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Root test fails, now what?

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data

    determine whether the series (1-1/n^(1/3))^n converge or diverge

    2. Relevant equations
    all the testing procedure



    3. The attempt at a solution
    So I did the root test first, but the limit on the inside is 1. I then tried the ratio test but then when I tried taking the limit, I ended up with e^(-infinity+infinity). Finally, I tried the test for divergence on the original series, but i turned out be zero so it might or might not converge. Now I'm totally stuck.
     
  2. jcsd
  3. Oct 14, 2013 #2

    D H

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    Did you try the comparison test?
     
  4. Oct 14, 2013 #3

    Mark44

    Staff: Mentor

    You should never end up with something like this ( -∞ + ∞), as it is indeterminate. I agree with what you got on the root test, so that's not a useful test, and the n-th term test seems to give a value of 0, so it doesn't tell us anything.

    I would take another look at the ratio test to see what the limit actually is.


     
  5. Oct 14, 2013 #4

    Ray Vickson

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    If
    [tex]t_n = \left(1 - \frac{1}{n^{1/3}}\right)^n,[/tex]
    you can look at ##L_n = \ln(t_n)## and use the series expansion of ##\ln (1-x)## for small ##x = 1/n^{1/3}##, to get the behavior of ##t_n## for large n. In fact, you can even get a simple upper bound ##u_n##, so that ## 0 < t_n < u_n##, and ##\sum u_n## is easy to analyze.
     
  6. Oct 15, 2013 #5

    Mark44

    Staff: Mentor

    The OP might not have been exposed to series yet...
     
  7. Oct 15, 2013 #6

    Ray Vickson

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    Series are not really needed here; it is enough to know that 1-x < exp(-x) for x > 0, so that ln(1-x) < -x for small x > 0.

    Anyway, if the OP has seen series the method could be useful; if he/she has not yet seen series it is perhaps not useful. However, I have no way of knowing what the OP has, or has not, seen.
     
    Last edited: Oct 15, 2013
  8. Oct 15, 2013 #7

    D H

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    I'm not seeing how that would help. If it's working toward the ratio test, the ratio of consecutive terms approaches 1 as n→∞, so that test is inconclusive.

    There is a very simple series that does converge that looks somewhat like this.
     
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