 Sorry, let me make one comment about the web design problems or the polar area problems like keeping these emails. It's important, it's critical that you think about what the integral represents so that if you're finding, you have some function in polar coordinates that looks like this and I want to find this area, I have to remember that when I do a polar integral, I'm adding up things that look like this and so you really have to think about what the integral represents geometrically. So for example, there's a problem where you're supposed to find the area between two circles that look like this, it is if you just write, integrate from where the two things intersect to where the two things intersect and take the squares and subtract them and zero because really what you're finding is you integrate from here to here and you get this area, minus this area, well that is equal so that you can do it. So this problem which probably a lot of you have not done so you don't have a clue what I'm talking about, if you solve, so this is like r equals cosine theta, r equals sine theta so I'm not telling you how to do it, I'm telling you what not to do, if you solve where these two things equal, then you get theta is pi over 4 where theta is pi pi over 4 and if you just do something silly like write the integral from pi over 4 to pi pi over 4, 1 half cosine squared theta minus sine squared theta, the answer is zero, that's not the answer to this question and that's the casualty that's represent is the area, cosine squared, this area minus this area, well it's zero, think about what's going on. Okay, so let's move back to infinite series or infinite sums, so as I said before, I will try to say the word sum instead of series, the book does not use the word sum, it uses the word series so if you, the reason I prefer to use the word sum is because it means adding, series means something slightly different in English than it means in math, so anyway. Last time, we saw that if we have something like this where r is a number and so it's value less than 1, it's a geometric series, or geometric gas, geometric series so this is 1 plus r, r squared, r cubed and on forever so what we saw last time is that this adds up to that so that means that if we see something that looks like obviously this 8 thirds plus 16 9ths minus 32 27ths plus what's next? 64, 81st minus blah blah blah, what would you think? How would I do this? So I'm not using the clicker so I can't ask this if it was a question, but that doesn't mean you get off from thinking, so I'm going to tell me how to move this problem. Okay, so what do you see that should happen? What's the numerator of the problem? 2 to the n? Well, but it's not every 2 to the n, you're going to need your clothes, that's not 2 to the 2 n. Yeah, okay, sure, this is not what this equals, we're closed. There's a negative, there's a negative, so is it this? I'm going to start with 0 to the 1. Well, of course it matters, but so let's just try and figure out what it is and then we'll turn it into something that looks like this, so this is not right, what's wrong with this, okay, or I can just, all right, so if you're saying do that, and what is n, sorry, 0, so the n is 0, we have positive 2, the n is 1, we have minus 3, so that seems to work, so okay, and this doesn't look like that, but how can we make this look like that, or do we have any hope of making this look like that? So what's the problem with this, why does this look like that, right? So how can I turn it into being the same power? Yeah? You move over here, what's wrong with you? Why the place that's in there? No, okay, fine, so go ahead. You can make 4 into minus 3 times n, you can split the power. Split the power, right, so that's what I mean. So write this as, so that's around this power of 2, so that I have minus 2 to the 2 times, minus 2 to the n over 3 to the n, all right, that's what you were saying, I hope? Okay, so then that is, well, minus 2 to the n is 4, and it's 4 times every single term, so I can just factor the 4 out, just by playing around with this stuff, I can factor each term out. Now if you're more comfortable working with the initial thing, you could factor the 4 out of everything up there first. The first term is not 1, you factor the 4 out of everything, so if you refer, you could also write this as 4 times 1, minus 2 thirds, plus 4 ninths, minus 8, 27ths, plus 16 eighths, minus 1, right, just factoring the 4 out of this, I get that. It's the same thing, so now this is easy, so what is this sum? The formula's on the board, it's just a matter of noticing and plugging in, how many of you are planning to major in engineering? Okay, so you should be really good at looking at formula, you know, and manipulating them and turning them into specific applications, engineers do that a lot. So what is this sum? What do you do with engineers? Come on, 3 fifths, how do you do that? Yeah, okay, so that's 3 fifths, why is that wrong? Because it's times 4, so it's 12 fifths. It's close, the bridge fell down, but okay. It's close, it's both nice, it's what it fell down for. Okay, so these kinds of things, they're not, you know, so horrible, so it consists of a lot of looking at it, manipulating it, turning it into something that you know, going from there. Notice that we can also do something like, if I had, maybe, what if it doesn't start at 0? Suppose I have something like, let's just do the same one, n equals 4 to infinity of minus 2 thirds, well, let's do this, one fifth to the