okay so we are in section 5 of chapter 9 we're still asking the same types of questions we're asking about some particular series now you have to understand what I mean when I say series or other word otherwise these lectures just are not clicking your that's just you're a step behind at every point remember a sequence you have to go back to 9 one is a let's just make it very simple a list a set of numbers ok that is described by some function now a series is going to be the sum of that sequence where you're starting to add things up it just so happens that we are interested in infinite series and so if you're going to add up an infinite amount of numbers you have to understand that you don't have an infinite amount of time to do so and so you need to start we we need other ways of arriving at these results so the next test that we're going to discuss is what we call the alternating series test and I want to stay here for just a second what we mean by alternating so then if not all the majority of all the sequences that you have seen up to this point have all been positive numbers right so like 1 than 1/2 then 1/4 etc all positive numbers what you're now going to see is when we say alternating meaning the sign is going to be alternating between negatives and positives or positives and negatives and there's there's two types of alternating series that you're going to see they're the only two types that you will see are either the odd terms are going to be the negative ones or the even terms are going to be the negative ones you're not going to see any type of you know negative negative positive negative you're not gonna see any it's always going to go back and forth between negative and positive negative and positive and if you have a second take a look at what I'm bracketing right now this negative one to the power of n or negative one to the power of n plus one this is what is making your your terms of your series right or of your sequences have alternate signs okay it's just so you basically if you have your general your general term your in term it's either going to be attached to positive 1 or negative 1 or positive 1 or negative 1 then that's why you're seeing the signs alternate back and forth now the way this test works it's very simple if the following two conditions are met you just get to say converges okay so what we're gonna look at is so we have two different series we have a sub N and then we have the following so we have negative 1 to the power of n negative 1 to the power of n plus 1 it but it's very simple basically the way this works is you're going to test if the limit of a sub n as n approaches infinity is equal to 0 if that may if that's the case check that's the first condition the second condition is basically stating that and I'll write this for you so for the first condition it's very it's fairly simple is as n goes to infinity a sub n is equal to 0 ok you take the limit and make sure the limit is equal to 0 now the second condition basically says that for for all if the a sub n plus one is smaller right or equal to then a sub n meaning every every a sub n plus one I mean the the the next term is always going to be smaller than the previous term for all in small it's smaller than or equal to but but it's so the the following term we'll always be smaller than previous term okay now to get around this negative because we know it's going to be alternating negative positive negative positive I'm gonna I'm gonna give you kind of a way of shortcutting this so you don't have it nothing nothing is so it won't be so cluttered so the best thing to do right now is to jump right into a question so notice it says determine the convergence or divergence of the following series okay so listen this part up at the top where it says I'll use the alternating series test that would be very kind of your professor if he or she were to put those into your directions but the question stands as follows right there's nothing there's you're not guaranteed that your professor is going to put use this test on this series so we at first we have to be able to recognize that there's something in here that's making your terms alternating alternate between negatives positives positive negatives etc and really what you're looking for is this negative one to the power of n plus one okay now what I'm showing you what I'm showing you here I'm gonna see if I can box this in a little bit better without taking up any space you can also rewrite that as follows right so instead of negative 1 to the power of n plus 1 times 1 over n you could write it as over in okay and now here's the part that I really want you to write down so I think you'll agree with me that this is this is fairly cluttered right this is this is cluttered to start trying to take the limit of but if you just you just accept that this numerator is always going to be 1 or negative 1 you will you agree with me there if you choose a 1 then it's negative 1 to the power of 2 if you choose a 2 then it's negative one to the power of three so it's this numerator is always going to be either one or negative one so what I'm telling you is before you take the limit just take the limit just assume that the general term is 1 over n I'll say that one more time because that will save you so much time the part that is alternating between negative and positive you can just for now assume that it's positive 1 henceforth okay so let's start we got a the idea is the conditions are not the conditions have to be met because you're in the you're going to make a statement of convergence notice that the statement make sure you don't push the test any farther this the test is basically saying that if the conditions are met you have convergence the test doesn't say anything about