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### Course: Integral Calculus>Unit 5

Lesson 10: Alternating series error bound

# Worked example: alternating series remainder

Using the alternating series remainder to approximating the sum of an alternating series to a given error bound.

## Want to join the conversation?

• okay, @ why does R sub k start with "k+1"? Shouldn't it just start with k? I have been trying to figure this one out but it just doesn't seem to make sense for me.
• The k term is the last term of the partial sum that is calculated. That makes the k + 1 term the first term of the remainder. This is the term that is important when creating the bound for the remainder, as we know that the first term of the remainder is equal to or greater than the entire remainder. Sal discusses this property in the previous video.
• Why does the remainder need to be less or equal to 0.001? Where does this 0.001 come from?
• It's the question that Sal asked. He said in the video to determine the k that has its remainder less than or equal to 0.001.
• so what happens if the first term you neglect is negative do we just take the absolute value of that term?
• Yes, because we are interested in the "margin of error," so we want it to be within x (like 0.001 in the above video) of the original sum. It could either be 0.001 less than or more than S (it could go either way) and hence the sign of the term you neglect does not matter.
• In this unit, we learn a lot about different tests on series convergence/divergence. But sometimes I have no idea to use which test to apply on the problem. How can i solve it? thanks.
• It is possible |Rk| = |a(k+1)| ? or should I just say |Rk| just less and not equal to |a(k+1)| ?
• Yes it is possible. We're talking about alternating series that satisfy the requirements of the alternating series test, one of which is that the series is decreasing. Here "decreasing" only means "never increasing." It doesn't mean that successive terms can never be equal. If we get to a point where all the terms after a(k+1) are zero, then yes, |Rk| = |a(k+1)|.
(1 vote)
• I am not sure...how this entire unit of infinite series is related to Calculus?
(1 vote)
• It's related to calculus as soon as you start applying the limit process to an "infinite" series. This means that as soon as you say, "ok, I am gonna sum up the terms of this sequence from n=1 to n=infiniy," u r applying a limit process. U r not necessarily saying that the sum of the infinite series is a definite number, but u r saying that it approaches (calculus comes in) to some number. U do this using limits. Calculus is the mathematics of change and u can intuitively associate the sum of the series changing as n increases in an infinite series. Hope this helps! :)
• Why does R start with k+1 and not just k? Where does the +1 come from?
• Remember that 𝑆(𝑘) is the sum of the first 𝑘 terms, while 𝑅(𝑘) is the sum of the remaining terms.
This means that the last term of 𝑆(𝑘) is ±1∕√𝑘, and thereby the first term of 𝑅(𝑘) must be ∓1∕√(𝑘 + 1)
• How do I know whether Rk is less than or less than/equal to the value of the next term in the alternating series??
• This video is very helpful in understanding the concepts.
However, it states several times that the technique shown will give you the minimum k that satisfies the requirements. That isn't true, is it? Instead, doesn't it give us the minimum k for which we know using this estimate that it satisfies the requirements? This technique only gives us an upper bound on the remainder. There could be (and probably is) a much lower k that also satisfies the requirements. For example, k=499999 seems to have < 0.001 error.
See Alternating_series#Approximating_sums on Wikipedia: "That does not mean that this estimate always finds the very first element after which error is less than the modulus of the next term in the series."

• how can i proof that error will be always smaller than first term in the remainder terms ?
• (1 vote)

## Video transcript

- [Voiceover] So we've got an infinite series here. Negative one to the n plus one over the square root of n. I always like to visualize this a little bit more, expand it out. So when n equals one the first term is negative one to the second power, so it's going to be positive. It's going to be one over one and so it's one minus one over the square root of two plus one over the square root of three minus one over the square root of four and then plus, minus, and we just go on and on and on forever. Now when we 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 with 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 after you've taken the first k terms. 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 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. Alright 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 is going to start with the k plus first term. So it's going to be negative one k plus one plus one, so it's negative one to the k plus two over the square root of k plus one. And then the next term I can just write plus negative one to the k plus three over the square root of k plus two. And it's just going to go on and on and on like that. Now what we know already or what we've seen an example of and we'll verify it or at least conceptual verification for us at the end of this video is 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 remainder R sub k... And some people refer to this as 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 one to the k plus two over the square root of k plus one. And of course we want that to be less than one thousandth. So that needs to be less than or equal to 0.001. And so now this setup right over here inspires you, I once again encourage you to pause the video and see if you can figure out what is the minimum k that satisfies this inequality. Well once again I'm assuming you've had a go at it. And the key realization here is that this negative one to the k plus two this makes this whole thing either positive or negative. The denominator is going to be positive. Well it's definitely going to be positive for all the ks frankly for which the principal root is going to be defined, but obviously these are all positive ks as well. So the numerator here just flips the sign. We flip between negative one or positive one, negative one or positive one. But if we take the absolute value whether it's negative or positive it's going to end up being positive. So this thing on the left hand side that's the same thing as just one over the square root of k plus one and then that's going to be less than or equal to 0.001. And now we just have to solve this inequality. Find out which ks satisfy this inequality. And so let's see, we can multiply both sides by the square root of k plus one. So square root of k plus one so we can get this out of the denominator. And let's actually multiple both sides times 1,000 because this is a thousandth and so we'll end up with a one on the right-hand side. So times 1,000, times 1,000. 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 1,000 is less than or equal to the square root of k plus one. We could square both sides and we get 1,000,000 is less than or equal to k plus one. And then we can just subtract one from both sides and we have 999,999 is going to be less than or equal to k. So 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 k if k has 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 999,999. Now let's convince ourselves that that remainder, that the absolute value of the remainder is definitely going to be less... 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 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 of the first 999,999 terms and we're claiming that this is going to be within one thousandth of this right over here. So this is going to be plus and so the first term of the remainder, of R sub k is going to be our millionth term. And so the millionth term, we just have to remind ourselves, is going to be negative one to the million and oneth power so negative one to the million and oneth power that's going to be negative one because that's an odd number. So it's going to be negative one over the square root of a million. So actually let me do it like this. So it's going to be minus one over the square root of a million. And then the next term is going to be plus one over the square root of a million and one. And it's positive now because this is gonna be negative one to the million and second power. And then it's going to be minus one over the square root of one million and two. And then plus, I'll just do two more terms. One over the square root of a million and three. And then minus one over the square root of a million and four 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, 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 0.001 because it's one over one thousand and we have this negative right over here. Now the first thing we could realize, 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, and we could keep going. So 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 get 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 that's going to be a negative value. This right over here is going to be a negative value. And we can say plus that, plus that, and so we're going to have a bunch of negative values together. So just like that by really looking at the parentheses a little bit differently we're able to say that 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 the absolute value of our remainder can't get any larger than one thousandth. As we add more and more and more terms here we're not going to get any further away from the actual sum than the first term.