n. What can I do with this? So this starts at n equals 4, so this begins, I don't even know, 1 over 5 to the 4 plus 1 over 5 to the 5 plus, blah, blah, blah, blah. Okay, you could do that. So this is the same as n equals 0 to infinity of 1 over 5 to the n minus, this song, well, let me just write it, minus the first three terms, four terms, 1 plus a fifth plus 25 plus 125th, so you could do that, so that's, I guess I'll just, I guess I'll write it over here, is the same as 1 fifth to the n, so that's 1 over 1 minus a fifth, which is 5 fourths minus some junk that I don't want to add, so it's 5 fourths minus whatever that junk is. I do that right? Four fifths, 5 fourths, yeah, so it's that. Another way you could do it, which is a little easier for me to actually do the arithmetic, so what's another way? A here is the first term, so you plug in 4 for n, 1 fifth to the n. Wait, say that again? A is the first term, so you're very certain there's 4, so you plug in n to 4, 1 fifth to the 4 to the first term, and then that over 1 minus a. Okay, yeah, so in other words, factor out the first term. Right, so this is the same thing as 1 over 5 to the 4th times 1 plus a fifth plus fifth squared plus blah, blah, blah. So this is 1 over 5 to the 4th, now this is the standard geometric sum, this is 1 over 5 to the 4th times 1 over 1 minus a fifth, which is whatever the heck it is, 1 over 5 to the 4th times 5 fourths is whatever that is, 1 over 125 times a. So I don't want to spend the whole class doing this kind of nonsense, but you should get comfortable with manipulating these things and playing around. Okay, so that's one thing we did last time. Another thing we did last time was the harmonic series. This is a very important fact, like it should add up to something in fact it gets as large as you like. The way I show that is by grouping the terms and noticing that I can make them, I'm computing the partial sum and then noticing that I can make them as large as I want, they just get large slowly. I'll come back to this later. I'll prove it a different way in a minute, several minutes. So let's look at some other kinds of a series. So let's think about, when we're thinking of, I mean, let's look at an example like 3 plus 8. You can start anywhere, how about 0? Should this converge or diverge? And how can we tell? So he thinks it should converge. Yes, right. So it actually diverges. Why is it diverged? So for n really big, this looks like, in other words, so what this squishy statement for n really big is I just move over here, the limit is n goes to infinity of n squared over 5 n squared plus 8 is a fifth. And so when I start writing it first level, so when n is 0, that's easy, that's 0. And when n is 1, that's 13. And then there's a lot of stuff that I want to, and then I wait until about the millionth term and it's something, there's this 0.2. So 0.2 is some stuff, 0.2 is some stuff, 0.2 is some stuff and so on. And there's infinitely many of these things. You add up a lot of 0.2 as you get a very big number. So more generally, this tells us that if, well in this example, this is not a proof, but it's easy to prove, if it doesn't matter what we sum over, if we have an infinite sum like this converges, then for sure it has to be 0. Or say in another way, not 0, not 0, let's say there's a limit. If the bit of the term is not 0, then it's important to realize that it doesn't mean if the limit is 0, then for sure it converges. I can only say if it's not 0, then it diverges. Is this clear? Pretty early because she already knew it. Is anyone confused about this? Okay. So what we're saying is if the terms don't get small, when you add a lot of them, you get it big. If I start adding a lot of things, each of which is not really honey, then I get a lot of stuff. The only way that I can add an infinite amount of things together and have something finite is if they get small. That's what this says. Yeah. I don't care about the 8 when n is 1,000. So when n is 1,000, if I have 1,000 squared over 5,000 squared over 5 times 1,000 squared plus a little about 2,000, a lot of that's good. I mean it's 0.2, it's actually 0.19999999 out of the blue. Very close to a 5th. So, you know, when we're talking about 50 cents in the federal budget, nobody knows. Or even you go to the grocery store and you buy a candy bar, they don't even bother with the penny. This is a penny. This is $10. Or $100. Okay, so one fact here is that when we add infinitely many things together ultimately those things have to be tiny and they have to be as tiny as they want so we want it to add up to something. Let me turn to another example. Suppose I have something like this. So I mean part of what I'm doing right now is I'm just putting a bunch of examples up and I'm messing around with them and showing that we can do something. We'll turn to developing some theory precinct but suppose I have something like this. So this is, so when n is one I get a half. When n is two I get a fifth. When n is three, six, one over four plus two, you're right, a sixth. When n is three I get nine plus three is twelve. When n is four I get sixteen plus four is twenty. And so anything I can do here. Notice that it has a chance of converging because the limit as n goes to infinity of this is zero. So it has a chance. Maybe it converges, maybe it doesn't. How could we see that it converges? In fact I can see what it has to do in the hand of the one. Yeah. Well if I replaced everything with one sixth then I would get an infinite number. So