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Prove that the harmonic series is divergent

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that the [tex]\sum1/n[/tex] is divergent.

    Does anyone know a simple proof for this. I understand that it does not converge intuitively but I'm not sure how to prove it in symbols.

    Thank you for your help.
    M

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 6, 2010 #2

    rock.freak667

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    Homework Helper

    The integral test works well for that one.
     
  4. Apr 6, 2010 #3
    Cauchy condensation, sum(a_n) converges <=> sum(2^k*a_(2k)) converges, only works for monotonically decreasing a_n though
     
  5. Apr 6, 2010 #4
    I tried that and it works really well, but my professor told us that we haven't properly defined what log is. Therefore we cannot use log to prove anything at this point. Which is a bummer. Is there another way to approach this beside the integral test?
     
  6. Apr 6, 2010 #5
    For Cauchy condensation?? You use log?? You can prove that test true by the comparison test. No logs required
     
  7. Apr 6, 2010 #6
    [tex]
    H = \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ...
    [/tex]

    Notice that 1/3 + 1/4 > = 1/4 + 1/4 = 1/2.

    And that that 1/5 + 1/6 + 1/7 + 1/8 > 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2

    So, in essence,

    [tex]

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

    The latter part diverges, of course.

    EDIT: This is the intuitive way. There is also an elementary, but rigorous way of doing it without the integral test.
     
  8. Apr 6, 2010 #7
    You pick one of the a_2^k terms and since it's decreasing,
    2^k*a_2^k
    <a_(2^k)+...
    +a_(2^(k+1)-1) , by summing over these you Get series comparison
     
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