can use on a test, but it is a nice organization of the Techniques and there are quite a few you can see front and back that once we're done there are quite a few Tests for convergence, so keep that handy. We had an example yesterday Kind of ended class with it, but here was the problem in question We decided that if we could determine something about this series Which is a whole lot like the one in question We could somehow compare it to the to this one Then maybe we could make a decision about the one in question based on the one that is similar to it But he is one that we know something about What did we decide is true about one over three to the end? that converges Unfortunately, it's not Valid if I say it converges or you say it converges. So we probably should say why it converges Why does this one which is very similar to the one? We're trying to figure out what it does What does what do we know about this one? Okay, what type of series is it? infinite Geometric series and what is the ratio in this one? I think we wrote this down yesterday in class the ratio is one-third If you don't like the form that that's in then that's just tough. No, we can convert it to another form Couldn't this be one-third to the end Right doesn't matter one to the end is still going to be one three to the end is what we want in the bottom anyway Doesn't that look a little bit more like an infinite geometric series? That we've got higher and higher powers of one-third. So pretty clearly the ratio is one-third So if the ratio is a third and it converges when the absolute value of the ratio is less than one That's all we need right this is enough Validation what kind of series is it? It's infinite geometric. What's the ratio and is the absolute value of the ratio less than one? Therefore all this stuff Results in this conclusion Now even in this form although it looks more geometric It doesn't look like the one that you'll see on that sheet the one on the sheet. I'll come back to this page So we've taken this and we made it a little more geometric looking And how about the one on the sheet isn't it supposed to or can't we somehow? Put it into this form right Let's see if we can actually put this of the series that we said was infinite geometric One over three to the end. What's the first term first term is one-third, right? so There's a you want to write it like that you can our Is also one-third and we could raise it to the n minus one As long as we start in at one and let it run to infinity is that What we want is that one over three to the end Probably should have brought a napkin or something like that. I'm going to draw all down my face here today, too That's the same one. We started with is that correct one over three to the end first term is one-third ratio is one-third So it is Able to be converted to that kind of standard generic form. You don't have to do that, but at least it is able to be done So if that was bothering you that this is what all geometric series look like and ours doesn't really have that look We can make it have that look. All right back to This problem. We're trying to decide about this Does it converge or does it diverge? How is the one in question related to the one that we already know something about? Term by term It's smaller isn't one over Three to the end plus five less than One over three to the end Correct larger denominator smaller value If this is true and we know for a fact that this one converges which we do Then the conclusion is according to the comparison test. This is the comparison test the one in question also Reason comparison test. So if something is already small enough to converge Something new is smaller than the one that already converges doesn't it make sense that it should also converge This one is already small. I've used this in main reasoning before so those of you that were in my class last semester This one is already small This one is even smaller What is smaller than small? Isn't it also small? Does that make sense? Yes, it does. That's to me If this is small and this is even smaller this one would also be in the small category So what's the other direction of the comparison test? What's the other side of it? smaller than small is so-called small What's the other end of that? How about larger than large? Isn't it also large? sounds good Somebody walked in the door and they're six foot six and you said wow, they're tall and then somebody walked in after that and they're seven foot three You'd say wow, they're tall too. Why because they're taller than the person that was already pretty tall taller than tall is tall and What about if someone were you know Like my mom you wouldn't believe my mom was five three. I was taller than her in third grade So my mom is small So she walks in the door And you say gosh, she's small and then someone else walks in and they're four foot nine You say gosh, they're also small because they're smaller than my mom and she was already small She had a quick hand though, buddy. I'm telling you she would reach up to me and I'd say something smart and I Felt it many times. So she was quick and But I deserved it every time every time she Wacked me in the mouth. I deserved it. I probably deserve more than that So smaller than small is small. So we have one that's convergent. We have one that's even smaller than that It also converges. We have one that diverges We have one that we know is even larger than an existing divergent series. It also diverges now the question is What about if it's? Larger than a convergent We can't Make a decision about that What about if it's a little bit smaller than a divergent? We can't make a decision about that So there is an area there where this test is not good enough to make a decision Nicole so if 1 over 3 to the end diverged instead of converged or whatever and The other part was 1 over 3 to the end minus 5 with that part diverged because it was Larger than that's exactly right. So that's the other part to this test So let's say that we had another series. I don't know if I have an example. I don't think I have one written down saw this That's not gonna work, but let's go with this just to show that it's not gonna work So we would try to compare this To 1 over 2 to the end This converges why it's an infinite geometric series the ratio is a half, which is less than 1 In absolute value in order to make a decision about this. We would have to know that this is smaller than The one that we already determined is convergent. Is that true? It's larger when you subtract 1 from the denominator The net value of that fraction is in fact larger than this right if you want to make a fraction smaller You make its denominator larger if you make the denominator larger you even instead Go in the other direction you've made the value of the fraction larger. So this is not true Therefore there is no decision in the comparison test So the comparison test Fails to give us a conclusion here now. There's one that's very similar that we're a few minutes away from Discovering as well called the limit comparison test, but just the good old-fashioned Comparison test isn't good enough to tell us what this one does What what would your guess be if this one converges? Don't you think this one also converges, but this test isn't good enough to tell us that So we'd have to get another test so let's say So we can get an example What is one that we know diverges? 