divergence okay so sometimes you just have to say you know sometimes you have to use another test okay or the test fails all right well let's start with the very first condition the very first condition was to take the limit of a sub n as n goes to infinity and I've already told you that we can just assume that a sub n is 1 over n as opposed to negative 1 over n plus 1 2 over in right because we know again that this is either going to be 1 or negative 1 so in in the limit this is not going to affect our result so this is quite easy this is this is quite easy to recognize that what's gonna happen is as you take as you take the limit as n goes to infinity of 1 over N well what do you you by now you know that that is equal to 0 now if that part up here was confusing as to you as to why you could why you could just use a 1 instead of a negative 1 then let me offer a quick explanation so we just used the positive numerator well are you can you imagine that this would be any different if this would be negative one over in would that result be any different at all no it would still go to zero so in this case we have met our first condition the limit as n approaches infinity of a sub n is equal to zero and now let's go to our our second condition and you should write your second condition it might be worth writing out like this you can it is you should try to stay formal but you should really try to tackle this the way it is within plus one and in and not and not use like you shouldn't do a term by term comparison but listen if if if you're out of if you're out of luck trying to use the formal approach then you might just list out the first few terms of a a sub n plus one in the first few terms of a sub N and that might be how you have to approach the problem right I mean I I would assume that it's not formal it's not a formal but sometimes you have to you know it's adds as a plus as opposed to keep just letting it be the question be blank sometimes you just have to go as far as you can go so what am I about to do right now well again I'm going to use this one over in I'm not gonna worry about this negative one part so let's go ahead and rewrite this so a sub n that's just the one over in now if I if I use in plus one okay you can you can now assume that what I'm gonna do is I'll start with a sub in but I'm going to be plugging in plus one in four in so this ends up being 1 over what 1 over n plus 1 now I need to be able to prove that for all n 1 over N is going to be greater than or equal to 1 over n plus 1 and how can you do that well fairly simple you can if it if it helps you you can you can cross multiply you can multiply by you can multiply by the LCD right so at this stage right here we could multiply by n times n plus 1 and let's see what that gives us so that gives us just in less than or equal to and over here would be in plus 1 now do you do you see that at this stage n plus 1 will always be greater than in right because you're always adding 1 so in other words you have you have met the criteria for the second condition do you have met the criteria for the first condition the second condition so then what can you conclude the conclusion is the series converges okay the series converges so series converges by alternating series test so you can imagine when I'm when I'm grading your work to just simply tell me the series converges without telling me how you know that is the equivalent of you telling me a true or false question is true without telling me how you know that it's true it's we are we are not I can't award guessing especially when you know you I just can't I can't no you need to arrive at the right solution the right answer for the right reason so this is you you use the alternating series test this is the work that I need to see to justify your conclusion okay but as you can see it's it's really not that bad it's just a matter of as long as you know how to take a limit and then as long as you can work work the following we're gonna we're gonna go ahead and go to an example of when let's say we'll use an example when the alternating series test does not apply okay so what they have let's kind of notice what they gave us so they give us this negative one to the power of n plus 1 times n plus 1 over N now this is the alternating series now what we've what we've done is we went ahead and produced some terms here so notice how it goes from a positive to a negative to a positive to negative to a positive now this is gonna be one of those scenarios where one of those conditions is going to fail so and then what I'm going to write is extremely important for when when this this test fails so all right one more time the it is okay if you want to use what I'm right now if you want to use that for a sub n that is perfectly fine but what I'm telling you is in the limit it's not gonna matter because this right here is either going to be negative 1 or positive 1 so get rid of it and just use just pretend like it's 1 just use the positive version of it for now ok alright so we are going to use an a sub n of just n plus 1 right because we're going to use the positive version of that of that alternating negative 1 so now it's going to be the limit within going to infinity of what of n plus 1 over n all right now notice that if you were to try to evaluate the limit as it sits we would get some type of in determined form so no worries it's at this stage is best to use l'hopital's rule so that you can reevaluate the limit so what I'm going to do is I'm going to take the derivative of that numerator and I'm going to take the derivative of that denominator are you ready so let's start well the derivative of the numerator is just 1 please make sure you you write right here so make sure you know that you're using