if I replace everything with one sixth I would have one half plus as many sixths as I can. Well then I would show what a diver is. Is that what you're saying? So you're saying replace these with the sixth. That's pretty soon to fail. You can try it. It's not going to work. So I can tell you at the beginning that this converges. It is obvious that it does converge. It's obvious that you don't have to do this stuff. It's not obvious. It shouldn't be obvious yet except for your refugees from 127. Okay. So it's not obvious even what to do here. So a trick is we can use partial fractions. Remember those? So maybe we can do something by rewriting these in a clever way. So if we have that, so that means that A times N plus one plus B times N equals one. Right? Using partial fractions. B is for integration. And so this means that we have A plus B is zero because I have an N here and N here and no N here. And A is one. So B, right? A is one. So B is minus one. So that means one over N squared plus N is the same thing as one over N minus one over N plus one. And so now, I guess I'll go back over here. So using that, write this this way. So those are the same by algebra. So now if we start writing out some terms, the first term is one minus. And then the next term is a half minus a third. And then the next term is a third minus a quarter. And then the next term is a quarter minus a fifth. And so on. And notice what's happening. I take a half, I throw a little bit away. I take one, I throw a little bit away. Then I add that little bit back again and I throw away a different amount. Then I add that amount back and I throw away a different amount. And so on. So this is one. And if I group these together, I get nothing. And if I group these together, I get nothing. And if I group these together, I get nothing. So this is one. This is called a telescoping series. Or maybe telescopic, I don't know. And it should be like, you know, in the pirate movies when they take out their things. And then they close it up again. So this expanded little big thing, which you could just collapse. So this trick or technique works sometimes, but not all that often. Once in a while, it's going to be a handy thing. Now usually, usually actually finding the sum is not so important. Sometimes it is. Finding it exactly is not so important. What's important is knowing whether it does have one or not. So a lot of our focus won't necessarily be on when these things add up, what do they add up to? Because I can use a computer to do that. If I want to know what this adds up to, I can find it to a thousand decimal places by just running a computer for a little while. But if it doesn't add up, I'll get an nonsense answer. So the real question, a lot of the time, is not what is the sum, but is there one? So almost more important than what the sum is is should I even bother to look? Serious. Less important. Less important. It's also harder. Well, it's harder to get exact and not so hard to approximate. So this is a very important question. So a lot of times we'll just focus on does this converge? Yes? No. And another thing that's important is to be able to convince somebody that they're not just flipping a coin and saying, yeah, it converges. So you have to be able to justify your answer. So, see, this is already an important thing. If the limit of the nth term is not zero, then it doesn't converge. We're done. We already have one of these kinds of things. And let's work on developing a couple others. So, and let's work on it with a specific example. The quantum series, sum of 1 over n, does not converge. Let's look at what I started with. So it squares plus a quarter plus a ninth plus 16 and so on. Now, does this converge? It has a hope. It has a hope of converging because certainly the term is zero. It also has a hope of converging because it looks kind of like this one. Looks a lot like 1 over n squared plus n. So, and in fact, just using this path, that one is, well, it's not quite good enough. Right? So this looks like whatever this means, and I'll make this precise in a few minutes. Well, no, probably next week. I mean, if we just didn't have this stupid end there, then it would be good. We know this one is one. So we would expect that probably this converges. But expecting that probably it converges is not good enough. We have to know that it converges. But how do we maybe know that this converges? Notice that we can't say, well, every term here is less than one of these because actually these are small. This guy is still here. The first term there is one. The first term here is a half. The second term there is a quarter. This term is six. So I can't say, well, this one's bigger than that one's with flushes with n because actually that one's bigger than this one. But they're kind of related. The trick is directly comparing isn't going to work. So we have to try something else. So what do we do a lot in this class? What have we been doing all semester? Integrating. Maybe integrals will have something useful to tell us. So let's draw a picture. Well, let me draw a picture. You don't have to draw a picture. I'm going to draw a picture and I'm going to think about what that picture is telling me. Let's take a bunch of rectangles with one, one over n squared for the various values of n. So I'm going to draw it bigger than it is. So here is a rectangle of height from zero to one of height one. And then next to it is a rectangle which I'll put between one and two of height one over two squared. And then