in this video we're going to talk about power series and let me give you an example of a power series here's one that starts from zero goes to infinity and a power series is basically an infinite series with the variable x in it and x is raised to the n power so this is a power series that is centered at zero so c is equal to zero for this one and here's another power series but this one is not centered at zero in this case the center is c now the first thing you need to be able to do is determine where it's centered at so let me give you some examples so for each of these examples determine where each power series is centered at feel free to pause the video and try now let's consider the first one on the upper left where is the power series sent to that so just by looking at it you could see that it's centered at c equals two now what about the second one on the bottom left another way you could find the answer is by setting the inside in this case x plus four equal to zero and solve for x so at negative four this whole thing becomes zero so it's centered at c equals negative four now for this one if you don't see a c value if it's just x to the n then it's centered at zero you can write this as x minus zero to the n and that's the same as x to the n now what about the last example if you don't see what it is immediately set the inside part 3x minus 2 equal to 0 and solve for x and so you should get x is 2 over 3 and so it's centered at that value now the next thing that we need to be able to do is we need to determine the radius of convergence and the interval of convergence of a power series and so you need to use the ratio test to do that so let's say if you use a ratio test you take the limit as n goes to infinity of u sub n plus 1 divided by u sub n now let's say if you get zero what does this mean these three scenarios you need to be familiar with this is the first one so if the ratio test gives you zero then it means that the series it converges for all x values it's always convergent so the radius of convergence that's r that's equal to infinity and the interval of convergence it's from negative infinity to infinity so what this means is that at any x value the power series will converge now the second situation is after doing the ratio tests let's say you don't get zero let's say this time you get infinity instead of zero so what does that mean now if you recall in order for a series to converge the ratio test have to give you an answer that's less than one so if it's greater than one or if it's infinity that means that it diverges for all x values except one particular x value it will only converge when x is equal to c at this point you're going to get a limit with some expression of n times zero so the whole limit is equal to zero and because according to the ratio test is less than one it's going to converge for that point but for everywhere else for all other x values that is not equal to c you're going to get infinity and so the series will diverge for all other x values but for this particular x value where x is equal to c it's going to converge and because it only converges at one finite number the radius of convergence is zero and the interval of convergence is only one number it's not a range of numbers so the interval of convergence is just the x equals c value or c itself so let me separate the first situation from the second let's move on to the third situation so for the third scenario if we use the ratio test we're going to get an expression in terms of x typically it's going to look something like this 1 over r absolute value x minus c which is less than 1. so it's going to vary but c could be four c could be zero so you may just see an absolute value of x r could be anything it could be one half it could be two but usually in this format and once you get to this point you wanna multiply both sides by r and keep in mind you want to set it less than one because you'll get this but you need to introduce this inequality because it only converges when a ratio test gives you a value less than one so this is something that you have to add the less than one part and then once you multiply both sides by r for this expression you're gonna get x minus c is less than r so whatever number you see at this point that is your radius of convergence now because of space i'm going to clear the board so let's start with this so this is where the series converges now whenever the absolute value of x minus c is greater than r then at that point the series it diverges but since we're looking for the radius of convergence we're not going to use this so we're just going to be focusing on this expression now to get rid of the absolute value symbol we could say that x minus c is less than r but greater than negative r now if you add c to both sides you'll get that x is between negative r plus c and r plus c so the interval of convergence becomes this it's negative r plus c comma r plus c now the last thing you need to do is check the endpoints because even though you have parentheses here sometimes you could have two brackets two parentheses a bracket and a parenthesis or parenthesis in a bracket so there's four possibilities here and you have to check your endpoints which i need to show you by example but that's how you could find the interval of convergence so now you know the three scenarios so let's work on some practice problems so let's say if we have the power series from zero to infinity of x to the n over n factorial go ahead and determine the radius and the interval of convergence for this power series so let's start with the ratio test the limit as n goes to infinity of u sub n plus 1 divided by u sub n and we need to see where it's less than one where the series converges so u sub n plus one all you gotta do is in this expression replace n with n plus one so it's x to the n plus one divided by n plus one factorial and you said that that's just the original expression x to the n over n factorial so let's take the limit as n goes to infinity of u sub n plus one so that's this and