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Does (-1)^n Converge?

  1. Apr 21, 2009 #1
    Intuitively I would think that [SUM of (-1)^n] or [SIGMA (-1)^n as n->infinity] Converges to 0 but apparently it doesn't according to the test for divergence ????

    Does it actually converge or diverge??? Why?

    Thanks
     
  2. jcsd
  3. Apr 21, 2009 #2

    dx

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    Hi zmike,

    It doesn't converge:

    1 = 1
    1 + (-1) = 0
    1 + (-1) + 1 = 1
    1 + (-1) + 1 + (-1) = 0

    It keeps oscillating between 1 and 0.
     
    Last edited: Apr 21, 2009
  4. Apr 21, 2009 #3
    thanks, just one more thing,

    so for ratio test, root test, test for divergence ALL work for alternating series? I am asking this b/c my textbook says they only work for positive sequences

    also can I seperate a sum? so can I take
    [sigma an] * [sigma bn] = sigma (an*bn)? If not, wouldn't that violate the properties of limits?

    Thanks
     
  5. Apr 21, 2009 #4

    dx

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    The ratio test applies only to positive series.

    For finite sums (∑a)(∑b) = ∑∑(ab) is true.
     
  6. Apr 21, 2009 #5
    (∑a)(∑b) = ∑∑(ab)

    What does the ∑∑ mean? take the sum twice? so is this also true with infinite sums?

    also, does this mean that if either a or b is divergent, I can conclude that the series is divergent?

    thanks
     
  7. Apr 21, 2009 #6

    dx

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    Sorry about the confusing notation. Latex is not working currently and I didn't know how to write it.

    Sum(i = 1 to n)[A_i] x Sum(j = 1 to m)[B_j] equals Sum(i = 1 to n)Sum(j = 1 to m)[A_i B_j]

    For infinite series, if both the series are convergent, this will still work. If either of them is divergent, then clearly the left hand side is not defined.
     
    Last edited: Apr 21, 2009
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