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Proof for Convergent of Series With Seq. Similar to 1/n

  1. Apr 6, 2016 #1
    1. The problem statement, all variables and given/known data
    [itex]\sum\limits_{n=1}^{\infty}\frac{n-1}{(n+2)(n+3)}[/itex]

    2. Relevant equations
    [itex]S=\sum\limits_{n=1}^{\infty}a_n[/itex] (1)
    [itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\gt 1\rightarrow S\ is\ divergent[/itex] (2)
    [itex]\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\lt 1\rightarrow S\ is\ convergent[/itex] (3)
    [itex]\sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ convergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ convergent[/itex] (4)

    3. The attempt at a solution
    I suspect that the given series is divergent, thus the series would not be convergent. Thus I take the contrapositive of statement 4 which comes out to be [itex]\sum\limits_{n=1}^{\infty}\mid b_n\mid\ is\ divergent\ \land\ \mid b_n \mid\ \leq\ \mid a_n \mid\ for\ every\ n\rightarrow \sum\limits_{n=1}^{\infty}\mid a_n\mid\ is\ divergent[/itex].
    Let [itex]a_n=\frac{n-1}{(n+2)(n+3)}[/itex] and [itex]b_n = \frac{1}{n}[/itex]. The statement [itex]\mid b_n \mid\ \leq\ \mid a_n[/itex] is never true in the domain [itex][1,\infty)[/itex], so we cannot say that [itex]a_n[/itex] is divergent. Any suggestions?
     
  2. jcsd
  3. Apr 6, 2016 #2

    Samy_A

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    That's a good idea, comparing with the harmonic series ##\sum \frac{1}{n}##.

    Hint: ##\frac{n-1}{n+2}## is close to 1 for large n.
     
  4. Apr 7, 2016 #3

    LCKurtz

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    @The-Mad-Lisper Have you studied the limit comparison test yet?
     
  5. Apr 13, 2016 #4
    Only after the fact, I managed to modify Nicole Oresme's proof of the divergent nature of a harmonic series.
     
  6. Apr 13, 2016 #5

    Ray Vickson

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    The terms of your sum are all positive, so it is enough to give a lower bound series that diverges.

    For ##n > 6## we have ##n-1 > \frac{1}{2}(n+3)## and ##(n+2)(n+3) < (n+3)^2##, so
    [tex] \sum_{n=6}^{\infty} \frac{n-1}{(n+2)(n+3)} > \sum_{n=6}^{\infty} \frac{1}{2} \frac{n+3}{(n+3)^2}
    = \frac{1}{2} \sum_{k=9}^{\infty} \frac{1}{k} [/tex]
     
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