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Alternating series test

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.10 (EK)
When a series alternates (plus, minus, plus, minus,...) there's a fairly simple way to determine whether it converges or diverges: see if the terms of the series approach 0.

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  • blobby green style avatar for user Maxim Nasurdinov
    From to - what is the second condition given for?
    Isn't it clear, that if lim(b_n) = 0 when n -> infinity, then {b_n} should be decreasing sequence only? How this sequence could be not decreasing if the limit of b_n (when n -> infinity) is equal zero?
    (16 votes)
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  • blobby green style avatar for user Steven Kessler
    The restrictions on this test seem redundant.

    from to , Sal gives three restrictions on the series:

    1) Bn ≥ 0 for all relevant n (namely positive integers n).
    2) lim as Bn→∞ = 0
    3) {Bn} is a decreasing sequence.

    Don't the first two rules imply the third?
    (6 votes)
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    • blobby green style avatar for user Andrew
      Consider the function f(n) = x*e^(-n). This will be our Bn.

      What do you get at n= 0, n=1, n=2?
      f(0) = 0,
      f(1) = e^-1 = 0.37,
      f(2) = 2/e^-2 = 0.27
      So the first 3 terms of the sequence are: [0, 0.37, 0.27].
      Notice the b2> b1 this function is not always decreasing! Yet all terms are greater than zero. And the limit as n→∞ = 0. Some functions like this rise a little bit, then fall back down as they go on. So the third rule is necessary.
      (8 votes)
  • ohnoes default style avatar for user Cyan Wind
    Does this test work for an increasing sequence?
    (3 votes)
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    • female robot grace style avatar for user J Helston
      This test is used to determine if a series is converging. A series is the sum of the terms of a sequence (or perhaps more appropriately the limit of the partial sums).

      This test is not applicable to a sequence.

      Also, to use this test, the terms of the underlying sequence need to be alternating (moving from positive to negative to positive and so on). Therefore the sequence would not be increasing but rather 'oscillating' between positive and negative terms.
      (8 votes)
  • piceratops ultimate style avatar for user Blake Kohrmann
    At -: Why is it necessary that the alternating sign be expressed by either (-1)^n or (-1)^(n+1). Aren't there a whole slew of other ways to produced an alternating sign? What about (-1)^(n+2), (-1)^(n^2), or any polynomial exponent with odd integer coefficients?
    (3 votes)
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    • leaf blue style avatar for user Stefen
      We are only talking about the form the series takes on. We know that it alternates, so the question is, is a negative term first, or a positive term. Given n goes from 1 to infinity, the first term of the (-1)^n series will be negative, and the first term of the (-1)^(n+1) series will be positive. That is all that is meant by the form of the series. Why make it any more complicated? It is an alternating series, either the first term is positive, or the first term is negative.
      (6 votes)
  • leaf green style avatar for user Abdo Reda
    At He says that we can use other techniques like the limit comparison test, I tired using the limit comparison test but I can't figure out another functions that behaves like this function and that I can figure out It converges ... can someone please tell me how :)
    (3 votes)
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  • leafers ultimate style avatar for user Daniel Russell
    While attempting some practice problems, I couldn't get the correct answer, and this came up as a hint. "This series meets all the conditions for the alternating series test and hence it converges. However, since we can show that ∑n=1∞ n+1n2 diverges by using a comparison test with ∑n=1∞1n. Thus the series converges conditionally." I do not understand what this means, What are the "conditions"?
    (2 votes)
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  • piceratops sapling style avatar for user Aditya Makesh
    Is this test also called the Leibnitz test for alternating series??
    (3 votes)
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  • piceratops sapling style avatar for user Travis47
    Can you use the ratio test on an alternating series?
    (3 votes)
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  • orange juice squid orange style avatar for user daca7474
    So I just tried the first practice problem in the "Alternating Series" challenge, and I found that one of the hints contradicts what Sal says in this video. At , Sal says, "Now once again, if something does not pass the Alternating Series Test, that does not necessarily mean that it diverges; it just means that you couldn't use the Alternating Series Test to prove that it converges."

