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Homework Help: Determining the limit of an infinite sequence

  1. Nov 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the limit of the sequence: an = (1+(5/n))2n

    2. Relevant equations

    L'hopitals rule, or at least that's what I'm thinking. Otherwise, general formulas for determining the limit of a sequence.

    3. The attempt at a solution

    an = (1+(5/n))2n

    Considering the behavior of the sequence as n goes toward infinity, (5/n) is a very small change in size, so it goes approximately to zero, leaving:

    an = (1)2n

    Which, as n goes to infinity, is essentially: 1, an indeterminate form. To resolve this, take the natural log:

    an = ln(12n) = 2n(ln(1))

    Rearranging it to give a L'hopital's rule acceptable form of 0/0, gives:


    And taking the derivative:

    0/(-1/2n2) = 0

    And raising to e to get back to the original value:

    e0 = 1

    Except this is clearly incorrect, as we determined in class that the sequence converted to something around 22,000, and not 1.

    Other than this, I tried not canceling the (5/n), taking the log, and then splitting it using the limit law:

    lim[f(x)*g(x)] = lim(f(x)) * lim(g(x))

    But this didn't work either, as the limit of 2n by itself diverges as n goes to infinity, and this sequence is known not to diverge.
  2. jcsd
  3. Nov 20, 2012 #2


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    Do you know what the following limit is?

    [itex]\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n[/itex]
  4. Nov 20, 2012 #3


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    As I think you have recognized, this is not valid. It's true that (5/n) is going to zero, but 1 + 5/n never EQUALS 1. It's always strictly larger than 1, and meanwhile the exponent, 2n, is getting larger and larger. The result very much depends on how fast the exponent is growing vs. how fast (5/n) is shrinking.

    Well, this doesn't get you anywhere, because ln(1) = 0. I'm not sure what you were trying to do after that.

    Do you know the following limit?

    [tex]\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n[/tex]
  5. Nov 20, 2012 #4
    As I recall, it's e, but I never saw how that was proved or derived. The instructor I had for first semester calculus said that it was true "by definition".

    If it helps, the convergence several classmates say they found for the problem I'm having trouble with was precisely e10.
  6. Nov 20, 2012 #5


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    Your classmates are right. And you can use l'Hopital. But apply it to (1+(5/n))^(2n). Not 1^(2n), that would be foolish.
  7. Nov 20, 2012 #6
    Excellent, I've got it now. Thanks for the assistance.
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