1 over n I'm going to start this at 2 for obvious reasons I think when I work my way back over here because I've got to start this one at 2 Otherwise, I'm in trouble right in the denominator with a zero So we know for a fact that this one diverges All we need to say is it's harmonic. We've already kind of validated that a couple of different ways What about 1 over n minus 1? Compared to 1 over in well by subtracting 1 from the denominator. We made it larger This one is already divergent This one term by term is larger. Therefore this one must also diverge By the comparison test So smaller than an existing convergent series also converges Larger than an existing divergent series Also diverges. Any other decision we try to make is not this test isn't going to work for us All right. Well, let's go back to this one where the regular comparison test failed Let's go to a test that's similar In as far as the name of it and in a sense the kind of the nature of our conclusions But we want to see since the kind of the baseline or simpler comparison test failed Let's try what's called a limit Comparison test. I think you'll see this on the sheet This is actually on the back side of the sheet So we're we're getting there We don't have to be concerned about which one is in the numerator and which one is in the denominator We're going to have the one in question. You can call it a sub n or b sub n It doesn't matter and then the one that we think is similar to it Give it whatever name you want to call it a sub n or b sub n. This is the one we know something about So this is known To be convergent the one in question we try to use the comparison test we kind of went through that It's not going to work because this term by term is not smaller But in fact a little bit larger, but we have an idea that it also converges so we'll take a limit of the one in question Over and you can transpose this doesn't matter which one you put in the numerator which one you put in the denominator Over the one that we know something about and we're really only concerned way out to the right What happens as we compare these two now? Let's go through what might happen. Let's say that We get a limit of one. I don't know what we're going to get but let's say we get a limit of one Doesn't that mean that way out to the right? There's practically no difference between the one in the numerator and the one in the denominator Is that what that means if the limit is one numerator and denominator are practically the same What if it's three? That means the numerator is three times the size of the denominator This one already converges. Does that mean this one would also converge? Yeah, just converges to a Number that's three times as large as this one So no matter what you get as long as you get a limit could be 11 or 17 They both do the same thing So if we know this one converges and we get a limit of one or three or 11 or 17 this one also converges or 111 or 117 so if you get an answer they both do the same thing So if we're dividing by that one will multiply by its reciprocal I don't know. What do you want to do to get a solution on that a couple of different ways? It's not a mystery that two to the end and Two to the end is not going to be a whole lot different when you get way out there to the right What's the difference between two to the one thousandth and two to the one thousandth? Not gonna matter too much isn't If you had six hundred trillion dollars, and you dropped one would you bother bending over to pick it up? Probably not so The numerator and denominator are about the same But let's actually kind of validate that how about if we divided through by two to the end What happens to this? One over two to the end term as in approaches infinity Disappears So the limit is one that means the numerator is practically the same series as the denominator So if the one in the denominator Converges the one in the numerator also converges. That's exactly right if it's a hundred That's fine. It still converges. It just converges to a number in all probability. That's about a hundred times as large So in this case they both converge Because the one that we knew something about converges the one in question Does the same thing if you get it an answer for the limit problem? Now what when would this test fail? I mean they all kind of have Bounds when they work and when they don't work. Can you think of what could happen here that we cause this test to fail? Okay, how about equal to zero what if this were equal to zero that means that the numerator is a whole lot smaller than the denominator If this already converged wouldn't that converge even more rapidly, right? possibly How about if the we didn't get a limit the limit does not exist Well if the limit does not exist we can't say they both converge or both diverge We don't make a decision on that one So we do want to get an answer to the limit problem does not exist is not really technically an answer So in this case if we know the one we know something about Chandler question You say it mattered Where you put the known and the unknown On top on bottom doesn't matter doesn't matter in this problem It clearly wouldn't matter because it was one we still get one over one In a problem where it's not one. Let's say it's one third Then that would mean the denominator is three times as large as the numerator If you transpose that you get the numerator being three times as large as the denominator. It doesn't matter Right, it's a ratio of the one in question to the one that we know something about Put it either one in either position numerator or denominator now it does say