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Question about the integral test

  1. Jul 24, 2011 #1
    1. The problem statement, all variables and given/known data

    We have to determine whether [itex]\sum 1/n^2 + 4[/itex]
    is convergente or divergent
    2. Relevant equations

    I'm trying to work the problem through trigonometric substitution. I was wondering if I could just determine that by the P-series test, the function 1/n^2 will always be larger than the other one, since p is greater than 1 in this case, both are convergent.
    3. The attempt at a solution
     
  2. jcsd
  3. Jul 24, 2011 #2

    tiny-tim

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    welcome to pf!

    hi salazar888! welcome to pf! :smile:
    yes, always 0 < 1/(n2+4) < 1/n2,

    the latter sum converges (from the p-series test), so so must the former (from the direct comparison test) :wink:
     
  4. Jul 24, 2011 #3

    Ray Vickson

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    Obviously, since the series [itex] \sum \frac{1}{n^2} [/itex] converges, the sum you wrote, [itex] 4 + \sum \frac{1}{n^2}[/itex] converges also.

    RGV
     
    Last edited: Jul 24, 2011
  5. Jul 24, 2011 #4
    The series 1/n does not converge. It's the harmonic series. You read it wrong. I get what you're saying though.
     
  6. Jul 24, 2011 #5

    Ray Vickson

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    I saw the error and edited it immediately.

    RGV
     
  7. Jul 24, 2011 #6

    HallsofIvy

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    But the series
    [tex]\sum\left(\frac{1}{n^2}+ 4\right)[/tex]
    does NOT converge!
     
  8. Jul 24, 2011 #7

    Ray Vickson

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    I agree, but that is not what he wrote. We all know he meant sum 1/(n^2 + 4), but he wrote sum (1/n^2) + 4, which is very different according to standard math expression padding rules. Since he was using 'tex' anyway, he should have been able to enter "{n^2+4}" as the second argument of the '\frac' command.

    RGV
     
  9. Jul 24, 2011 #8
    Yes it was my fault. I've only been on the forum for a couple of days. Thanks for the help guys. I will improve at typing the commands.
     
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