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Finding the divergence or convergence of a series

  1. Nov 27, 2011 #1
    Ʃ ,n=1,∞, (2/n^2+n)

    Does this series converge or diverge?

    Im not sure how to start can i use the comparison test here?
     
  2. jcsd
  3. Nov 27, 2011 #2

    LCKurtz

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    You can begin by making sure you stated the problem you meant to state. Is that expression (2/n2)+n, which is what you wrote would mean, or 2/( n2+n)?

    What have you tried? What are your thoughts about it?
     
  4. Nov 27, 2011 #3
    its 2/((n^2)+n)
    and I'm thinking it converges because you are adding an infinite amount of numbers that are continually getting smaller and smaller
     
  5. Nov 27, 2011 #4

    LCKurtz

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    Both convergent and divergent series have an infinite number of terms that get smaller and smaller, so that observation is of no value. For example, do you know the p series
    [tex]\sum \frac 1 {n^p}[/tex]
    and for what values of p it converges or diverges?

    Do you know how to use the comparison tests? What might you compare your series with and why?
     
  6. Nov 27, 2011 #5
    if p is greater than 1 the series converges
    im not very good at the comparison test and how to find what to cop are it to
     
  7. Nov 27, 2011 #6

    LCKurtz

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    The first step in a comparison test is to decide by examining your series whether you think it probably converges or diverges. This is usually decided by previous experience with previously analyzed series such as a p series. The next step depends on your expectation of convergence or divergence.

    If you think your series might converge, try to find a larger series that you know does converge. Then you have a series smaller than a known convergent series so it converges.

    If you think your series might diverge, see if you can find a smaller series that you know diverges. If your series if larger than a known divergent series, it must diverge.

    Your problem has similarities to a p series about which you know convergence or divergence. So think about that.
     
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