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### Course: AP®︎/College Calculus BC > Unit 10

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, @2:02why 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.(14 votes)
- 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.(20 votes)

- Why does the remainder need to be less or equal to 0.001? Where does this 0.001 come from?(7 votes)
- 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.(18 votes)

- so what happens if the first term you neglect is negative do we just take the absolute value of that term?(5 votes)
- 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.(4 votes)

- 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.(2 votes)
- It is possible |Rk| = |a(k+1)| ? or should I just say |Rk| just less and not equal to |a(k+1)| ?(3 votes)
- 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! :)(8 votes)

- Why does R start with k+1 and not just k? Where does the +1 come from?(3 votes)
- 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)(4 votes)

- How do I know whether Rk is less than or less than/equal to the value of the next term in the alternating series??(4 votes)
- 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."

Am I missing something or is the video wrong about this?(3 votes) - how can i proof that error will be always smaller than first term in the remainder terms ?(3 votes)

## 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.