then divided by u sub n so let's simplify the expression x to the n plus one we could say that's x to the n times x to the first power n plus 1 factorial is n plus 1 times n factorial and then using the expression keep change flip we're going to change division to multiplication and we're going to flip the second fraction so it's n factorial over x to the n so we can cancel an n factorial and we can also cancel x to the n and so we're going to be left over with so just i'm continuing from here just keep that in mind i'm not saying this is equal to this here so this is equal to the limit as n goes to infinity and then we have absolute value 1 over n plus 1 times the absolute value of x so basically separated x over n plus one into those two parts now as n goes to infinity what happens to one over n plus one that's going to go to zero and zero times x is zero now what do we say will happen if we get a limit of zero if the limit is equal to zero which is always less than one then that means that the series it converges for all x values regardless of what x is x is a fixed number by the way so if you choose x to be 2 or 8 or 25 or negative 4 all of those numbers all of those fixed values multiplied by zero is equal to zero so for any x value that you choose the series will converge and so what does this mean so for this situation i believe we defined it to be what was a scenario number one i think that was scenario number one and so when a limit equals zero the radius of convergence is infinity and the interval of convergence is negative infinity to infinity because x could be anything and the series will converge x can be a thousand it will still converge and so this is the answer anytime you get a limit that's equal to zero that's all you need to write now let's work on another example this time we're going to have the series of n factorial times x to the n feel free to pause the video if you want to try so let's start with the ratio test now u to the n plus one that's going to be n plus one factorial times x to the n plus one and then we're going to divide it by u to the n which is the original expression that we see here so that's u to the n so now let's simplify so now we have the limit as n goes to infinity and then n plus 1 factorial that's n plus 1 times n factorial and x to the n plus 1 that's x to the n times x to the first power divided by these two so now we can cancel x to the n and we can cancel n factorial and then i'm going to separate n plus 1 and x so what i now have is that the limit as n goes to infinity of n plus one my absolute value doesn't need to be so big and then times the absolute value of x so you can move this to the front because it's not affected by n so you can say that this is the absolute value of x times the limit as n goes to infinity of n plus one and n is always positive so we really don't need the absolute value symbol there but as n goes to infinity what happens to n plus one n plus one also goes to infinity now what does this mean the absolute value of x times infinity that means that the limit goes to infinity and what do we know when the limit goes to infinity this is the second scenario that means that the radius of convergence is zero and the interval of convergence is our c value so what is c so at what number is the function centered at there's no x minus c to the n it's x minus 0 to the n which is x to the m so we can say that is centered at zero so when x is zero it's going to converge now let's understand why so if x is zero we're going to have the limit as n approaches infinity n plus one times zero n plus one times zero is zero so then the whole limit will equal zero if x is zero now if x is anything else but zero let's say if x is a half one half times the limit as n goes to infinity of n plus one well this is going to go to infinity and if you multiply that by a half then the whole thing is going to go it's still going to be infinity half of infinity is infinity so if x is any other value then 0 the limit will go to infinity and so it's going to diverge but when x is zero then the limit goes to zero and by the ratio test that's less than one it converges so anytime you do the ratio test if you get infinity as your output that means that it's only going to converge when x is equal to c in this case your c value is zero so it converges when x is equal to zero because the limit will go to zero and so your answer for this situation is the radius of convergence will be zero and your interval of convergence is your c value it's when x is equal to c so in this case your interval of convergence is simply zero because c is zero now let's try another similar but different example so consider the series from one to infinity of n factorial times two x minus one raised to the n factorial so go ahead and determine the radius of convergence and the interval of convergence so as always we're going to start with the ratio test so u to the n plus 1 that's going to be n plus 1 factorial times 2x minus 1 to the n plus 1 factorial divided by n factorial times 2x minus 1 to the n so that's our use of n so this is u sub n plus 1 and this is the original u sub n expression that we see here so go ahead and simplify it so n plus one factorial we can write that as n plus one times n factorial and two x minus one to the n plus one that's two x minus one to the n times 2x minus 1 to the first power and so we could cancel 2x minus 1 to the n and n factorial and so now what we have left over is the limit as n goes to infinity and then we have n plus one and also two x minus one so what happens when n goes to infinity n plus one also goes to infinity so we're gonna have infinity times two x minus one now where is the power series centered at what is our c value to find our c value we need to set the inside equal to zero so if you set two x minus one equal to zero x is one half and so c is one half now let's think about what we have if x is anything but one half it's going to diverge let's say if x is five if we plug it in here two times five minus one that's nine nine