l'hopital's rule right here's right so go ahead and and and write it out so that you don't so you don't confuse yourself or and and wonder later when you're looking at your notes what you used right so now if I take the derivative of the denominator that's also 1 right I'm just gonna write if f of X is equal to n plus 1 I'll write f of n that way we keep our that's then f prime is equal to 1 if G of n is equal to n then G prime of n is equal to 1 I just wanted you to understand how I arrived at this so really what we have is the limit as within approaching infinity of one well the limit of a constant is just a constant right it is it is irrelevant what n does because you're going to stay at 1 so this is equal to 1 this fails the condition the conditions that the limit of a sub n must equal 0 right so it fails in addition 1 if it fails condition 1 stop there's no need to go to condition 2 it's already failed the condition 1 so what what can we say you can say alternating series test fails what did I not say I the alternating series test will tell you if it converges when it fails that doesn't mean it diverges that is not what the test is saying make sure you don't push the scope of the of the tests the tests simply says well or what the test tells you what you can say if conditions are met the conditions weren't met so the alternating series test fails you need another test all right so the last part of this section and is what we call the the alternating series remainder and this is actually a very very cool theorem so let me let me see if I can let's take a look at this setup here okay and I want to make sure you understand what everything means inside of in the setup of this equality so the S is the actual sum all right if you could find the true sum of the series that would be s now S sub n is your current partial sum all the way up to n right so if you added up the first five terms then that's you have XS up five that's your that's your current sum s would be your tiresome all the way to infinity now a sub n plus one would be the value of your next term so let's again let's make sure we understand so a sub five would mean that you plug five in or let's let's think let me I'm gonna I'm gonna do it this way what is the difference between S sub 5 and a sub 5 I'm gonna hold there I'm gonna ask you that question again I'm gonna hold this so what is the difference between s sub 5 and a sub a sub 5 well s sub 5 is the first 5 terms added up a sub 5 is the value of your fifth term alright so what this says is a sub n plus 1 is just the value of the next term so meaning if you're at N and plus 1 would be the value of the next term now our R sub n is the remainder okay so let's see if you understand what let's see let's just focus on this part right here okay so if you have if you know what something adds up to be that's s - what you've added up already will eat whatever's left over is the remainder if you in other words if you add the remainder plus your current sum you would be able to get to the truesung with me so the actual sum that's this s - the current sum is going to be equal to that remainder now what is important is this because obviously you're not going to know the true sum that's the point of this little theorem but what we are gonna know is now let's look at this part we are going to know is that the remainder is always going to be less than or equal to your next term so you it's you can think of it as you you have a boundary you have an upper bound for your air ok cuz you're gonna you're gonna deal with some error but you least have an upper bound for your error so the remainder is always less than the value of the next term ok so now let's see if we can put this to use now give you some step by steps here so I'm gonna already assume the conditions have been met we've already we've already let's assume that we've already established that so I'm gonna give you some a playbook here steps 1 step 2 step 3 ready so the first the first thing you want to do is find s sub in now let's take a look at how many terms that we're given we're given 1 2 3 4 5 6 so you're given 6 terms so what you're looking for is S sub 6 ok so S sub 6 and S sub 6 would be the the ideas you would add up the first six terms right so this would be what remember factorial you've already gone over factorial so I'm gonna go over that again so this would be the equivalent of 1 minus well 2 factorial is 1 over 2 times 1 so that's 1 minus 1/2 plus 3 factorial is 3 times 2 times 1 so 1/6 minus 124 plus 1 over 120 - I believe 6 factorial is 720 I'll confirm that really quickly yep so minus one over 720 okay so you let your calculator do this heavy lifting here so the the current sum is 91 over 144 right approximately zero point six three one nine four okay so step one I think you'll agree with me that's they since they give you some terms you just you're gonna add up your first six terms now step two step two is to find the value of the next term what is the next term well you have to know what your current term is before you can talk about your next term so let's you agree with me that it stops right here this is a sub 6 one over six factorial so the next term would be a sub seven okay so you just you would plug in a fact or a 7 in for in so you would have it would be a positive now so this would end up being 1 over 7 factorial which is 1 over 7 factorial is 5040 okay so I'm gonna box in S sub 6 I'm gonna box in a sub 7 alright now remember on the previous on the previous page where I list out listed out that the remainder of the theorem for the remainder let's establish what you have inside here so do what you have the true some no obviously not you didn't add up all of these terms do you have the sum all the way up to the sixths