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Divergent Series question

  1. Sep 26, 2011 #1
    (1) Using the Archimedean definition of divergence, prove that if [itex]\sum_{i=1}^{\infty }x_{i}[/itex] diverges to infinity, then so does either [itex]\sum_{i=1}^{\infty }x_{2i}[/itex] or [itex]\sum_{i=1}^{\infty }x_{2i+1}[/itex].

    (2) Show an example where [itex]\sum_{i=1}^{\infty }x_{i}[/itex] diverges to infinity but [itex]\sum_{i=1}^{\infty }x_{2i}[/itex] does not.

    So this is what I have for (1):

    If [itex]\sum_{i=1}^{\infty }x_{i}[/itex] diverges then for all N, there is a number m such that [itex]N < \sum_{i=1}^{\infty }x_{i}[/itex] diverges for all [itex]k\geq m[/itex]. Then, [itex]\lim_{k \to \infty }\left | \frac{x_{k+1}}{x_{k}} \right | > 1[/itex] or [itex]\lim_{k \to \infty }\left | \frac{x_{k+1}}{x_{k}} \right | = \infty[/itex] and [itex]\lim_{k \to \infty }\left | \frac{x_{k+1}}{x_{k}} \right |[/itex] will become larger than 1 at some point. There exists an m such that [itex]\left | \frac{x_{k+1}}{x_{k}} \right | > 1[/itex] when [itex]k\geq m[/itex]. This means that when [itex]k\geq m[/itex], [itex]x_{k} < x_{k+1}[/itex] and [itex]\lim_{k \to \infty }x_{k} \neq 0[/itex]. [itex]x_{k} < x_{k+1}[/itex] implies that either [itex]x_{k} < x_{2k}[/itex] or [itex]x_{k} < x_{2k+1}[/itex], depending on whether or not the series is alternating. So either [itex]\sum_{i=1}^{k }x_{i} < \sum_{i=1}^{k }x_{2i}[/itex] or [itex]\sum_{i=1}^{k }x_{i} < \sum_{i=1}^{k }x_{2i+1}[/itex], and we have a number m such that [itex]N < \sum_{i=1}^{k }x_{i} < \sum_{i=1}^{k }x_{2i}[/itex] or [itex]N < \sum_{i=1}^{k }x_{i} < \sum_{i=1}^{k }x_{2i+1}[/itex] for all [itex]k\geq m[/itex]. Therefore, [itex]\sum_{i=1}^{\infty }x_{2i}[/itex] or [itex]\sum_{i=1}^{\infty }x_{2i+1}[/itex] diverges.

    Is this right?

    I can't think of an example for (2). Any help?
    Last edited: Sep 26, 2011
  2. jcsd
  3. Sep 26, 2011 #2
    I don't seem to understand the first line of your proof, but I think that's my fault: I've never heard of the "archimedean definition of divergence", so never mind me.

    But I wanted to comment on (2): it's always handy to think of the most extreme case, i.e. one where [itex]\sum x_{2i}[/itex] is zero, what's an obvious candidate? Use the left-over freedom to construct a divergence sequence.
  4. Sep 27, 2011 #3

    Can anyone look over my proof and tell me if there are any logical errors?
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