times infinity is infinity so whenever x let's say if x does not equal one half infinity times whatever that value is is going to be infinity and so the series is going to diverge whenever x is anything but one half but what happens when x is a half well if you plug in one half into this expression we know we're going to get zero two times a half is one minus one is zero and so zero times n plus one is zero thus the whole limit is going to go to zero which means that the power series converges so anytime you get infinity as a result it's only going to converge when x is equal to c that is where the power series is centered at so for this problem the radius of convergence is zero because it's only one number where this series converges one x value so the interval of convergence is x equals c which is just one half so it's not always zero it depends on where the power series is centered at so this is the answer for this problem so anytime you get infinity the radius of convergence is zero and the interval of convergence is where the power series is centered at that's where x equals c so you just put in your c value now let's work on another example problem so let's say we have the power series from 0 to infinity of x raised to the 2n divided by 2n factorial so go ahead and work on this problem as always let's start with the ratio test so u sub n plus 1 that's going to be x to the two times n plus one so replace n with m plus one and then two n factorial is going to become two times n plus one factorial and then we're going to divide it by u sub n so u sub n is just x to the 2n over 2 n factorial so right now what we have is x to the two n plus two and on the bottom we also have two n plus 2 factorial and then we're going to multiply by the reciprocal of the other fraction so this is going to be 2n factorial over x to the 2n now x to the two n plus two we can separate that into x to the two n times x squared two n plus two factorial so let's see if you have eight factorial that's eight times seven times six and if you want to you can stop at five factorial so you have to keep subtracting one to get the next number so if we start with the first number two n plus two to get to the next number we need to take two n plus two and subtract it by one so it's going to be two n plus one and then if we subtract two n plus one by one then we'll get two n we wanna stop here because we can cancel it with the two n factorial on top so now we can cancel x to the 2n and 2n factorial so what we now have is the limit as n goes to infinity and then we have since n is positive i don't need an absolute value expression anymore so this is going to be 2n plus 2 times 2n minus 1 and then times x squared which is also always positive so what happens to this expression when n becomes very large when it goes to infinity as n goes to infinity that fraction goes to zero and so it's gonna be zero times x squared which is zero for all x so regardless of what x is if it's 20 1000 negative 50 if you multiply by zero it's going to be zero so this means that the series converges for all x values so anytime you get an answer of zero for the ratio test that means that the radius of convergence is going to be infinity and the interval of convergence is negative infinity to infinity and so that's it for this problem now let's try another one so consider the power series the square root n times x minus 1 raised to the n go ahead and try that one so once again let's start with the ratio test as always u to the n plus 1 that's going to be the square root of n plus one times x minus one raised to the n plus one divided by the original expression u sub n so this is equal to the limit as n goes to infinity now both of these expressions they're inside the square root so i can write it as one expression the square root of n plus 1 over n now x minus 1 to the n plus 1 that's going to be x minus 1 to the n power times x minus 1 to the first power and so i can cancel these two so what i have left over is the limit as n goes to infinity square root n plus 1 over the square root of n or just i'm going to write like this and then times x minus 1. so what happens to n plus 1 over n when n goes to infinity when n becomes very large the one becomes insignificant so you're gonna get one n over one n and so it's going to become one and the square root of one is one so we're gonna have one times x minus one now this is all still inside the absolute value expression so we have the absolute value of x minus one and whenever you have let's say an answer in terms of x after the ratio test set that answer less than one because you want to find out for what values of x it converges so we can rewrite this expression we can get rid of the absolute value expression by saying that x minus one is less than one but greater than negative one so now if we add one to all three sides we can see that x is less than one plus one which is two but greater than negative one plus one which is zero so therefore the interval of convergence is from zero to two or at least it appears that way we still need to check the endpoints so we can adjust it shortly now what is the radius of convergence if you recall once you have this form if there's no number in front of the absolute value then whatever number you see here is the radius of convergence in this case it's one and you could see that we have x minus one or x minus c so c is one for this example it's centered at one and you can see it here too that's x and minus c now let's check the end points so let's start with zero we're gonna see if it converges when x is equal to zero so starting with the power series replace x with zero so we're gonna have the square root of n times zero minus one to the n power or negative one to the n power so does the series converge in this form what would you say well a simple way is to try the divergence test the limit as n goes to infinity for the square root of n times negative 1 to the n power as n goes to infinity the square root of n what's going to happen to that that's going to be the