term yes do you have the value of the next term yes so let's go ahead and do you have obviously do you if you don't have the room if you don't have the true sum then you don't have the remainder so in this case what you're gonna do though because the nice part about this theorem is it allows us to make the statement that look whatever s minus s sub six is we know it's going to be less than or equal to the next term that's what's nice about this ready so I'm gonna go ahead and plug in so that would mean s minus ninety one over 144 is less than or equal to one over five thousand and forty all right so the way you will the way you can think about that is that I'll write I'll capture this right down here and then I'll come back and finish that up so the true some so s this is what you're gonna be stating s is gonna be somewhere between what well it's going to be I'm gonna go ahead and use some decimals here okay just so you can well no I won't do that so the true sum is going to be somewhere between S sub six plus or minus a sub seven okay so I'm gonna go and write that so it's going to be 91 over 144 minus 1 over 5,000 40 and 91 over 144 plus one thousand five thousand 41 over five thousand four all right well let's just get our let's just approximate here so the true sum is going to be somewhere between zero point six three one seven four and zero point six three two one four does that make sense that all you're doing is taking you're taking S sub six plus your next term then S sub 6 minus the value of your next term and the theorem tells us that the true sum we don't know it but we know it's somewhere between there that's very very nice way of having some type of upper bound I'll show you a better much even better application of this exact same theorem okay so assume that whatever application that you're doing okay whatever you're trying to model assume that you are you're willing to tolerate an error less than point zero zero one okay so you you are willing to tolerate some error and whatever it is that you're trying to do and your as long as the air is is less than 0.001 so the question is determine the number of terms required to approximate the true sum of the series with an error less than 0.0 0.8 a sub 2 a sub 3 a sub 4 and you can keep playing this game and then you can add them this this plus a sub 1 plus a sub 2 or a sub 1 plus a sub 2 plus a sub 3 or a sub 1 plus a sub 2 plus a sub 3 plus a sub 4 know what it but what if you didn't want to play this game all the way to infinity what if instead you said look as long as I know that I'm less than this amount of error then I am willing to just add up the first three terms if I know that I'm let if even though that's not the true sum as long as I know that I'm less than the error that I'm required so what we're saying is that that remainder needs to be that we start off with the part that the remainder is less than or equal to the value of the next term so I want to know how many terms do you have to add up before your point zero zero one within the true of the true some do we have to add up three of them four of them ten of them how many alright well let's let's see if you can follow me here so first off let's establish that this numerator is going to be negative 1 or positive 1 always so we're just gonna use the positive 1 for a sub n so 1 over N to the power 4 so the first job because you're trying to you're trying to solve for n how many how many numbers do I have to add up so that I'm within 0.0 0.1 plus 1 this would be the next term you're gonna insert n plus 1 in 4 in so I'm gonna do that right now sub n plus 1 in 4 in all right so that gets me to 1 over n plus 1 to the power of 4 okay now now we're gonna take this okay and set it to be less than your error that you're willing to accept okay so we're going to set the result from step 1 that is 1 over n plus 1 to the power 4 we're gonna set the result of step 1 to be less than because you don't want it to be equal to you want to be less than the error that you're willing to accept 0.001 okay step 2 is done step 3 is to solve for N and when you do you'll know how many terms you have to add up so that you're close enough okay so I'd I mean this is just baby algebra to solve for in so alright so the idea is you can basically what let me just write any write what I have here you're gonna end up with you're going to end up with so multiply both sides but I'll just do it for you so I'm going to multiply both sides of this of this inequality by n plus 1 to the power 4 that's my first step okay let's write that so that would be one less than 0.001 times n plus 1 to the power of 4 ok now this next step we're going to divide by point zero zero one here and point zero zero one here alright so what does that look like so now one divided by point zero zero one is a thousand so you have a thousand is less than n plus 1 to the power of 4 now you're going to go ahead and take the you're gonna an index of force are going to take the fourth root of both sides I'm gonna write that fourth roots a little bit better so you can see it and that'll leave you with well if you take the the fourth root of something to the power four you're just left with that something n plus one now subtract one all right so that in that case n is going to be the fourth root of a thousand minus one basically n is approximately 4.6 terms what did you just do well you would if you just added up a sub one plus a sub 2 plus a sub three plus a sub four that's not enough to be within the margin of error but if you add up one more term a sub five right four point six terms so if you just go out five terms you are guaranteed to be within a one basically point zero zero one of your true sum I hope that makes sense I'll see you for the next videos