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Direct Comparison Test

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that:

    [itex]\sum \frac{3}{n^{2} + 1}[/itex]

    converges from n = 1 to ∞

    2. Relevant equations

    If Ʃbn converges, and Ʃan < Ʃbn.

    Ʃan also converges.


    3. The attempt at a solution

    [itex]\sum \frac{1}{n^{2}}[/itex] converges

    [itex]\sum \frac{3}{n^{2} + 1}[/itex] = 3 * [itex]\sum \frac{1}{n^{2} + 1}[/itex]

    [itex]\sum \frac{1}{n^{2} + 1}[/itex] < [itex]\sum \frac{1}{n^{2}}[/itex] for all n from 1 to ∞.

    Therefore [itex]\sum \frac{1}{n^{2} + 1}[/itex] converges.

    Therefore [itex]\sum \frac{3}{n^{2} + 1}[/itex] also converges.

    The problem I am having is if the 3 remained in the summation.

    [itex]\sum \frac{3}{n^{2} + 1}[/itex] is not less than [itex]\sum \frac{1}{n^{2}}[/itex] for all n from 1 to ∞.

    Why does placing the 3 outside the summation make the problem work?
     
  2. jcsd
  3. May 6, 2013 #2

    SammyS

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    If you leave the 3 inside the sum, you will have to find a different converging series to compare it to. The method will still work.
     
  4. May 6, 2013 #3
    So that means I would need to compare the series to [itex]\frac{3}{n^{2}}[/itex] instead.

    I can prove [itex]\frac{3}{n^{2}}[/itex] through the integral test. Though speaking of the integral test, it doesn't seem like the test works for the original series since I am getting tan-1(∞). So would the integral test be inconclusive in this case?
     
  5. May 6, 2013 #4

    Dick

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    arctan(∞), which is what I assume is what you mean by tan^(-1)(∞), is finite.
     
  6. May 6, 2013 #5

    Ray Vickson

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    You are making it WAAAAAY too complicated: it is a simple fact---easily proved---that if c is a constant, then ##\sum_n c a_n = c \sum_n a_n##, in the sense that if one side converges, so does the other (and the two sides are equal), and if one side diverges so does the other.
     
  7. May 6, 2013 #6

    SammyS

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    No. You would not necessarily have to compare to [itex]\displaystyle \sum_{n=1}^\infty\frac{3}{n^{2}}\ .[/itex]

    There are many other possibilities.
     
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