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# Alternating series remainder

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.B (LO)
,
LIM‑7.B.1 (EK)

## Video transcript

let's explore the infinite series so we're going to start at N equals one and go to infinity of negative one to the n plus one over N squared which is going to be equal to let's see when n is one it's going to be this is going to be positive it's going to be one and you can go minus one over two squared so it's minus one-fourth plus 1/9 minus 1/16 plus one over twenty-five I'm actually going to go pretty far minus 1 over 36 plus one over 49 minus 1 over 64 and it's pretty good I'll stop there and of course we keep going on and on and on and it's an alternating series plus/minus just keeps going on and on and on and on forever now we know from previous tests and in fact the alternating series test this satisfies the constraints of the alternating series test and we're able to show that it converges what we're doing now is actually trying to estimate what things converge to so we want to converge so it would not converge we want to estimate what this value is and we're going to do that by doing a finite number of calculations by not having to add this entire thing together and so let's estimate it by taking let's say the partial sum of the first four terms so let's take these four terms right over here and let's call that so that's going to be S sub four and then you're going to have a remainder which is going to be everything else so all of this other stuff I don't want even the brackets to end that's going to be your remainder the remainder to get to your actual sum or whatever is left over when you just take the first four terms so this is from the fifth term all the way to infinity and we've seen this before so the actual sum is going to be equal to this partial sum plus this remainder well we can calculate this this is going to be let's see common denominator here C 9 times 16 is 144 so it's it's going to be 144 and then that's going to be 144 minus 36 over 144 plus 16 over 1/4 4 minus 9 over 144 let's see that is 144 30 negative 36 plus 16 is negative minus 20 so it's 124 minus 9 is 115 so this is all going to be equal to 115 over 144 I didn't even need a calculator to figure that out plus some remainder plus some remainder and so if we could figure out some bounds on this remainder we will figure out the bounds on our actual on our actual sum we'll be able to figure out well how far is this away from this right over here and there's two ways to think about it so let's let's let's look at it so the the first thing I want to see is I want to show you that this remainder right over here is definitely going to be positive and I actually encourage you to pause the video and see if you can prove to yourself that this remainder over here is definitely going to be positive so I'm assuming you've had a go at it let's write the remainder down so actually I'll just write it actually I'll write it up here so R sub 4 is 1 over 25 1 over 20 I don't even have to write it separately I can show you in just right over here that this is going to be positive how do I show that well we just pair let's just create put some parentheses in here and just pair these terms like this so 1 over 25 minus 1 over 36 1 36 is less than 1 over 25 this one's positive this one's negative so this is positive then you have a positive term subtracting from that a smaller negative term so this is going to be positive and so eat you if you just pair all of these terms up you're just going to have a whole series of positive terms so just like that we have established that R sub 4 or R 4 we could call it is going to be greater than 0 so R 4 is going to be greater than 0 and now the other thing I want to prove is that this remainder is going to be less than the first term that we haven't calculated that the remainder is going to be less than one over 25 and once again I encourage you to pause the video and see if you can put some parentheses here in a certain way that will convince you that this entire infinite sum here this remainder is going to sum up to something that's less than this first term once again I'm assuming you've had a go at it so let's just write it down so I'll do that same pink color so our remainder when we just use when we take the partial sum of the first four terms so it's 1 over 25 and the way I'm going to write it instead of writing minus 1 over 36 I'm going to write minus I'm going to put the parentheses now between the second around the second and third term so there's going to be 1 over 36 minus 1 over 49 and then we're going to have minus 1 over 64 minus actually the next term is going to be 1 over 9 squared 1 over 81 and then minus and we keep going like that on and on and on on and on and on forever now notice what happens this this term right over here is positive we have a larger we have a smaller number being subtracted from a larger number this term right over here is positive so we're starting with 125 and then we're subtracting a bunch of positive things from it so this thing has to be less than 1 over 25 so R sub 4 is going to be less than 1 over 25 or we could even write that as R sub 4 is less than 0.04 0.04 same thing as 1 over 25 and actually this this logic right over here is is is the basis for the proof of the alternating series test this probably this should make you feel pretty good that hey look this thing is going to be greater than 0 and you could and it's it's increasing the more terms that you add to it but it's bounded by a bounded from above it's bounded from above at 1 over 25 which is a pretty good sense that hey this thing is going to converge but that's not what we're going to concern ourselves with here here we just care about this range some are is the sum of these two things and so the entire sum is going to be less than 115 115 over 144 plus the upper bound on our 4 so plus 0.04 and it's going to be greater than it's going to be greater than well it's going to be greater than our partial sum plus 0 because this remainder is definitely greater than 0 so you could just say it's going to be greater than our partial sum and just like that just doing a calculation that I was able to do with hand we're able to put pretty nice bounds around this infinite series infinite series and let's let's now get the calculator out just to get a little bit better sense of things so if we say 115 divided by 144 that's 0.7 9 8 6 1 repeating so this is zero point seven nine eight six one repeating is less than s which is less than this thing plus point zero 4 so let me write that down so so plus point zero four gets us to point eight three eight six one repeating eight three eight six one repeating so actually I could have done that in my head I know I resorted to a calculator zero point eight three eight six one repeating and just like that we just just a calculation we're able to do on by hand we were able to come up with a pretty good approximation for s and the end the the big takeaway from here and we're going to build on this but this was really to give you the intuition with a very concrete example is is when you have an all training series like this the type of alternating series that satisfies the alternating series test where you can write it as negative 1 to the N or negative 1 to the n plus 1 times a series of positive terms that are decreasing and whose limits go to zeros and approaches infinity not only do those things not only do those things converge but you can estimate your error based on the first term the you are not including now this was a one example there's going to be it's going to be different depending on the weather the first term is negative or positive and we're going to introduce the idea of absolute value there the magnitude but the big takeaway here is that the magnitude of your error is going to be no more than the magnitude of the first term that you're not including in your partial sum
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