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Convergence problem (nth-term test)

  1. Jan 17, 2013 #1
    Show that the sum of (n/(3n+1))n from n=1 to ∞ converges.

    The book solves this with a comparison test to (1/3)n, but I'm making a mistake with an n-th term test somewhere.

    an = (n/(3n+1))n
    Take ln of both sides, then use n = 1/(1/n) to setup for l'Hopital's rule.
    ln an = ln(n/(3n+1)) / (1/n)
    ln an = -n/(3n+1)
    ln an = -1/3
    raise both sides
    an = e(-1/3)

    Which would mean divergence, right? Since the nth term does not equal 0?

    But this problem converges
  2. jcsd
  3. Jan 17, 2013 #2


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    But that is neither an ##\frac\infty \infty## or ##\frac 0 0## form.
  4. Jan 17, 2013 #3
    Ohh, ok

    1) If [itex]\sum[/itex] an converges, then an → 0

    2) [itex]\sum[/itex] an may or may not converge if an → c, c being any real number.

    3) If an as n → ∞ fails to exist, then [itex]\sum[/itex] an diverges.

    e: thank you very much, this was driving me nuts
  5. Jan 17, 2013 #4


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    If ##a_n## does anything but converge to 0, the series diverges. If it does converge to zero, the series may or may not converge. That's all there is to it.
  6. Jan 17, 2013 #5
    Double thanks, my mistake is with l'Hopital's rule.

    f(a) = g(a) = 0 as the limit of n → a must be true.
    The top function:
    does not go to 0 as n → infinity, so l'Hopital's rule cannot be applied.
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