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Comparison test to determine convergence

  1. Nov 19, 2011 #1
    Use a comparison test to determine whether the series [itex]\sum[/itex] (n+1)/(n[itex]^{2}[/itex]+n+1) diverges or converges.

    I started out by simplifying the series to 1/n+1 and then from there I compared it to 1/n, which converges. 1/n is greater than 1/n+1 so based on the comparison test, the original series should also converge, is this correct? I also tried a limit comparison test and got n/n+1 which equals 2 which would mean that both Ʃa and Ʃb converge. I am kind of shady on my series and and getting very confused with this question.
    thanks!
    1. The problem statement, all variables and given/known data



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    3. The attempt at a solution
     
  2. jcsd
  3. Nov 19, 2011 #2
    You could use this inequality [itex]n^2+n+1\le n^2+2n+1= (n+1)^2\quad\forall n\in\mathbb{N}[/itex] so:

    [itex]\frac{1}{(n+1)^2}\le \frac{1}{n^2+n+1}[/itex]

    multiplying both sides for [itex]n+1[/itex], we have:

    [itex]\frac{1}{n+1}\le \frac{n+1}{n^2+n+1}[/itex]

    So

    [itex]\sum_{n=0}^\infty\frac{1}{n+1}\le \sum_{n=0}^\infty\frac{n+1}{n^2+n+1}[/itex]

    but
    [tex]\sum_{n=0}^{\infty}\frac{1}{n+1}=\sum_{m=1}^{\infty}\frac{1}{m}=\infty[/tex]

    therefore [itex]\sum_{n=0}^\infty\frac{n+1}{n^2+n+1}=\infty[/itex]
     
    Last edited: Nov 19, 2011
  4. Nov 20, 2011 #3

    HallsofIvy

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    First, what do you mean by "simplified the series to 1/n+ 1". "Simplifying" normally reducing to something equal by canceling, say. That is not the case here. Second, the series [itex]\sum (1/n)[/itex] does NOT converge.
     
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