 I guess I should just pick up where we left off. So if you remember we developed a whole pile of tests for when infinite sums converge. I don't want to list them all here, I listed them all last time. We have, so I'll just say them, we have the integral test. We have a couple forms of comparison tests. We know about geometric series and T series. We have the regular comparison test, the limit comparison test. We know about alternating series, we have this alternating series plus, and we have the ratio test. There's a few other more sphere ones that we want to cover, which is fun. That gives us sort of an arsenal of techniques that when you have an infinite sum, you want to decide whether it converges or you can't do it. So now we sort of, and if you think about it, so say I wanted to know, so let's restart with something easy here. So we know this converges because it's a P series. What did I want to be here? Okay, but suppose that I multiply this by, say, series, we would want to actually converge. Let's start. And suppose that I wanted to look at instead something like, I don't know, 2 to the n over n squared. So what tests, what would I use to determine whether this converges or not? Sorry? The ratio test would maybe be useful. So we can try the ratio test. So that means I want to limit the limit as n goes to infinity of the next term over the current term. And if this is, so I'm really comparing this to a geometric series, so if this ratio of the next to the first one is less than one, it will converge faster than the geometric series with that ratio, so that will be good. So in this case, I want to look at the limit as n goes to infinity of 2 to the n plus 1, that's already this way, over n plus 1 square. And I want to divide that by 2 to the n over n squared. So that's a n plus 1 divided by a n. And then I do some algebra. So, well, let me just rewrite it in a step that I would usually write. The n squared goes to the top of that, so that's the same thing. So I can simplify a little more. So that equals, so this divides that leaving a 2 on the top. And then this, well, that's just the limit I can do. So this is the limit as n goes to infinity of 2 times n squared over n plus 1 square. And then as n goes to infinity, this goes to 1, so this is 2. So what does that tell me? It converges because 2 is bigger than 1. But if instead of the 2 I put a hash there, then this would converge. We could use comparison tests. This is less than 1 over 2 to the n. Or we could use the ratio test. Do I need to go through the ratio test? Or do people need to see the details of why this converges? Or is it clear to everyone? It's clear to her. Anyone else? Anyone else who needs me to work this out? She does. She does. Okay. So you could use ratio test. If you use the ratio test, it's exactly the same. The limit would be a half, which is less than 1. So I won't do the ratio test because it's all here. Just everywhere you see a 2, right, a half. Well, not everywhere because this 2 is the half. Or you could compare and compare 1 over 2 to the n n squared to 1 over n squared. This is less and it converges. So by comparison or whatever. So now if we think about this, what if I change this 1 half to, I don't know, a 3 quarters. Same business. It still converges. 3 quarters is less than 1 or that's less than that and so on. So I don't have to keep doing the same problem over and over and over again. I could just decide once and for all I could ask about this series instead of having a half there or a 2 or anything. I could just do it with an x. So this is just, so I can do this once and for all and say for what x, namely enough room, does this converge. So instead of talking about just one infinite sum, I can talk about lots. This is also a function. But let's save that for a minute. Let's just say for what x, so for some x values this converges for other ones it doesn't. Now the answer is for absolute x less than 1. Let's just go through the process here. So the ratio test would be useful here. So I'll just go through the work again. I use the ratio test and I want to look at the limit where n goes to not a, to the infinity of the next term or the current term. So that would be, here I have x to the n plus 1 divided by n plus 1 squared. I want the ratio so that's a n plus 1 and I want to divide it by x to the n divided by n squared. Absolute value. So now I invert and clean things up. So that gives me, I'll just write the intermediate step. And now I can cancel stuff like crazy. So I'll have a whole lot of things. Divide this into this and I'm left with an x. And then this limit would be 1, but I'll write that. So this divided by that gives me absolute x and then I have n squared over n plus 1 squared. And when I take the limit as n goes to infinity, this goes to 1. And I get absolute x as the ratio of the next term to the current term. And so this convert by the ratio test, this will converge if x is less than 1 in absolute value. It diverges. You can read about Tico Brachy. It diverges. This converges if this is less than 1. In fact, it converges absolutely because we looked at the series of absolute values. Because the ratio is less than 1. So this is faster than a geometric series. And it diverges if the absolute value of x is bigger than 1. And if the absolute value of x equals 1, we have to look more. We have no information when the absolute value of x is 1. So we have to try something else. But we got a lot of information. We learned about a whole pile of sums from this here. And so now we have to check what happens. So we have two cases. If x is plus 1, because when absolute