Main content

## Integral Calculus

### Course: Integral Calculus > Unit 5

Lesson 10: Alternating series error bound# Alternating series remainder

AP.CALC:

LIM‑7 (EU)

, LIM‑7.B (LO)

, LIM‑7.B.1 (EK)

By computing only the first few terms of an alternating series, we can get a pretty good estimate for the infinite sum. See why.

## Want to join the conversation?

- What is an "error", in this case?(11 votes)
- R is the error, and it is added to your partial sum. The total sum will be between the partial sum, and the partial_sum + error. The error makes it that the answer is not exact. But you can get a small enough enough error by calculating a partial sum of more terms. Because the error will always be less then the first term that didn't make the cut for the partial sum, the bigger the number of terms you choose for the calculation of the partial sum, the smaller the error.(15 votes)

- how do we know how many terms to take? For instance, in the above video Sal took first four terms of the series. The approximation (or the error) can be made "better" if more terms are included? Please do help.(10 votes)
- Similar with estimation with integrals, taking a higher number to test with will increase the accuracy of the answer, however, I think for this purpose Sal just chose a lower number so the math was less complex. I suppose it depends on what you're looking for. If you're in a test and crunched for time go with a smaller number to make it simpler. If you're doing something in physics and aerospace engineering maybe kick it up a few numbers. :P(8 votes)

- What does a negative remainder mean?(7 votes)
- it means that the estimate is greater than the true value.(0 votes)

- Is there any method by which we can find the exact value of infinite series ?(3 votes)
- Some series have what is called a "closed form", where we can express them as a finite number of functions nested together, like ∛(2-√(5/3)).

However, in the grand scheme of things, these series are quite rare. It's because of careful cherry-picking on the part of teachers and textbook authors that you usually see sums that can be written neatly like this. Proving*that*a series converges is usually easier than finding out*what*it converges to.

But for series that*do*have a closed form, we often have to play it by ear. Geometric series can be expressed as a/(1-r), which is proven on Khan Academy. There are telescoping series, which are a type of alternating series where almost every term is subtracted from itself, leaving one or two terms and a bunch of zeroes.

There are other techniques for computing series, many of which can be found in places like solutions to IMO problems.(8 votes)

- Alternatively, if we chose to estimate the alternating series by S5 + R5, we could make the case that R5 is negative by the same logic of pairing each remaining term where a5 is more negative than a6, etc. Can we not use the integral test in this case, and does this mean that we must be mindful of the index chosen for partial sum and remainder in estimating alternating series?(5 votes)
- Let me know if you ever figured out this answer bcuz I have to take a BC calc test on it in a few days lol(1 vote)

- I am not sure why Sal didn't compare and contrast what he did in this video to what he did in the previous video. To see the difference, I computed the upper and lower bounds and came up with a completely different range. Sal's upper bound was way better than mine; his lower bound was way worse than mine.

My upper bound was 115/144+ 1/4 (not +1/25)

My lower bound was 115/144 + .1/5 (not + 0)

While the inequality holds, it appears that using the integration method, I ended up with a worse upper bound. So knowing the the |error| is <= the magnitude of the first non-included term, why would one use the integration method?(1 vote)- Over time "previous video" changes due to Sal's amazing productivity, but if you are talking about the two videos on using integrals to place bounds on an infinite sum, then I don't think that method applies when you have an alternating series as you do in this video. He specified in the first of the pair of videos that the series needed to be
`continuous, positive and decreasing`

to use integrals as a bounding method. You can see that if you try to graph this series---you cannot apply**that**integral method because your values are jumping alternately above and below the x-axis. Every other value is negative. So, instead, we can use the tricks he has shown us to bound this wildly leaping creature.(8 votes)

- okay so I jumped ahead like a loooot, but I do have a question:

how come

a - b + c = a - (b - c)

thanks!!(1 vote)- If we start with a-(b-c), then we can distribute the negative sign to the terms inside the parentheses. So, a-(b-c) = a+(-b+c) = a-b+c.(5 votes)

- If the series at0:44started at n=0, would the value at n=0 be the first term or does the first term only exist at n=0?(1 vote)
- IF the series started at n = 0, we would have a real problem: the first term would be
`undefined`

, and therefore the sum would be undefined. ¹⁄₀ is a pretty effective showstopper without even getting into the (-1) multiplier(4 votes)

- At4:55when you distribute parenthesis for R4, why is it 1/25 - (1/36 - 1/49) - (1/65 - 1/81)...? Shouldn't it be 1/25 - (1/36 + 1/49) - (1/65 + 81)...? Why are you changing the signs inside the parenthesis?

