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Proving the divergence of a Harmonic Series

  1. Oct 31, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that Hn converges given that:

    [tex]H_{n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}[/tex]




    3. The attempt at a solution
    First I supposed that the series converges to H:

    [tex]H_{n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\geq1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{n}[/tex]

    Which implies that [tex]H_{n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\geq1+H_{n}[/tex]

    Which contradicts with the series converging. Hence the series doesn't converge. I know there is another proof where you set H to be greater than an infinite sum of 1/2's but I wanted to think of something else. Is this correct? Any help would be appreciated.
     
    Last edited: Oct 31, 2013
  2. jcsd
  3. Oct 31, 2013 #2
    On the track; not quite right.

    To prove something diverges -- which this does -- you need to show that it gets greater than any arbitrary integer N . There are various proofs of the divergence of the harmonic series, but the most common one begins like this:

    1 +1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + .... 1/16 ) + ...

    If you can show that the bracketed terms are all larger than 1/2, then you are adding up infintiely many 1/2's. So you can always get larger than any N.
     
  4. Nov 1, 2013 #3
    Yes that's the proof i'm talking about because the first bracketed terms are greater than 1/4+1/4=1/2
    and the next 4 terms are greater than 1/8+1/8+1/8+1/8 and so on and so forth and at the end you have a sum that's greater than an infinite sum of 1/2. That's it but I was trying to see if the proof I made up was on the right path.
     
  5. Nov 1, 2013 #4
    Let's look at an example: ##H_6## = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 > 1 + 1/2 + 1/4 + 1/4 + 1/6 + 1/6 = 1 + 1/2 + 1/2 + 1/3 = 1/2 + ##H_3##. And checking one step further we can see that ##H_8## > 1/2 + ##H_4##. This is not what you wrote for the general case.

    In general then you have ##H_n## > 1/2 + ##H_m## where m = n/2 if m is even and m = (n-1)/2 if n is odd, i.e. m = ##\lfloor{n/2}\rfloor##

    In and of itself this is not a proof. You can extend it into a proof by setting up a recursion:

    ##H_n## > 1/2 + ##H_{\lfloor{n/2}\rfloor}## > 1/2 + 1/2 + ##H_{\lfloor{n/2}\rfloor/2}## , etc. Perhaps you would like to write up this approach.

    You came up with a good idea, and that shows me you are interested in mathematics and able to be creative about it.
     
  6. Nov 1, 2013 #5
    Or how about if I restate the given information and say prove that the harmonic series:

    1+1/2+1/3+1/4+... diverges.

    Then now i can suppose that the harmonic series converges to H and we let:

    H=1+ 1/2+ 1/3+ 1/4+ 1/5+ 1/6+ 1/7+ 1/8+...≥1+ 1/2+ 1/4+ 1/4+ 1/6+ 1/6+ 1/8+ 1/8+...

    which equals 1+1/2+1/2+1/3+1/4+... = 1/2+H .

    And the inequality holds because 1/3+ 1/4≥ 1/4+ 1/4 and 1/5 + 1/6 ≥1/6+1/6 and so on and so forth.

    So we supposed that the harmonic series converges to H yet we found that H≥1/2 + H , which is a contradiction so therefore the series diverges. I think this is set up much better.
     
  7. Nov 1, 2013 #6
    Actually, you don't need a contradiction. You are looking at ##H_n##. Whether there is an H or not you can show ##H_n## > 1/2 + ##H_{n1}## > 1/2 + 1/2 + ##H_{n2}## > ... where each nk is more or less half the preceding one.

    If you want to show ##H_n## gets greater than some integer M, you just have to pick n big enough to generate 2M 1/2's. I think that's probably n = 2##^{2M}##.
     
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