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Divergence of a sequence

  1. Nov 8, 2012 #1
    I'm trying to understand divergence of a sequence (not series). What methods can I use to prove divergence? I know that convergence can be proven using various methods, such as squeeze theorem and sum, difference, product and quotient rule etc.

    Could I use the following to prove divergence?

    If [itex] a_{n} [/itex] is a sequence of real numbers, [itex] f(n) = a_{n} [/itex] and [itex] \lim_{n→∞} f(n) [/itex] does not exist, but is not equal to ∞ or -∞, does [itex] a_{n} [/itex] necessarily diverge?

    If [itex] a_{n} [/itex] is a sequence of real numbers, [itex] f(n) = a_{n} [/itex] and [itex] \lim_{n→∞} f(n) = ∞ [/itex], does [itex] a_{n} [/itex] necessarily diverge?

    These two ideas will greatly facilitate my understanding of sequence divergence.

  2. jcsd
  3. Nov 8, 2012 #2
    Yes to both questions.
  4. Nov 8, 2012 #3
    Hey micro, but what about the sequence [itex] a_{n} = sin(2πn) [/itex]. It is the case that
    [itex] \lim_{n→∞}f(n) [/itex] does not exist, yet the limit of [itex]a_{n}[/itex] converges to 0, right??

  5. Nov 8, 2012 #4
    The limit [itex]\lim_{n\rightarrow +\infty} f(n)[/itex] does exist and is zero. (I assume that n is always an integer)

    However, if you extend f to [itex]f(x)=\sin(2\pi x)[/itex] for [itex]x\in\mathbb{R}[/itex], then the limit [itex]\lim_{x\rightarrow +\infty} f(x)[/itex] doesn't exist.
  6. Nov 8, 2012 #5


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    Why do you say that?

    Edit: ninjaed
  7. Nov 8, 2012 #6
    micromass, I'm sorry I think I misphrased my question. When I refer to f(n) in my original post, I refer to it as a function with domain ℝ as opposed to [itex]a_{n}[/itex] which I take to be defined only for natural numbers.

    Given this clarification, which of the following original statements is true and why?

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