Where the answer is finite and greater than zero in In the convergence case zero probably would mean that they would both converge I think that would be problematic in the divergence Because zero times the Divergence that that could be problematic. So let's just throw zero out. We don't want to get zero either That's another way this test could fail We want to get something some number other than zero and we certainly don't want the limit to not exist What's this one do? one over three to the end Converges infinite geometric series the ratio is the third which is less than one in absolute value Here's one that looks a whole lot like it You have to decide am I going to be able to use the comparison test or Do I need to use the limit comparison test? What would make that decision for you here? Comparison because we add something to the denominator. Don't we make it smaller? Right, this one's convergent this one term by term is even smaller So the regular comparison test is going to work here The regular comparison test is not going to work here But we still have a pretty good idea because this one looks a whole lot like this one This one converges. We have a good idea. This one's also going to converge so we would Kind of skip the comparison test and go right to the limit comparison test And see what happens way out to the right and we're going to get Probably one again, right? Which means that if this one converges so does this one so it's a little More sensitive of a test. I guess it's going to pick up more values than the just the straightforward comparison test Little bigger net I should have started class with this today, but it actually just came to my mind Add a couple of questions at the end of class yesterday And I know what we have yet to do in this section And it would probably be a good thing to address that question that came up right after class yesterday See if I can the first one was one over n squared Which we now know that converges because it's a p-series and p is larger than one So we had a picture And actually a couple people had the same question after class So there may have been more of you that had this question that didn't ask it in class yesterday Let's write out the first few terms of this series the first term would be one and then one over two squared one over three squared One over four squared So we know it converges This is when we were doing the integral test. We also integrated one over x squared From one to infinity the first term I'm going to write what I wrote yesterday And then I'll address the question that came up after class I Wrote the first term of this Series out here, which is one That block is one wide and one tall Here should be the second term The f of two if you come up to the curve itself or the curve is one over x squared So it'd be one over two squared, which is one fourth. So that's one wide and one fourth tall So there's the second term This is one ninth tall in one wide So we actually started the integration at one that's where this starts and We said because the integral one over x squared from one to infinity Converged That the series associated with it even though it didn't capture all the boxes We had an extra box of area one So we kind of wrote this out to the side and said we're going to add in that one block or box To what we get for the area under the curve. I Think we also got one for this if I remember correctly The important thing is because this converged The series associated with it also converged if the integral converges So does the series by the integral test Somebody brought up the question. Well, why didn't you just? Instead of having this one by one block out here by itself Why didn't you move everything over to the right by one? Did not what that was the question that came up Okay, and we'll do that because it makes sense based on what we have yet to finish in this section that We see why that although it's plausible idea why it's we don't want to do that So the rationale is we're able to find the area under this curve because it converged Therefore we should be able to add one fourth to one ninth to one sixteenth to one over twenty five It's underneath the curve if the area under the curve itself converged Certainly the area of the blocks underneath the curve that didn't even make it up to the top We've got some missing pieces. They will also converge and then we were missing one We'll just add it back in that's not going to change convergence or divergence So the question came up. Why not? Come out here to one Why don't we just shift everything over? So instead of having this one by one block out here by itself Why don't we draw the one by one block right here? There's that block. We're just going to shift it over to here This block the one fourth block. We're going to shift over to here the next block Why do we not want to do that? Because we're over the curve Under the curve. Yes that converged, but we're adding some on to that now Does that maybe take something that was convergent? We add a little triangle on each of these now? Could it possibly diverge that's possible? So that's back to the This is convergent. Therefore. This is small. We want to stay smaller than small, right? We don't want to be larger than small. We don't know what that means My mom's five three. Someone walks in that's taller than her. Does that all of a sudden make them tall? No, they could be five four five three and an eighth Okay, larger than small doesn't make you large That's what we want to steer clear of so that's why we have to take this as it is Even though we don't capture the one by one block. That's fine. We can add it on at the end Let's take all the rest of the terms as long as they stay under the convergent curve. That means the series is convergent the other question or issue was What was that curve? Just one over in this one we know Diverges the integral associated with it also diverges which that's kind of the basis of the integral test So if this one's already in trouble and is divergent What do we want to happen? With the blocks that we kind of associate with this particular diagram We don't want them to come underneath the curve. That'd be smaller than large I don't know what smaller than large is might be small might be large might be medium We don't have a medium category So we want to do what with the blocks that we put in here Could we put them over here put the one by one block here? We could we I mean we get to draw them wherever we want to draw them so we could draw