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Convergence/Divergence of an Infinite Series

  1. Oct 8, 2016 #1
    1. The problem statement, all variables and given/known data
    To Determine Whether the series seen below is convergent or divergent.
    2. Relevant equations
    ∑(n/((n+1)(n+2))) From n=1 to infinity.
    3. The attempt at a solution
    Tried to use the comparison test as the bottom is n^2 + 3n + 2, comparing to 1/n. However, this does not work as the first series is "smaller" than the harmonic series and thus it can't be proved to be divergent. I also tried to separate it into partial fractions but I don't know how to prove it is divergent from there. Thanks a lot for any help.
  2. jcsd
  3. Oct 8, 2016 #2


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    2017 Award

    Staff: Mentor

    Is it still smaller if you start the harmonic series at six? And does this make the harmonic series convergent?
  4. Oct 8, 2016 #3


    Staff: Mentor

    Have you seen the limit comparison test?

    Also, this is definitely not a precalculus problem, so I have moved it to the calculus section.
  5. Oct 8, 2016 #4
    Thanks for the response. So to prove using comparison just start harmonic series at six. Makes sense, I hadn't thought of that.
    Referring to Mark44's response, no, I haven't heard of the limit comparison test, however, I will look it up.
    Thanks heaps for your time and input.
  6. Oct 9, 2016 #5

    Ray Vickson

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    I don't see how starting the harmonic series at six solves the problem, unless you also do some other things at the same time.

    To me, the simplest solution is to note that for ##n \geq 1## we have ##n/(n+1) \geq 1/2##, so that
    $$t_n \equiv \frac{n}{(n+1)(n+2)} \geq \frac{1}{2} \frac{1}{n+2}.$$
    That means
    $$\sum_{n=1}^N t_n \geq \sum_{n=1}^N \frac{1}{2} \frac{1}{n+2} = \frac{1}{2} \left[\frac{1}{3}+ \frac{1}{4} + \cdots + \frac{1}{N+2} \right] $$
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