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Calc II - Alternating Series Test/Limits

  1. Aug 8, 2014 #1
    Hello PF,

    I've got a homework question I'm having some trouble with regarding series, particularily alternating series.

    The question asks you to test the series for convergence or divergence for an alternate series by using the A.S.T. :


    (-1)n-1e2/n
    n=1


    2. Relevant equations

    The Alternating Series Test states that a series, an, is converging if
    (a) an+1 ≤ an (i.e. the series is decreasing) for all n, and
    (b) the limit of the an = 0 as n → ∞

    3. The attempt at a solution

    In my attempt I have confirmed (a) by using the first derivative test to show that the series is decreasing:

    Let f(x)= e2/x
    f'(x) = -2e2/n / x2

    which is going to be negative for all x > 1, thus satisfying the first condition.

    The part where I am having trouble is confirming that the limit of e2/x equals/does not equal 0 as x approaches infinity (the second requirement for the A.S.T.).
    Looking at the graph of e2/x, the limit seems to be approaching 1, but I do not know to prove this/which method to use.

    It's been some years since I have taken Calc I and as a result, I'm having some diffculty with computing limits, though I am understanding the concepts of convergance/divergence well enough.

    Thanks for reading and for any assistance that you might be able to give!
     
  2. jcsd
  3. Aug 8, 2014 #2

    Ray Vickson

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    Is the function ##f(w) = e^w## continuous in the variable ##w##? For the case of ##w = 2/x## what is the limiting value of ##w## as ##x \to \infty##? So, what is the limit of ##e^{2/x}##?
     
  4. Aug 8, 2014 #3
    f(w) = e^w is continuous.
    For 2/x, it is not continous at 0. But it is continuous from 1 to infinity.
    So, the limit is 1?
     
  5. Aug 8, 2014 #4

    Ray Vickson

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    You tell me.
     
  6. Aug 8, 2014 #5
    Well, as 2/x becomes infinitely small, e goes to 1. I see that now, after doing some calculations.
    However, I'm just not sure why it doesnt go to zero instead. I seem to be missing some crucial fact about e..
     
  7. Aug 8, 2014 #6
    Oh, goodness,

    I understand now, as 2/x becomes increasingly small, it basically becomes 0.
    e^0 = 1.
     
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