    The problem hint, however, says, "...Since lim n→∞ an = 1 ≠ 0, the series fails the Alternating Series Test and therefore diverges."

    Help? Can the Alternating Series Test be used to prove DIVERGENCE, or only conclusively prove CONVERGENCE?
    (3 votes)
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  • piceratops ultimate style avatar for user declanhoban
    How can one use the limit comparison test in this example? I found that it is inconclusive to use 1/n as a comparison, but I don’t know what else to try.
    (2 votes)
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Video transcript

- [Narrator] Let's now expose ourselves to another test of convergence, and that's the Alternating Series Test. I'll explain the Alternating Series Test and I'll apply it to an actual series while I do it to make the... Explanation of the Alternating Series Test a little bit more concrete. Let's say that I have some series, some infinite series. Let's say it goes from N equals K to infinity of A sub N. Let's say I can write it as or I can rewrite A sub N. So let's say A sub N, I can write. So A sub N is equal to negative one to the N, times B sub N or A sub N is equal to negative one to the N plus one times B sub N where B sub N is greater than or equal to zero for all the Ns we care about. So for all of these integer Ns greater than or equal to K. If all of these things, if all of these things are true and we know two more things, and we know number one, the limit as N approaches infinity of B sub N is equal to zero. Number two, B sub N is a decreasing sequence. Decreasing... Decreasing sequence. Then that lets us know that the original infinite series, the original infinite series, is going to converge. So this might seem a little bit abstract right now. Let's actually show, let's use this with an actual series to make it a little bit more, a little bit more concrete. Let's say that I had the series, let's say I had the series from N equals one to infinity of negative one to the N over N. We could write it out just to make this series a little bit more concrete. When N is equal to one, this is gonna be negative one to the one power. Actually, let's just make this a little bit, let's make this a little bit more interesting. Let's make this negative one to the N plus one. When N is equal to one, this is gonna be negative one squared over one which is gonna be one. Then when N is two, it's gonna be negative one to the third power which is gonna be negative one half. So it's minus one half plus one third minus one fourth plus minus and it keeps going on and on and on forever. Now, can we rewrite this A sub N like this. Well sure. The negative one to the N plus one is actually explicitly called out. We can rewrite our A sub N, so let me do that. So negative, so A sub N which is equal to negative one to the N plus one over N. This is clearly the same thing as negative one to the N plus one times one over N which is, which we can then say this thing right over here could be our B sub N. This right over here is our B sub N. We can verify that our B sub N is going to be greater than or equal to zero for all the Ns we care about. So our B sub N is equal to one over N. Clearly this is gonna be greater than or equal to zero for any, for any positive N. Now what's the limit? As B sub N, What's the limit of B sub N as N approaches infinity? The limit of, let me just write one over N, one over N, as N approaches infinity is going to be equal to zero. So we satisfied the first constraint. Then this is clearly a decreasing sequence as N increases the denominators are going to increase. With a larger denominator, you're going to have a lower value. We can also say one over N is a decreasing, decreasing sequence for the Ns that we care about. So this satifies, this is satisfied as well. Based on that, this thing is always, this thing right over here is always greater than or equal to zero. The limit, as one over N or as our B sub N, as N approaches infinity, is going to be zero. It's a decreasing sequence. Therefore we can say that our originial series actually converges. So N equals 1 to infinity of negative one to the N plus over N. And that's kind of interesting. Because we've already seen that if all of these were positive, if all of these terms were positive, we just have the Harmonic Series, and that one didn't converge. But this one did, putting these negatives here do the trick. Actually we can prove this one over here converges using other techniques. Maybe if we have time, actually in particular the limit comparison test. I'll just throw that out there in case you are curious. So this is a pretty powerful tool. It looks a little bit about like that Divergence Test, but remember the Divergence Test is really, is only useful if you want to show something diverges. If the limit of, if the limit of your terms do not approach zero, then you say okay, that thing is going to diverge. This thing is useful because you can actually prove convergence. Once again, if something does not pass the alternating series test, that does not necessarily mean that it diverges. It just means that you couldn't use the Alternating Series Test to prove that it converges.