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Prove that the series converges

  1. Nov 22, 2012 #1
    1. The problem statement, all variables and given/known data

    If [itex]\sum(an)[/itex] converges, and an>0, prove that [itex]\sum\sqrt{an}/n[/itex] converges

    2. Relevant equations



    3. The attempt at a solution

    I'm trying to use the comparison test, since an>0. So I have to prove that an>[itex]\sqrt{an}[/itex]/n. But I keep getting stuck here because an approaches zero, so an is not greater than [itex]\sqrt{an}[/itex]
     
  2. jcsd
  3. Nov 22, 2012 #2

    haruspex

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  4. Nov 22, 2012 #3
    I think so!

    If [itex]\sum\sqrt{an}/n[/itex] diverges, then the limit as n approaches infinity of [itex](\sqrt{a(n+1)}/(n+1))/\sqrt{an}/n[/itex] is greater than or equal to 1, which implies that the limit as n approaches infinity of [itex]\sqrt{a(n+1)}/\sqrt{an}[/itex] is greater than or equal to 1, which implies that the limit as n approaches infinity of a(n+1)/an is strictly greater than 1, which implies divergence.
    A contradiction, thus the above series is not divergent.

    I think that's it. If so, thank you so much!
     
  5. Nov 22, 2012 #4

    haruspex

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    No, that isn't what I had in mind. [itex](\sqrt{a(n+1)}/(n+1))/\sqrt{an}/n[/itex] might not converge to a limit.
    Thinking about it some more, I don't think the "comparison test 2" can help after all. The problem is that the ratio of consecutive terms of an could go the 'wrong' way occasionally. E.g. a2n+1 = a2n/10 = a2n-1/5.
    But I think I see how to do it. Try partitioning the an series into two sums, one consisting of those terms ≤ 1/n2, and one consisting of those > 1/n2
     
  6. Nov 22, 2012 #5

    LCKurtz

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    Here's a different hint. Since$$
    0\le (a-b)^2= a^2-2ab+b^2$$it follows that$$
    ab\le \frac{a^2+b^2}{2}$$holds for any numbers ##a## and ##b##. See if you can use that.
     
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