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Convergent sequence property and proving divergence

  1. Mar 18, 2012 #1
    I feel like I'm missing something obvious, but anyway, in the text it states:

    lim as n→∞ of an+bn = ( lim as n→∞ of an ) + ( lim as n→∞ of bn )

    But say an is 1/n and bn is n. Then the limit of the sum is n/n = 1, but the lim as n→∞ of bn doesn't exist and this property doesn't work...


    Second thing, I was reading an example of how to prove a sequence is divergent, specifically the sequence [(-1)n], and the text proves it by contradiction. "Take any positive number [itex]\epsilon[/itex]<1. Then the interval (a - [itex]\epsilon[/itex], a + [itex]\epsilon[/itex]) has length less than 2. Therefore, it is not possible to have both odd and even terms of sequence in this interval, therefore the sequence [(-1)n] cannot converge to a."

    Why does not being able to possess both odd and even terms in that interval mean that the sequence doesn't converge? It seems like a random leap of logic to me...
     
  2. jcsd
  3. Mar 18, 2012 #2

    jgens

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    The theorem should properly be written as "If [itex]\lim_{n \to \infty}a_n[/itex] and [/itex]\lim_{n \to \infty}b_n[/itex] exist, then [itex]\lim_{n \to \infty} (a_n + b_n) = (\lim_{n \to \infty} a_n) + (\lim_{n \to \infty} b_n)[/itex].

    Now notice [itex]n < n + n^{-1}[/itex]. So [itex]n + n^{-1}[/itex] diverges as [itex]n \to \infty[/itex].

    A sequence [itex]\{x_n\}_{n \in \mathbb{N}}[/itex] converges to [itex]x[/itex] if and only if for every open neighborhood [itex]U[/itex] of [itex]x[/itex], there exists an [itex]N \in \mathbb{N}[/itex] such that [itex]x_n \in U[/itex] whenever [itex]N \leq n[/itex]. The example shows that regardless of which [itex]x \in \mathbb{R}[/itex] you choose, there is an open neighborhood of [itex]x[/itex] which does not contain infinitely many of the [itex]x_n[/itex]. This means that [itex]\{x_n\}_{n \in \mathbb{N}}[/itex] does not converge.
     
  4. Mar 18, 2012 #3
    Thanks for the prompt response!

    Okay, so to try to simplify it/put it in more understandable phrasing (for my own sake): for a sequence to be proven divergent, I just have to show that for any x there is an open neighborhood of x such that the sequence does not have infinitely many terms contained in that neighborhood?
     
  5. Mar 18, 2012 #4

    jgens

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    That is correct. More explicitly a sequence [itex]\{x_n\}_{n \in \mathbb{N}}[/itex] converges to [itex]x[/itex] if and only if for all [itex]0 < \varepsilon[/itex] there exists [itex]N \in \mathbb{N}[/itex] such that [itex]N \leq n[/itex] implies [itex]|x_n - x| < \varepsilon[/itex]. This means that the sequence [itex]\{x_n\}_{n \in \mathbb{N}}[/itex] does not converge to [itex]x[/itex] if and only if there exists [itex]0 < \varepsilon[/itex] such that for all [itex]N \in \mathbb{N}[/itex] there is some [itex]N \leq n[/itex] such that [itex]\varepsilon < |x_n - x|[/itex]. This is all just the standard [itex]\varepsilon,\delta[/itex] stuff that you learn in basic calculus classes.

    Now let's look at the non-convergent case again. What it is saying is that we can find an [itex]0 < \varepsilon[/itex] such that no matter how big we choose [itex]N[/itex], we can always find an [itex]N \leq n[/itex] with [itex]x_n \notin (x-\varepsilon,x+\varepsilon)[/itex]. From here it is pretty easy to work out this the open neighborhood [itex](x-\varepsilon,x+\varepsilon)[/itex] does not contain infinitely many of the [itex]x_n[/itex].
     
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