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Series - comparison test

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to show that:

    [tex]\sum_{n=1}^\infty \frac{n^{2}}{2^n}[/tex]

    converges. I know I can compare it with the larger convergent geometric series:

    [tex]\sum_{n=1}^\infty \frac{1.5^{n}}{2^n}[/tex]

    Which is larger for all terms for n> 13.

    My question is, I found this "13" through trial and error. Is there any concrete way of determining when bn becomes larger than nc?

    Thanks!
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 20, 2009 #2
    You could use induction to show that it works for all n>13. However, using the ratio test or root test would be much easier for this series.
     
  4. Oct 20, 2009 #3
    Yeah the problem actually asks me to put an upper bound on Sn, so I needed to compare it to a geometric series with a known sum. So trial and error is the only way to go to get that 13?
     
  5. Oct 20, 2009 #4
    Finding that 13 by trial and error isn't important; what is important is that you can show that it's true for n>13.
     
  6. Oct 20, 2009 #5
    But I need to know at which point it becomes true to collect a partial sum...
     
  7. Oct 21, 2009 #6
    Ok, so basically the problem asks me to place an upper bound on:

    [tex]\sum_{n=1}^\infty \frac{n^{2}}{2^n}[/tex]

    So what I did was use this:

    [tex]\sum_{n=1}^{13} \frac{n^{2}}{2^n} + \sum_{n=13}^\infty \frac{1.5^{n}}{2^n}[/tex]

    as my bound.

    Is there a more methodical way than trial and error to figure out at which point the geomtric series is larger than the original one?
     
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