square root of infinity which is basically infinity and then the only difference is it's going to alternate in sign it's going to switch from negative 1 to 1 so our answer is going to be positive infinity negative infinity and so forth it's always going to alternate so we could say that the limit doesn't exist because it doesn't equal one value sometimes it will equal positive infinity sometimes it will equal negative infinity now if the limit doesn't exist then we say that according to the divergence test the limit diverges so if it diverges then it doesn't converge when x is equal to zero so we're not going to include zero as an endpoint so we're going to leave it with parentheses it doesn't include zero we're not going to use brackets if it converged then we would use a bracket at that endpoint so now let's test the other endpoint when x is two so we're going to have the square root of n times 2 minus 1 to the n power which is 1 to the n and 1 to the n is just 1 so this is what we have now using the divergence test this is going to go to infinity so it still diverges so 2 should not be included in the interval of convergence now let's plot the interval of convergence for this problem so it starts at zero and it goes to two now it doesn't include zero or two so we're gonna have an open circle at those points so anywhere in this region where x is between 0 and 2 the power series will converge now the power series is centered at one so this is the c value and the radius of convergence is basically the distance from the center to the end of the interval so this distance is r and so we can see why r is one because between zero and one that distance is approximately one so r is typically one half of the length of the interval so if it goes from zero to two then if you take the difference between zero and two which is two divided by two that's your radius of convergence and so that's it for this problem hopefully this gives you a better understanding of it let's try this power series let's say it's x minus 3 to the n divided by n so let's start with the ratio test so u sub n plus 1 that's going to be x minus 3 raised to the n plus 1 divided by n plus 1 and then we're going to divide that by u sub n which is x minus 3 to the n over n so then we can write x minus 3 to the n plus 1 as x minus 3 to the n times x minus 3 to the first power and n plus one that's not going to change we're just going to leave that as n plus one now we're going to change division to multiplication and then flip the second fraction so we have this the only thing we can cancel right now is x minus 3 to the n power and so we're left with the limit as n goes to infinity of the absolute value or we could just say 1 over n plus 1 times the absolute value of x minus 3. so as n goes to infinity actually wait i'm forgetting something i can't forget this n so let's put that on top that would have been a very big mistake so as n goes to infinity what happens to n over n plus one the 1 is insignificant so it becomes n over n which becomes 1. so this is 1 times the absolute value of x minus 3. and so that's less than 1. so we can say that x minus 3 is between 1 and negative 1. now if we add 3 to both sides then x is between 4 and 2. and the midpoint of two and four is three so we can see that it's centered athering so c is three now what is the radius of convergence what is that equal to well if you take the distance between the midpoint of the interval which is three and one of the endpoints the difference between three and two is one so the radius of convergence is one or once we had it in this form the absolute value of x minus three is less than one we have it in the form of x minus c the absolute value of that is less than r so you can see at this point r is one so now we need to check the endpoints so let's start with two so when x is two will the power series converge or will it diverge so 2 minus 3 that's negative 1. so we're going to have negative 1 to the n power over n so notice that this is the alternating harmonic series so we need to employ the alternating series test ast so let's start with the divergence test so we're going to take the limit as n goes to infinity of a sub n a sub n is everything except the negative 1 to the n power so we'll just focus on one over n and that goes to zero so it passes the divergence test this tells us that it may converge or it may diverge now for the alternating series test the second thing we need to show is that a sub n plus 1 is less than or equal to a sub n so in this case for the alternating series test a sub n is one over n so just keep that in mind it's you take out the negative one to the n power now a sub n plus one that's one over n plus one one over n is always greater than one over n plus one so the two conditions of the alternating series test has been met so we could say that the series converges by the alternate series tests so we can say that x is equal to or greater than 2 so 2 is included now we need to see if 4 is going to be included in the interval of convergence now let's substitute x with four so we're gonna get the series four minus three to the n which is one to the n and one to the n is just one so we're gonna have 1 over n so this is the divergent harmonic series and based on a p-series test you can see that p is equal to 1. so if p is equal to 1 it's going to diverge in order for the p series to converge p has to be greater than one then it will converge if it's less than or equal to one it diverges so since p is exactly one this particular series diverges so 4 is not included so we can write the interval of convergence like this so it's going to be a bracket at 2 and a parenthesis at 4. so plotting it on a number line we have the center at c and the endpoints are 2 and 4. so we're going to have a closed circle at two and an open circle at four and here is our c value the power series is centered at three and so this is the radius of convergence which is one and so that's it for this problem so anywhere between two and four if you plug it in the series will