them over here Why would we not want to do that? Well, we put the one by one block there and we start our area right here Then the next block is what? One wide and one half tall Now we've got both of these The next one is one wide and one third tall Now we've got a problem because this area Was divergent Now we're underneath that We're missing some pieces. Does that take what was divergent and possibly make it convergent? We don't know So what would be a better picture than this is the picture that we actually did yesterday Why don't we take our one by one block here for one half by one block here? In our one third by one block over here Now we can all of a sudden make a conclusion because what this was divergent Now we've got even more than that. It must also diverge. Does that make sense? So in case you had that same question But didn't stay after class yesterday to address that that's why we want the figure to look like this And not like this No conclusion can be made here. We can make a conclusion here Same thing with the other diagram. We can start it, but I don't think we can finish this Let's say we have an infinite series But we don't let it run to infinity because it's We can't physically write them all out and there's no other shortcut To getting the sum we don't have an infinite geometric series. So we can't find the sum But we know it's a convergent series. We want to approximate The sum so maybe we add the first 10 terms or the first 20 terms Or one example in the book talks about adding the first 100 terms of the series together You probably would expect a pretty high level of accuracy if you added the first 100 Terms of a series together But it's not going to be exact because you're missing some you're missing all the ones Beyond the 10th term or beyond the 100th term So when we do stop And have a finite number of terms S is the sum of all Terms S sub n is the sum of n Terms obviously there's a difference So just for notation purposes The authors address this the difference between the sum of all the terms And the sum of what we find the first 10 or the first 30 or the first 100 is the remainder Associated with the sum of the first n terms. So that's r sub n So these diagrams are related I think you'll see to what we just looked at in terms of how we draw the figure So we want to know kind of an upper bound for the remainder or error And a lower bound we're not going to be able to if we If we have the sum of the first 10 terms and we could find the exact error Then we'd be able to add it or subtract it to our Sum and we'd have the exact sum. Well, we're not going to be able to get the exact error We're going to get an upper bound Something out here And a lower bound So our error or remainder is somewhere in that vicinity. So we know kind of How bad we are we're really most of the time concerned with this one. How far off are we in terms of Being away from the exact sum Here's our first picture that we want to look at So we want to categorize this remainder or this error Let's kind of start with this and see if the picture makes sense Let's pick up where we left off So we're leaving off at n So we found the sum of the first n terms And we have an answer for that Add term one to term two to term three all the way up to term 10 So we have the sum of the first 10 terms So we're stopping at 10 We're stopping at n What are we missing well, we're missing the first term we're missing is the n plus first term So we come over to n plus one We go up to the curve. This is the curve that basically is the same function as the Descriptor of the series itself Then we would add in the next term. We're not going to these are just the terms we're missing And so on Is it possible to find the area under the curve? All the way out to infinity It is It not in in all cases, but if it is We take the area under the curve from in which is where we left off all the way out to infinity Isn't that more than the terms we're actually missing? Here's the area under the curve This whole thing There's what we're really missing This block this block this block and this block So isn't the area under the curve from n to infinity Actually more than the terms we're missing It is because we've got these little things that are area under the curve. We don't even care about them They're not terms in this series So is it true that the area under the curve f of x Integrated with respect to x from in which is where we left off All the way to infinity That area is more than what we're really missing. Is that correct? We're missing the blocks the area under the curve is actually more than the blocks So there's our upper bound for our error So it's the remainder or the error is bounded above by the area under the curve From in which is where we left off all the way to infinity The other side of this I think you can see why I wanted us to go back and look at the other situations It's very similar to this The other end of this What's the lower bound for this remainder? Well now instead of starting it in Why don't we go over one place whatever n was we stopped at 10 we added the first 10 terms together Well, let's go over to 11 Let's go over to the n plus first term. So we come over here to n plus one Come up to the curve. There's the n plus first term that we are missing There's the n plus second term That we're missing. There's the n plus third term that we're missing So can we go from n plus one? All the way out to infinity And capture kind of a lower bound for the remainder or the error Well, kind of but didn't we miss some? Right Isn't the blocks The sum of the blocks that we missed it's actually more than the area under the curve So it's a lower bound For the remainder or error estimate because really there's there's more to it than just the area under the curve We're missing these little triangular shapes as well So there's our lower bound We'll start with this next page tomorrow, but let me Put these two together So there's our upper bound for the remainder or error There's our lower bound So the actual error The terms that we missed by stopping at n equals 10 or n equals 100 Is actually somewhere between there Neither one of them captured it exactly one. We were Underneath the curve. We were missing some of the error The other We were above the curve and we had too much so it's actually somewhere in between So we'll start with this tomorrow and then we'll look at a couple of examples of how we can stop at the 10th term Figure out how far our answer might be from the actual sum