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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.

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  • starky ultimate style avatar for user Darren Leung
    What is an "error", in this case?
    (12 votes)
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    • piceratops ultimate style avatar for user Andrei-Lucian Șerb
      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.
      (16 votes)
  • male robot hal style avatar for user M@y$h
    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.
    (11 votes)
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    • leafers tree style avatar for user Jacob Rodgers
      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
      (9 votes)
  • piceratops ultimate style avatar for user TheSoundOfDrums11
    What does a negative remainder mean?
    (8 votes)
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  • blobby green style avatar for user Arpan Chatterjee
    Is there any method by which we can find the exact value of infinite series ?
    (4 votes)
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    • leaf green style avatar for user kubleeka
      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.
      (9 votes)
  • starky ultimate style avatar for user t r
    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?
    (6 votes)
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  • blobby green style avatar for user William Lucas
    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)
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    • purple pi purple style avatar for user doctorfoxphd
      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)
  • boggle yellow style avatar for user andy
    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)
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  • leaf grey style avatar for user Jim Halpert
    If the series at started 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)
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  • piceratops seedling style avatar for user C2Rane
    At when 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)
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  • purple pi pink style avatar for user Carolyn Dewey
    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)
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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.