value of x is 1, we have two choices, plus 1 and minus 1. So if x is plus 1, the series is the sum of 1 over n squared. And we know about that. If x is minus 1, the series is the sum of minus 1 to the n over n squared. Well, actually, this converges. And this converges too. Because by the alternating series test, or because this actually converges absolutely. So that forces this. So this converges by the alternating series. So the terms goes to 0. It's an alternating series. It's decreasing. So we're OK. Am I going through these tests too fast? Or people are mostly comfortable with them. So that tells us that, in fact, I don't need to have some value in here. The original series, x to the n over n squared, converges for x between minus 1 and 1 inclusive. So this is called the interval of convergence. So maybe I should write it as an interval. This is called the interval. Because we have a series that depends on a variable x. And we say, for x in this interval, it converges. And for x not in this interval, it diverges. OK? These intervals always work. Well, every interval has a center. Because an interval is from here to here. So there's always a midpoint. And the distance from the center to the edge of the interval is called the radius of convergence. So sometimes you encounter problems where somebody says, tell me the radius of convergence. So the radius is past the length of the interval. So here the radius of convergence is 1. OK. So why is this useful? Well, this is useful for two reasons. For one reason, it just allows us to talk about a whole class of infinite sums at once. But also, this is a function. Right? When we write, I'll just use this same example for a minute, x to the n over n squared. And let's say n becomes infinity. This is, but when n is 1, I get x over 1. Plus x squared over 4. Plus x cubed over 9. Plus x to the fourth over 16. This is like a polynomial with infinitely many terms. So this gives us a new kind of function. Well, maybe it's new. Maybe it isn't. But this gives us another kind of a thing that we haven't dealt with before. This could be some function of x. But this function of x only makes sense for certain ranges of x. This doesn't make sense if x is 3. It only makes sense when x is less than or equal to 1 in absolute value. Because this adds up to a number. We can evaluate the function for x is between plus and minus 1. But for bigger than plus or minus 1, it blows up. This is such a function. It's called the power series. So in general, the power series is something that looks like, well, it's something that looks like it has a first term. It has a second term. Maybe usually they will use a c. I'll use c. It looks like a polynomial with infinitely many terms. So it's not really a polynomial because polynomials stop somewhere. They have the highest power. But the power series, the terms go up to infinity. Now, sometimes these might be written like... So this is also a power series. Let me just write it. 2 plus 2x minus 1 over... Not over anything yet. Plus... Why did I put a 2 there? I don't know. I'll just leave the 2. 2x minus 1 squared over 2 plus 2x minus 1 cubed over 6 plus 2x minus 1 cubed over 4 factorial, blah, blah, blah. This is also a power series. The center, it's just shifted. If I expanded this, then all of these numbers would turn into these c's. I could take this x minus 1 cubed is x cubed minus 3x squared plus 3x minus 1 and I can put those x squared to this one and turn it into this one. But sometimes it comes up more naturally in a form like this. So in general, the power series looks something like some... And it starts somewhere. Usually it's 0, but not always. Some coefficient. And then x minus something to the nth power. So it's a series. It may be centered at some point, a, that a might be 0 by 9 and so on. These are both examples of the power series. And while this just seems like I'm playing a little game here, these functions are very useful. And they're very useful in approximation, but they're also just useful in ways to describe things in their own right. The issue is that the power series doesn't always converge. So it only has a small, well, it could be a big, domain of definition. But it has a specific domain of definition. It's not necessarily defined for all x's. Notice that it always converges at the center. The power series, if x equals a, then we only get the constant that corresponds to it. Everything has to be 0. In this case, when x is 0, this converges to 5. So the power series always converges at at least one point, the center. And then maybe it has some... So we have sort of several cases. So it converges at x equals a always. And then it converges on some interval around this number a. This interval could be infinite. So then r is infinity. So r here is called erroneous convergence. And maybe, at these n's, it might not converge. I can give you examples where it does and examples where it doesn't. Perhaps these points are included, perhaps not. So let me work with a few examples. So you did one, not a very interesting example. Suppose I have something like... Say I have something like this. This can start at 0. And I want to determine for what axis... Should I start with an easier one? Or is this okay? So I want to know for what axis does this converge? So sometimes this is phrased as determined convergence. So almost always in power series, unless you recognize it as one that you already know. And what we'll be doing for the next week or two, with a little interlude in the middle for the exam, is looking at functions as power series. And dealing with them and seeing that in