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Homework Help: Question about Alternating Series Test

  1. Apr 27, 2012 #1


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    The problem statement, all variables and given/known data
    After reading a few topics on this forum, i just realized that i had misunderstood the sequence v/s series theorem when it concerns the alternating sign. So, i went back to my notes, and i'm surprised to see that there is no mention of a series test. There is only a sequence test for alternating sign where i need to take the modulus. I guess my tutor skipped that part, or i lost my sheets. Either way, i'm now trying to learn the concept by myself.

    The attempt at a solution

    Here is what i've understood:
    The AST is used to test for a series' convergence.

    The first step is to check if: [tex]\lim_{n\rightarrow \infty} a_n = 0[/tex]
    The second check is if: [tex]a_{n+1} \leq a_n[/tex]
    I got those equations online (youtube), so i hope these are correct. However, i have several doubts:

    What if the first check is OK but the second check is not. Does that mean that the series diverges?

    In the example that i've seen in some video, i see that if the first check fails, then it automatically diverges. But i'm wondering if that's always the case?
  2. jcsd
  3. Apr 27, 2012 #2

    Ray Vickson

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    It does not matter if [itex] a_{n+1} \leq a_n[/itex] for all n, but only that it happens for all sufficiently large n (that is, it might fail for the first few terms, but must happen from all n beyond a certain finite value).

    As to your second question: the condition you cite is a sufficient condition for convergence, not a necessary one. For example, the series [itex] \sum (-1)^n t_n, [/itex], where [tex] t_n = \left\{ \begin{array}{cl}1/n^2,& n \text{ even} \\
    2/n^2 ,& n \text{ odd} \end{array} \right.[/tex]
    is convergent, but does not satisfy the inequality.

  4. Apr 27, 2012 #3


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    I'm not sure... So, the AST is synonymous to the nth-term test for divergence, in that, the latter can only be used for testing divergence but says nothing about the series' convergence. Similarly, the AST is only for testing convergence, and it cannot account for divergence. Correct?

    In your example, i don't understand why n is odd or even? What's your point?
    Last edited: Apr 27, 2012
  5. Apr 27, 2012 #4


    Staff: Mentor

    I wouldn't say that the alternating series test is synonymous to the n-th term test, but they are similar in a way.

    The alternating series test says that under certain conditions, an alternating series converges. It does not mention divergence.

    The nth term test says that under certain conditions, a series diverges. It does not mention convergence.

    Ray's example with different terms for odd or even n shows a series for which lim tn = 0, but the series is not monotonically decreasing.

    An important point about these theorems is the idea of what an implication means in mathematics. Many calculus students are not used to thinking logically, so the fine points about mathematical logic go right over their heads.

    Let's look at an example. Suppose I tell you, "If you mow my lawn this Saturday, then I'll give you $100."

    As long as the condition (you mow the lawn) is met, you are justified in expecting to be paid. If the condition is not met, you cannot reasonably conclude that I will pay you, and you also cannot conclude that I won't pay you. (Out of the kindness of my heart, I might just give you $100.)

    My point is - make sure that the hypotheses (conditions) of a theorem are met, then the conclusion inescapably must follow. If the hypetheses are not met, anything can happen.

    Case in point: nth term test, with series Ʃ (1/n). lim an = 0. The hypothesis of this theorem is not met, so the theorem does not apply.
    Last edited: Apr 27, 2012
  6. Apr 27, 2012 #5


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    I think i finally understand the conditions for which AST applies. The 'lawn mowing' example is a very practical way of untangling the ideas associated with the theorem.

    Thank you very much for your help, Mark44 and RGV. :smile:
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