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Homework Help: Help with convergence test?

  1. Mar 16, 2016 #1
    • Member warned about not using the homework template
    Which of these converge?

    1. 5n/ (2n-1)

    2. e^n/ n

    3. e^n/ (1+e^n)


    1) lim n-> ∞ 5n/(2n-1) = 5n/2n = 5/2 ≠ 0 so diverge?

    2) Change n to x
    e^x/ (1 + e^x)
    Antiderivative: ln |1 + e^x|
    lim t->∞ of ln |1 + e^x|
    ln |1 + e^∞| - ln |1 + e^0|
    Got stuck here

    3) Help :(
  2. jcsd
  3. Mar 16, 2016 #2


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    Science Advisor
    Gold Member

    By converge, do you mean limit exists as n->∞ ? If not, what do you mean? You don't give any information on what limit rules you are familiar with and allowed to use. Please add more such information.
  4. Mar 16, 2016 #3


    Staff: Mentor

    Are these sequences of numbers or are they infinite series? The first problem, interpreted as a sequence, converges to 5/2, as you said.
  5. Mar 17, 2016 #4


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    Homework Helper

    I will assume that you are looking for whether or not the sequences converge as n goes to infinity.
    As Mark pointed out, you correctly found that the limit of #1 exists as n goes to infinity, so the sequence converges. If you are looking for the sum of the terms, i.e. the series, then your rule about the limit not being equal to zero would be appropriate.
    For #2, it looks like you were working on #3.
    Also, when you do the integral test, it implies you are looking for the infinite sum, i.e.:
    ##\lim_{n\to \infty} \sum_{k = 1}^n \frac{e^k}{k}. ##
    If that is what you are looking for, then none of these (series) converge.

    However, it looks like these are better suited as problems about sequence convergence, so let's go with that.

    For #1, the limit exists, so the sequence converges.
    For #2, you can try a growth rate test.
    ##x_n = \frac{e^n}{n}, x_{n+1} = \frac{e^{n+1}}{n+1}.##
    ##G=\frac{x_{n+1}}{x_n} = \frac{n e^{n+1}}{(n+1)e^n}##
    If the limit as n goes to infinity of G is greater than 1, then the sequence is not converging.

    For #3, you can apply the same logic you did for #1, and just keep the terms of the highest order.
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