power series we represent known functions that way but also given functions that way. So here we want to know when does this function here make sense? For what axis does this make sense? So that means we treat this as an infinite sum. We check when it converges. So almost always the first thing that we want to do is do the ratio test because that gives us a lot of information without a whole lot of work. So we look at the limit of infinity of the next term by the nth term. Take the absolute value. We want to know when that ratio is less than 1. In this case, just to save a little space I'm actually going to do 1 over a n because it's easy. So a n plus 1 is what happens when I put an n plus 1 there and then 1 over a n I just flip it and that's the limit that I want to do. And so n factorial n plus 1 factorial is n plus 1 times n factorial. So this will cancel with that leaving an n on the bottom. Yeah. Oh, I changed the plus 2 minus. Yeah. Thank you for fixing your sign. Okay. And so now I can reduce this a little bit and I'll have a 2x minus 1. This one will cancel all but one of those. And then this will cancel all but one of those. So I have on the top 2x minus 1 and on the bottom I have n plus 1. Remember it's n that's going to infinity. So this one's easy. What's this limit? Zero. No matter what x is, this goes to zero. So I don't have to do any more work because for any x this limit is zero. That tells me so this ratio, this is kind of a stupid example but it's easy. This ratio is zero x can so for all x the ratio is less than one. So that tells me that this converges for every x. So the interval of convergence uh leaving a throne we put it here. The interval of convergence is all x. So this is the whole real y. Minus infinity plus infinity. So this is zero. So that one was an easy one. Let's tweak it a little bit and make it a little harder. So suppose that instead of having that I want to have let's put the why did I lose it? Let's put a power there. So let's put like a 5 to the n on the top. We'll take the same thing and tweak it a little bit. So let me point out a lot of times power series involve factorials because they arrive from taking derivatives. But we'll see that later. So let's take almost the same series and just multiply it by 5 to the n. That won't help, will it? Uh and the fifth is the same. What happened to my example that was interesting? Sorry. Let's see here. So what happens with that one? Let me not make it n factorial. Let me just make it n squared. So now what will happen to this? Well, I don't know. We have to check, right? So we do the same business. So I do the ratio test and so when I take n plus 1 I get 5 to the n plus 1 2x minus 1 to the n plus 1 divided by n plus 1 square. And 1 over a n is that guy flipped so I get an n squared on the top and a 5 to the n 2x plus 1 to the n bottom. And so now I want to reduce here and so this 5 to the n will divide into this 5 to the n plus 1 leaving me a 5 on the top. This 2x plus 1 to the n will divide into that 2x plus 1 to the n plus 1 leaving me a 2x plus 1 on the top minus, well, I keep wanting it to be plus. Should we just change it to plus? We want it to be minus. And then I'm left with n squared over a plus 1. And now I take the limit the 5 doesn't go anywhere the 2x minus 1 doesn't go anywhere the limit is n goes to infinity of n squared over a plus 1 squared is 1. So this is 5 times 2x minus 1 absolute value and we want this so this is just some number and I want to know when will this ratio be less than 1. If the ratio is less than 1 it diverges. If the ratio is bigger than 1 it diverges and if it equals 1 I have to work a little harder. So we just do some arithmetic here multiply both sides by 5 get rid of the absolute value so the absolute value and add 1 to both sides so I add 1 to both sides I get 4 fifths and 2x and 1 and 1 fifth is 6 fifths and then I divide by 2 x is between 2 fifths and 3 fifths So let me finish it so this tells me part of the question if x is between 2 fifths and 3 fifths provided I can screw up this series converges absolutely if it is bigger than 3 fifths or smaller than 2 fifths the series diverges and we still have the question what effects is 2 fifths or what effects is 3 fifths but we still have to look at the cases the end points when x equals 2 fifths and when x equals 3 fifths if we use the ratio test on those the ratio is 1 and it gives us no information we still have to handle those two end points cases so pick one of them if x is 2 fifths then well 2x minus 1 is well let me just do it so that's 4 fifths minus 1 over n square 4 fifths minus 1 is minus 1 fifth 5 to the n times minus the fifth to the n is minus 1 to the n and this is a converging series this is an alternating series so this is an alternating series and it converges since we let x equal 3 fifths then the series becomes a plus one on the top and this also converges these series so that means that my interval of the convergence includes both ends 2 fifths 3 fifths that tells us what happens when x is 2 fifths and what happens when x is 3 fifths so let me just remark let me use this problem a little bit so I won't do it, I'll just say it if this is an n instead of an n square what will be different? we'll get the same interval because you can just look everything's going to cancel nicely but here I will have an n which is minus 1 to the n over n that's still converging but here I will have a 1 over n which diverges so if that's an n there instead of an n squared the interval of convergence will be 2 fifths included and 3 fifths not included so let me just I'll write it down so that's