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Current time:0:00Total duration:10:21

AP.CALC:

LIM‑7 (EU)

, LIM‑7.B (LO)

, LIM‑7.B.1 (EK)

so we've got an infinite series here negative one to the n plus one over the square root of n and I always like to visualize this a little bit more expand it out so when N equals one the first term is negative 1 to the second power is going to be positive is going to be 1 over 1 so it's one go minus one over the square root of two plus 1 over the square root of 3 minus 1 over the square root of 4 and then plus minus and we just go on and on and on forever now when we've looked at convergence tests for infinite series we saw things like this this passes the alternating series test and so we know that this converges let's say it converges to some value s but what we're concerned about in this video is not whether or not this converges but estimating what this actually converges to we know that we can estimate this by taking a partial sum let's say S sub K this is the first K terms right over here and then you're going to have a remainder so plus the remainder well after you've taken the first K term so this actually starts with the k plus first term and so what I am concerned about or what I want to figure out is how many what's the minimum number of terms what's the minimum K here so that my remainder so that the absolute value of my remainder is less than or equal to 0.001 and I encourage you to pause the video based on what we've seen in previous alternating series estimation situations to see if you can figure this out all right I'm assuming you've had a go at it so let's just remind ourselves what R sub K looks like so R sub K R sub K is going to be it's going to tart with the k plus first term so it's going to be negative 1 k plus 1 plus 1 so it's negative 1 to the k plus 2 over the square root of K of k plus k plus 1 k plus 1 and then the next term I could just write plus negative 1 to the k plus 3 over the square root of k+ 2 and it's just going to go on and on and on like that now what we know already are what we've seen an example of and we'll verify it or at least get a conceptual verification for us at the end of this video is well this the this the absolute value of this entire sum the absolute value of this entire sum is going to be less than the absolute value of the first term so let me write that down the absolute value of this entire sum of this entire remainder R sub K and says you know some people refer to this too is kind of the alternating series remainder property or whatever you want to call it but the absolute value of this entire thing is going to be less than or equal to the absolute value of the first term negative 1 to the k plus 2 over the square root of k plus 1 and of course we want that to be less than one thousandth so less than that needs to be less than or equal to zero point zero zero one and so now this set up right over here inspires you once again encourage you to pause the video and see if you can figure out what is the minimum K what is the minimum K that satisfies this inequality well once again I'm assuming you've had a go through go at it and the the key realization here is that this negative 1 to the k plus 2 this makes this whole thing either positive or negative the denominator is going to be positive 4 well it's definitely going to be positive for all the case frankly for which this the principal root is going to be defined but we obviously these are all positive K's as well so the numerator here just flips the sign we flip between negative 1 or positive 1 negative 1 or positive 1 but we take the absolute value then we know or it's whatever it's whether it's a negative or positive it's going to end up being positive so this first this thing on the left hand side this thing on the left hand side that's the same thing as just 1 over the square root of K the square root of K plus 1 and then that's going to be less than or equal to that needs to be less than or equal to zero point zero zero 1 and now we just have to solve we have to just solve this inequality find out which K days satisfy this inequality and so let's see we can multiply both sides by the square root of K plus 1 so square root of K plus 1 so we can get this out of the denominator and let's actually multiply both sides times a thousand because this is a thousand and so we'll end up with a 1 on the right hand side so times a thousand times a thousand and what we're going to be left with is these cancel out on the left hand side we don't have to flip the sign because we just multiplied by positive things so we have a thousand is less than or equal to the square root of K plus 1 we could square both sides and we get 1 million 1 million is less than or equal to K plus 1 and so and then we can just subtract 1 from both sides and we have 999,999 is going to be less than or equal to K so keep K just has to be greater than or equal to 999,999 and remember we want the smallest K that satisfies these conditions so the smallest gift cans to be greater than or equal to this the smallest K that satisfies this is K is equal to 999,999 so let's write that down so the smallest K for which this is true is going to be K is equal to 900 99,999 now let's convince ourselves that that remainder that the absolute value of the remainder is definitely going to be less is definitely if we if our K if we take the partial sum of the first 999,999 terms that this remainder is actually going to be less than this I'm telling you so far we've just kind of worked from the premise that it will be but let's actually let's actually look at that and feel good about it and once again encourage you to pause the video and try it on your own so let's just rewrite this again but I'm going to expand out R sub K so we're saying that it's all going to converge to s and we're going to take the partial sum the partial sum of the first 999,999 terms and we're claiming that this is going to be with in 1000 of this right over here so this is going to be plus and so the first term is going to be the of the remainder of R sub K is going to be our millionth is going to be our millionth of our millionth term and so the millionth term we just remind ourselves it's going to be negative 1 to the million and one power so negative 1 to the millionth and 1 power that's going to be a negative that's going to be negative 1 because that's an odd number so it's going to be negative 1 over the square root of a million so lecture let me do it like this so it's going to be it's going to be minus 1 over the square root of a million a million and then the next term is going to be plus 1 over the square root of a million a million and 1 and it's positive now because this is going to be the millet to the million and negative 1 to the million and a million in two second power and then it's going to be minus 1 over the square root of 1 million 1 million and 2 and then plus plus I'll just do two more terms 1 over the square root of a million a million and 3 and then minus 1 over the square root of a million and 4 and then plus minus it just keeps going on and on on and on and on forever now I want to convince ourselves that this thing is going to be this the absolute value of this thing is going to be less than one thousandth it's going to be essentially less than the absolute value of this first term so how do we think about it so this first term we already know this right over here is this first term right over here is negative negative 0.001 because it's 1 over 1,000 and it's going to be have this negative right over here now the first thing we could realize if we just if we just put parentheses like this if we just put parentheses like this we see that this is going to be positive this term is larger than this term this term is larger than this term we could keep going so we're going to have we're start wheat the remainder kind of starts at negative one thousandth and then it just adds a bunch of positive terms so it's not going to get any more negative than this so we know that the remainder can't get more negative than this but let's also verify that it can't you know get forget more than a positive thousandth either and the way that we can do that is to just look at the parentheses in a different way if we do it this way if we do it this way that's going to be this is going to be a negative value this right over here is going to be a negative value and we can just keep so we could say Plus that Plus that and so we're going to have a bunch of negative values together so just like that we really looking at the parentheses a little bit differently we're able to say this is definitely going to be negative this remainder is definitely going to be a negative value but it can't be any more negative than our first term so that tells us that our remainder is not going to get any fur is not going to be any more can't get any the absolute value of the remainder can't get any larger than one thousandth we're not going to as we add more and more and more terms here we're not going to get any further away from from the actual sum than we were then then the I guess we could say then then the first term