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Convergence/Divergence of a Series

  1. Apr 8, 2012 #1
    Determine whether or not the series is convergent or divergent. If it is convergent, find its sum.

    1) 1/3 + 1/6 + 1/9 + 1/12 + 1/15 + ...

    2) E (n-1)/(3n-1)

    Okay, I'm having trouble with determining the convergence/divergence. My book states that, "If the series

    E Asubn

    is convergent, then lim as n goes to inf. asub n=0."

    and it also states that, "If lim as n goes to inf. asubn does nto exist or if lim as n goes to inf. does not equal 0, then the series

    E Asubn
    is divergent.

    Here's where I'm having trouble. I set up the sum series for the first question as:
    E Asubn 1/(3n)

    Then I took the limit as n goes to inf. of 1/(3n) and got 0. So by the book, wouldn't this make it convergent? However my book has the answer as being divergent.

    Then for the second question, I took the limit as n goes to inf. of (n-1)/(3n-1) to be 1/3. So wouldn't this make it convergent as well? The book has the answer as being divergent.

    If someone would articulate the convergent/divergent test a little better it would be greatly appreciated. Thanks!
  2. jcsd
  3. Apr 9, 2012 #2
    The theorem says:

    IF the series converges, THEN the limit is zero.

    You argue, that the limit is zero, thus the series converges. But that's the converse implication and it does not hold.

    For example:
    IF my name is Alex, THEN my name begins with an A.

    What you do is: my name begins with an A so my name must be Alex.
    This is of course invalid.
  4. Apr 9, 2012 #3
    Okay thanks, that makes sense now. But if I find that the limit is 0, then it can be convergent or divergent. How do I tell from there?

    E 1/3n

    lim 1/3n = 0
  5. Apr 9, 2012 #4

    Char. Limit

    User Avatar
    Gold Member

    Try comparing it with the series [tex]\sum_{n=1}^\infty \frac{1}{n}[/tex].
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