Also later in the video, why can you say S<(115/144)+R4? Where did that come from? Isn't the sum EQUAL to (115/144)+R4? Why did you add an inequality?(2 votes)- At4:55, it's 1/25 - (1/36 - 1/49) - ... because Sal essentially flipped the signs here. Let's treat this - at the beginning of - (1/36 - 1/49) as -1 to get -1(1/36 - 1/49). If we distribute this, we get -1/36 -(-1/49) = -1/36 + 1/49, which is the original.(1 vote)

- Is a series that includes (-1)^(n-1) considered an alternating series? The terms alternate between positive and negative, but it was not mentioned in the alternating series test video.(2 votes)

## Video transcript

- [Voiceover] Let's explore
the infinite series. 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,
this is going to be positive. It's going to be one. This, you go minus one over two squares, is minus 1/4 plus 1/9 minus 1/16 plus 1/25 ... I'm actually going to go pretty far ... Minus 1/36, plus 1/49, minus 1/64. Yeah, that's pretty good. I'll stop there. 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, in fact, the alternating series test, that 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. We want to estimate
what this value, S, is. We're going to do that by doing a finite number of calculations, by not having to add this
entire thing together. Let's estimate it by taking, let's say, the partial sum of the first four terms. Let's take these four
terms right over here. Let's call that, that's
going to be S sub four. Then you're going to have a remainder, which is going to be everything else. 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 actually sum, or whatever's left over when you just take the first four terms. This is from the fifth term
all the way to infinity. We've seen this before. 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, see, nine times 16 is 144. That's going to be 144, and then that's going
to be 144 minus 36/144, plus 16/144, minus 9/144. Let's see, that is 144,
negative 36 plus 16 is minus 20, so it's
124 minus nine, is 115. This is all going to be equal to 115/144. I didn't even need a
calculator to figure that out. Plus some remainder. Plus some remainder. So, if we could figure out
some bounds on this remainder, we will figure out the
bounds on our actual sum. We'll be able to figure out, "Well, how far is this away
from this right over here?" There's two ways to think about it. Let's look at it. 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. 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. I'm assuming you've had a go at it. Let's write the remainder down. Actually, I'll just write it ... Actually, I'll write it up here. R sub four is 1/25. Actually, I don't even have
to write it separately. I could 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 put some parentheses in here, and just pair these terms like this. 1/25 minus 1/36. 1/36th is less than 1/25. 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. So, if you just pair all these terms up, you're just going to have a whole series of positive terms. Just like that, we have established that R sub four, or R
four, we could call it, is going to be greater than zero. R four is going to be greater than zero. 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 1/25. 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. I'll do that same pink color. Our remainder, when we
take the partial sum of the first four terms, it's 1/25. The way I'm going to write it, instead of writing minus 1/36, I'm going to write minus, I'm going to put the parentheses now around the second and third terms. This is going to be 1/36 minus 1/49. Then we're going to have minus 1/64 minus ... Actually, the next terms is going to be one over nine squared, 1/81. 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 smaller number being subtracted from a larger number. This term right over here is positive. We're staring with 1/25, and then we're subtracting a bunch of positive things from it. This thing has to be less than 1/25. R sub four is going to be less than 1/25. Or, we could even write that as R sub four is less than 0.04. 0.04, same things as 1/25. Actually, this logic right over here is the basis for the proof of
the alternating series test. This should make you feel pretty good, that, "Hey, look, this
thing is going to be "greater than zero," and it's increasing, the more
terms that you add to it. But it's bounded from above. It's bounded from above at 1/25, 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. The sum is the sum of these two things. So the entire sum is going to be less than 115/144 plus the upper bound on R four. Plus 0.04, and it's
going to be greater than, it's going to be greater than, it's going to be greater than
our partial sum plus zero, because this remainder is
definitely greater than zero. 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 get pretty nice bounds around this infinite series. Infinite series. Let's now get the calculator out, just to get a little bit
better sense of things. If we say 115 divided by 144, that's .79861 repeating. This is 0.79861 repeating, is less than S, which is less than this thing plus .04. Let me write that down. Plus .04 gets us to .83861 repeating, 83861 repeating. Actually, I could have
done that in my head. I don't know why I
resorted to a calculator. 0.83861 repeating. And just like that, just a calculation we're
able to do by hand, we were able to come up with a pretty good approximation for S. And the big takeaway from here ... We're going to build on this, but this was really to
give you the intuition with a very concrete example, is when you have an
alternating series like this, the type of alternating
series that satisfies the alternating series test, where you can write it
as negative one to the n, or negative one to the n plus one, 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
that you're not including. Now, this was one example. It's going to be different depending on whether the first term
is negative or positive, and we're going to have 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.