converge it doesn't converge at 4 but it converges at 2 and anywhere between 2 and up to 4 but not including 4. so let's work on one more example problem consider the power series from 1 to infinity of negative 1 raised to the n plus 1 times x minus 4 raised to the n divided by n times 9 raised to the n so go ahead and determine the radius and the interval of convergence for this power series so let's start with the ratio test u sub n plus one that's gonna be negative one to the n plus one plus one so everywhere you see an n replace it with n plus one and then it's going to be divided by u sub n the original expression negative 1 to the n plus 1. i'm going to separate it like this negative 1 to the n plus 1. well this is m plus 2 really i'm going to separate it as negative 1 to the n plus 1 times negative 1 to the first power and then x minus 4 to the n plus 1. i'm going to write that as x minus 4 to the n times x minus 4 to the first power n plus one i can't really change that i'm just gonna leave it the way it is nine to the n plus one i could write that as nine to the n times nine to the one then change division to multiplication and then flip the second fraction okay so now let's simplify what we have these two we can cancel we can cancel x minus four to the n and also nine to the n power now let's write these two together so this is the limit as n goes to infinity n over n plus one and then we have absolute value negative one times x minus four now as n goes to infinity or let's not forget about the nine on the bottom so let's put this over nine as n goes to infinity what happens to n over n plus one this becomes one a thousand over a thousand and one that's about one and then the absolute value of negative one over nine we can take that out of the absolute value expression and write it as one over nine so what we have now is one over nine times the absolute value of x minus four and this is going to converge when the ratio test is less than one so now what we need to do is multiply both sides by nine and so these will cancel and so now we have x minus four is less than nine so notice that this is the form x minus c is less than r so therefore we could see that r is 9 that's the radius of convergence we could also see that c the series is centered at four so c is four so now let's determine the interval of convergence so we can say that x minus four is less than nine but greater than negative nine now let's add four to both sides nine plus four is thirteen negative nine plus four is negative five so this is what we now have now let's rewrite the original series which is negative 1 to the n plus 1 times x minus 4 to the n divided by n times nine to the n so just by looking at it you could see that the power series is centered at c equals four because of the x minus four in it now let's check the end points so if we plug in negative five will it converge or diverge so let's replace x with negative five so we're going to get the series negative minus 4 that's negative 9 to the n so we're going to have negative 1 to the n plus 1 and then negative 9 raised to the n over n times 9 to the n now negative 9 to the n divided by 9 to the n we can write that as negative 9 over 9 to the n which becomes negative 1 to the n and so if we multiply negative one to the n with negative one to the n plus one then that becomes negative one to the two n plus one so this gives us the series negative one to the two n plus one divided by n so what can we do with negative one to the two n plus one so when n is one two times one plus one that's going to be three and negative one to the third power that's negative one so we're gonna have negative one over one now the next term when n is two two times two plus one is five negative one to the fifth power is still negative one and then when n is three two times three plus one is seven negative one to the seven is still negative 1. so as you can see we don't have alternating signs so we can reduce the series to this we could say that it's simply negative 1 to the n and we can move the negative to the front so this becomes negative and then we have the divergent harmonic series so according to the p series p is one and because p is one we could say that the series is divergent so negative five is not included so we're going to have a parenthesis at negative 5. so the interval of convergence is negative 5 to 13. now let's check the endpoint 413. so when x is 13 what's going to happen so this is going to be 13 minus 4 which is 9. so we're going to have negative 1 to the n plus 1 one times nine to the n over n times nine to the n so these will simply cancel and now we have the alternating harmonic series we need to perform the alternating series test so let's start with the divergence test so the limit as n goes to infinity for a sub n which is 1 over n that's going to go to zero remember to ignore this so it passes the divergence test at this point the series may converge or it may diverge if it doesn't equal zero then it diverges now we need to show that a sub n plus one is less than or equal to a sub n so a sub n is one over n and a sub n plus one is one over n plus one so that's true so by the alternating series test we say that the series converges which means that 13 is included so the interval of convergence includes 13. so we're going to use a bracket for that now let's plot the solution on a number line so it's going to start from negative 5 and it's going to end at 13 and it's centered at 4. so 4 is right in the middle so we're going to have an open circle at negative 5 and a closed circle at 13. so this is the interval of convergence whenever x is between negative 5 and 13 the power series will converge and so in the middle we have our c value it's centered at x equals 4 and the radius of convergence is nine so 13 minus four is nine and four minus negative five that's also nine and so that's it for this video hopefully it gave you a good understanding of power series and how to find the interval of convergence and the radius of convergence thanks for watching