next to it I will put a little rectangle of height one-ninth. And then I will put a little rectangle of height one-sixteenth. Already my scale is off. But I have a bunch of rectangles that look like this. Why did I do this? I don't know. This is math class and we do weird stuff in math. What do we know about this area? What function is it related to? What have I done? This point, this point, this point, this point, this point, this point, all live on some curve. They live on the curve y equals one over x squared. This area here is certainly less than some integral. What integral? Somebody said it. Yes. It's less than the integral one over x squared from one to infinity. So it's less than, I guess I'm going to write it here, the integral, well you can't see as I write it, the integral from one to infinity of one over x squared dx. But we already dealt with those kinds of things. Well, we did it. Some people got it right. So we know that, I'll put it in quotes, we should know that the integral from one to infinity of one over x squared dx, this is the limit, as m goes to infinity of that integral, which is a number. So this is, why can't I do it? x is a minus two, so that's one over x negative from one to m limit. So that's, so we know for sure it's less than one. So we know then that this adds up and it adds up to something less than one. So now if we want to figure out what it adds up to, you can just take out your computer and start adding and you get a lower bar. Oh yeah, somewhere I screwed up, didn't I? Where did I screw up? Oh, because I did zero, no. What did I do wrong? Oh shoot, it's certainly bigger than one, isn't it? Yeah, it's a minus one over x squared, this is certainly less than, yes, stupid. What did I do wrong? I did wrong. Notice that the integral here, I started at one, so this rectangle wasn't included in the integral. The integral is this integral, but my sum has a one more. The integral only gives me this part. So we'll see that. Oh, it's less than two, thank you. It's close. But it certainly converts. In fact, if we want to get it bigger than something, you can put the boxes on the other side. But a terrible purpose technique, unlike, say, this telescoping something that only works if you're lucky and stuff cancels out for any integral. So let's see what we can do. So notice that this trick, this technique that I used here, didn't really use anything special about the specific series other than all the terms were positive. And I compared what the series is doing to what the integral is doing. So it doesn't really matter what the function looks like as long as all the terms are positive. So let's see if I can write that. So if the terms are always positive, and if also, well, let's see, and also my terms are some function, I'll rewrite this in a nicer way than that. Then if the integral, I want to go infinity and it doesn't really matter where I start. I'll start with one. If f of x dx converges, so does the sum. We can draw this picture. The sum is an approximation of the integral. In the picture, f of x is decreasing than the sum plus the signature. So the sum and the integral are closely related. It goes the other way, too. If the integral does not converge, neither will the sum. So let's look at that. So let me do the one that we know diverges. And then I'll say it again in a nicer way. The one that we know diverges. So we already know the sum of one over n diverges. We already did that by cleverly putting terms together. Let's compare that. We also know that the integral, well, I hope we know that this diverges when you do this integral, you get the log and the limit as x goes to infinity of the log is not zero. It flows up. So we did this before in class, I don't know, a month ago. We saw this diverge. You won't remember that. Let's relate these by picture again, but not the same picture. So here's the graph. y equals one over x. Let's put the points where my scale is off but I don't care. Here are the points. One half, one third, one fourth, and so on. Let's again put little rectangles on there. We don't have to start our little rectangles at zero. Let's just shift them over by one. Put them here. Now I'm putting the rectangles on the other side. So this area is bigger than the integral. But what is this area? Each of these I'm choosing to be one y. This is one plus this area is a half. This area is a third. This area is a fourth. And so on. That's my sum. So by thinking about it this way, we see that the sum of one over n is bigger than the integral of one over x. So let's put this on the other side. The integral pushes it up. And this is different. Now I didn't prove this but you can turn this into like a real mathematical proof. But let me just restate it the same way. This is something called the integral x. We just want to make sure I didn't leave out something. Write it a little more carefully. So f of x is a continuous decreasing function, decreasing function f of x. And we have f of n equals a of n. In other words, that's what our series is made up of. Then if the integral, it doesn't matter where you start from somewhere to infinity from f of x dx converges, so does the sum. If the integral of f of x dx diverges, so does the sum. Well, this is sort of the story that we've seen I see right now. This is the story that we've seen from the beginning. Integrals are just special sums. So if it's infinite, if it's improperly convergent, so does that. If it diverges, so does that. Okay, I'll see you my day. Have a nice weekend.