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Convergence of a series

  1. Dec 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Use the limit comparison test to determine whether the following series converges or diverges.

    [itex]\sum{\frac{2^n}{3^n - 1}}[/itex]

    Where the sum is from n = 1 to n = ∞.

    2. Relevant equations

    The limit comparison test:

    Suppose an>0 and bn>0 for all n. If the limit of an/bn=c, where c>0, then the two series [itex]\sum{a_{n}}[/itex] and [itex]\sum{b_{n}}[/itex] either both converge or both diverge.

    3. The attempt at a solution

    So I get that the point is to find another series that you know diverges or converges that will also make the above limit easy to evaluate. All I can think to do is to compare it to something like [itex]\sum{1/3^n}[/itex] or [itex]\sum{2^n}[/itex], the first of which results in the limit not existing and the second of which results in the limit being zero. I think it's obvious that the series should converge, as the denominator is always equal to or greater than the numerator. But I'm out of ideas on what other sums I could use in the limit comparison test.
     
  2. jcsd
  3. Dec 15, 2012 #2
    Compare it to
    [tex]\sum \frac{2^n}{3^n}[/tex]

    [tex]\frac{2^n}{3^n}\cdot\frac{3^n-1}{2^n}[/tex]
    [tex]\frac{3^n}{3^n} - \frac{1}{3^n}[/tex]
    [tex]1- \frac{1}{3^n}[/tex]
    With limit n->infinity, this = 1
     
    Last edited: Dec 15, 2012
  4. Dec 15, 2012 #3

    sharks

    User Avatar
    Gold Member

    Not quite, although the limit of the ratio is equal to 1. There are a few mistakes, in my opinion:

    First, the ratio is wrong. It should be ##u_n/v_n## or in the OP's equation ##a_n/b_n##. You've done the opposite. (A condition is missing in this case: c<∞)

    Second, the limit is obtained using L'Hopital's rule.

    Third, [itex]\sum_{n=1}^{\infty} \frac{2^n}{3^n}[/itex] is in fact a geometric series.

    It should be easy now to form the final conclusion.
     
    Last edited: Dec 15, 2012
  5. Dec 15, 2012 #4
    Doh! That should've been my next try. I got too caught up trying to do the same thing I was doing for the preceding questions. Thanks for the help guys :)
     
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