If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Worked example: alternating series

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.10 (EK)
Example using the alternating series test to determine which values of a variable will make the series converge.

Want to join the conversation?

  • leaf green style avatar for user Sanchit Mehra
    I understood how you took p/6 < 1, but can we not get the same result using the ratio test? That is, will the ratio test not be applicable here?
    (6 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user robshowsides
      Good question! The answer is almost, but not quite! Remember that the ratio test says that you should find (the absolute value of) the limit of the ratio of successive terms, and if THAT is < 1, then FOR SURE the series converges, but that if THAT is EQUAL to 1, then the test is INCONCLUSIVE. So in this case, it's true that the ratio test would say, "IF p/6 < 1, THEN it is certain that the series converges". BUT, the ratio test also says, "IF p/6 = 1, then I'm sorry but I can't say for sure what happens." Thus, the ratio test would not be able, alone, to choose between the first and second answer choices.
      (7 votes)
  • blobby green style avatar for user saud.aljaffer
    I didn't understand what if the question didn't say find the POSITIVE values, what if he just wanted the values.

    Does that imply that p could be negative?

    and if it was negative then how would they know that it's decreasing so we can apply the alternating series test because bn must be decreasing
    (4 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user CMcIntyre
      Any values of p must not be negative for it to be an alternating series. By definition according to the Alternating Series Test, all of the b_sub_n terms (which are (p/6)^n in this case) must be greater than 0. The part about the positive values in the question was just thrown in as a hint.

      If you DID consider p values that are negative, then (p/6)^n could be factored as (-1)^n * (-p)^n. Then the original problem would be:
      (-1)^n+1 * (-1)^n * (p/6)^n
      = (-1)^(2n+1) * (p/6)^n
      = (-1) * (p/6)^n (since 2n+1 is always odd)
      = -(p/6)^n

      And that, is not an alternating series.
      (5 votes)
  • hopper jumping style avatar for user Gavin
    Can someone help me verify if this is true?

    When using the Alternating Series Test (AST), do I need to look at the absolute values of the terms and see if they converge to confirm that the series is absolutely convergent? If they don't converge, would the series be conditionally convergent?

    Also, to prove if the absolute value of the series is convergent, what tests do I need to use? I've used the ratio and geometric tests, but are there others that are easier?

    Thanks!
    (2 votes)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user Hemang Rajvanshy
      No, you don't need to look at the absolute values. For instance the harmonic series (1 + 1/2 + 1/3 + ...) is divergent whereas the alternating series (1 - 1/2 + 1/3 - 1/4 + ..) Converges.

      You can also use the integral test to prove convergence for the absolute values. The type of test used varies with the nature of the series.
      (2 votes)
  • spunky sam blue style avatar for user pittkor
    Cant we include 6?
    If P is 6, then doesn
    t it converge to 0?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Sainath Rakesh
    How to prove that the positive terms of an alternating series forms a Divergent series?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      They don't necessarily form a divergent series. This is only true if the series is conditionally convergent, which means that the series converges if you add all the terms, but diverges if you add the absolute values of all the terms.
      (2 votes)
  • leaf green style avatar for user britaonomake
    Why can't p=0? (it can be greater than but not equal to).
    When explaining the alternating series test in a previous video, Sal wrote B sub n can be greater than or equal to zero.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user julee
    It says in the video that p can not be 6 because it wouldn't be monotonically decreasing. But isn't monotonically decreasing mean that that the proceeding term is less than or equal to the previous term? In order to use the alternating test, bn must be monotonically strictly decreasing, not just decreasing. 1 wouldn't work b/c it is a geometric series and |r|>= 1 is divergent.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user Peiman Kazemi
    At , Why does Sal write exponents just considering n? Does he neglect "plus one" we have next to the exponent "n"?

    In other words, shouldn't the series start with exponent (p/6)ˆ2, (p/6)ˆ3...
    (1 vote)
    Default Khan Academy avatar avatar for user
  • starky sapling style avatar for user 20leunge
    Does choosing between (-1)^(n+1) or (-1)^(n) matter when testing convergence or divergence?
    (0 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] What are all positive values of p such that the series converges? So let's see. We have the sum from n equals one to infinity of negative one to the n plus one, times p over six to the n. So there's a couple of things that might jump out at you. This negative one to the n plus one, as n goes from one to two to three, this is just going to alternate between positive one, negative one, positive one, negative one. So we're gonna have alternating signs. That might be a little bit of a clue of what's going on, and actually, let's just write it out. This is going to be, see when n equals one, this is going to be to the second power, so it's gonna be positive one. So it's gonna be p over six, and when n is two, this is gonna be to the third power, so it's gonna be minus p over six squared, then plus p over six to the third power. And I could even write to the first power right over here. Then minus p over six to the fourth power, and we're gonna just keep going plus, minus, on and on and on and on forever. This is clearly, this is a classic alternating series right over here, and we could actually apply our alternating series test. And our alternating series test tells us that if this part of our expression, the part that is not alternating in sign, I guess you could say, if this part of the expression is monotonically decreasing, monotonically decreasing, which is just a fancy way of saying that each successive term is less than the term before it, and if we also know that the limit of this as n approaches infinity, that also has to be equal to zero. So the limit as n approaches infinity of p over six to the nth power also has to be equal to zero. Under what conditions is that going to be true? Well, to meet either one of those conditions, p over six has to be less than one. If p over six was equal to one, if, for example, p was six, well then we wouldn't be monotonically decreasing. Every term here would just be one. It would be one to the one, one squared, on and on and on. And if p is greater than six, well, then every time we multiply by p over six again, we would get a larger number over and over again, and the limit for sure would not be equal to zero. We could say p over six needs to be less than one, and so multiply both sides by six and you get p needs to be less than six. And they told us, for what are all the positive values of p? So we also know that p has to be greater than zero. P is greater than zero and less than six, which is that choice right over here. Once again, we're not gonna say less than or equal to six because if p was equal to six, this term is gonna be one to the n, and so we're just gonna have, this would be one. It would be one minus one plus one, on and on and on and on forever. So definitely like that first choice.