1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Absolute Value Theorem

  1. Jul 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the limit of this sequence:

    (-1)n(2n3)/(n3+1)

    2. Relevant equations



    3. The attempt at a solution

    At first glance I would attempt to use the absolute value theorem which gives an answer of 2 as n approaches infinite. My question is, when the theorem fails ( limit =/ 0) then does that mean the sequence is divergent? I haven't seen any documentation that explicitly states an is divergent if the absolute value theorem doesn't hold.
     
  2. jcsd
  3. Jul 25, 2013 #2

    Mark44

    Staff: Mentor

    By taking the absolute value, you're essentially stripping off the factor of (-1)n, which isn't legitimate to do. If you write down 5 or 6 terms of the sequence, you should get an idea of what it is doing.
     
  4. Jul 26, 2013 #3
    From an analysis point of view assume that there is some limit L. Then, by definition, for any epsilon there must exist some N such that for all n>N. Take a small epsilon, less than the limit if there was no (-1)^n, and then show that the next term is larger than said epsilon.
     
  5. Jul 27, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, that's problemlematic with this limit!

    What I would do is divide both numerator and denominator by [itex]n^3[/itex]. That gives
    [tex](-1)^n\frac{2}{1+ \frac{1}{n^3}}[/tex]

    If it were not for that "[itex](-1)^n[/itex]" we could immediately say that the limit, as n goes to infinity, is 2. But what does that say about the limit with that "[itex](-1)^n[/itex]"?
     
  6. Jul 27, 2013 #5
    The question basically boils down to [itex]\lim_{n\rightarrow\infty}(-1)^n[/itex]. Does this limit converge? It was pointed out that with this term removed, the limit converges to the value of 2. This term switches between -1 and +1 every time we increase [itex]n[/itex]. Intuitively what does that mean?
     
  7. Jul 27, 2013 #6

    pasmith

    User Avatar
    Homework Helper

    That's because an arbitrary [itex](a_n)[/itex] might still converge. If [itex]|a_n| \to L > 0[/itex] then the three possibilities are that [itex]a_n \to L[/itex], or [itex]a_n \to -L[/itex], or [itex](a_n)[/itex] diverges, having the property that for any [itex]\epsilon > 0[/itex] there exists [itex]N \in \mathbb{N}[/itex] such that for all [itex]n \in \mathbb{N}[/itex], if [itex]n \geq N[/itex] then either [itex]|a_n - L| < \epsilon[/itex] or [itex]|a_n - (-L)| < \epsilon[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Absolute Value Theorem
  1. Absolute value (Replies: 7)

  2. Absolute value (Replies: 10)

  3. Absolute Values (Replies: 2)

Loading...