what I was meaning to do but I copied wrong it doesn't matter so 3 fifths does not work it diverges for x equals 3 fifths if I change it from an n square root n you can stand out of the way just think about this calculation here changing this n squared to an n to the 1 half what changes? does anybody have a clue what I've been doing for the last 40 minutes? it converges? why? okay so let's first look at this sign suppose I change this n squared to a square root so this is square root n here square root n plus 1 here everything else is the same this is a square root n here square root n plus 1 here everything else is the same when I take the limit this square root n over square root n plus 1 is going to be still one so I still get this so I get the same interval so this calculation is identical so I still have x between 3 fifths and 2 fifths the question is what happens when x is 3 fifths or when x is 2 fifths so when x is 3 fifths let's look at this calculation this is a square root same business square root what happens to x is 1 over n over square root n so I hear diverges anybody votes for converge that's the other choice they vote for converge they're right it's an alternating series as n goes to infinity the limit is 0 so it converges but in this case if we change this to the square root we have 1 over square root n so it will not include 3 fifths but it will include 2 fifths let me point out one other little modification that we can make here some people get a little lost in this part because it's annoying you have these inequalities yeah why do you think it's not it's a new problem that's exactly the same as the old problem it's just like you learn what the derivative of x squared is and now we learn what the derivative of x cubed is and then we learn what the derivative of x to the 23rd is they're all the same problem but they look different until you realize they're not so I'm just trying to do 3 different problems all at once so that you can see what the variations are without spending 15 minutes going through the same problem which is the same so I'm just trying to save a little time to show you more examples of variations okay I can do another problem I mean that's fine let me do another problem that is not this one but again has this this funny business here all the same yeah let's try it let me make it 2 more so let me do another problem that is slightly different with time and I'm going to do this one a slightly different way make it a little more complicated I had a 2 there how about a 3x minus 2 2n over I put a minus 1 okay so there's just some random collection of junk and again in these power series the technique is always you use the ratio test first and just because it's annoying to write 3x minus 2 or 3x minus 2 all the time I'm going to let you be 3x minus 2 and make my life a little easier I'm just going to make a substitution so that I don't have to worry about this mess and so by the ratio test the limit is n goes to infinity a n plus 1 times 1 over a n that's the value which is so here I have absolute value so since it's an absolute value minus 1 to the n do you need me to write it yeah minus 1 to the n plus 1 n plus 1 this is now u so I don't have to write it u to the 2n plus 2 square root n plus 2 because I'm increasing the index by 1 times this guy flips square root n plus 1 over minus minus 1 to the n n u to the 2n so all I did was write down what the ratio is now when I take the absolute so I divide this into this I get a minus 1 I'm taking the absolute value so I can forget about the minus 1 entirely if you like you can write this as 2n plus 1 it's just 2n plus 2 however you want to write it it's fine so I'm simplifying that this guy and this guy leading to minus 1 absolute value kills it this guy over this guy I will just write it down plus 1 so I take care of that this guy over this guy I will just write it down n plus 1 over square root of n plus 2 this guy over this guy so it's u to the 2n plus 2 divided by u to the 2n gives me a u squared so doing all of my simplifications a lot of stuff cancels and now I can take the limit and as I take the limit for n large this is 1 and this is u squared so I want to know remember u is 3x minus 2 I just used it so I didn't have to write it that much but let's still continue so I want to know for what u is this good so when is well it's the absolute value which is 2 to the square I suppose I could have let m be 3x minus 2 squared that would have been easier so for what u is u squared less than 1 because that's when the ratio test converges well that's when u is between plus and minus 1 but u is 3x minus 2 so the center is 3x minus 2 so that's the same as 3x minus 2 between plus and minus 1 and now I go through the same nonsense again I add 2 to both sides and I divide by 3 and so x is between 1 third and 1 now I still have to I still have to see what happens when x is 1 third and what happens when x is 1 when u is 1 and u is minus 1 so for u equals plus 1 my series becomes where is it minus 1 to the n n 1 to the 2n power over square root n plus 1 it diverges because the limit is minus 1 to the n n goes to infinity so this diverges the limit of a n is not 0 I did something wrong didn't I no I guess not and for u equals minus 1 I get the same nonsense I get minus 1 to the n n to the n over square root of n plus 1 which also diverges so that means that my interval of convergence is just x between 1 third and 1 and does not include 1 third and 1 okay so this material is on the exam you know slowly it's on the web assign that's due Wednesday do a little more with it on Monday and I will review