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I have a problem with the Harmonic Series, please School me

  1. Dec 4, 2007 #1
    Looking at the Harmonic series [tex]\sum_{k=1}^{\infty}\frac{1}{k}[/tex]
    I at first justed accepted my teachers assertion that it diverges. I still do, not to mention that I trust my textbook (and Nicole Oresme!).

    However in seems counterintuitive to me (and since it is math, I guess that is okay) .

    It just seems that by virtue of the fact that [tex] \lim_{\k\rightarrow \infty}\frac{1}{k}=0[/tex] that at some point we would be adding zero to the sum.

    Someone please help to clarify this.

    Thanks,
    Casey

    p.s. I don't know why the "k" is not showing up under the limit sign:(
     
  2. jcsd
  3. Dec 4, 2007 #2

    Dick

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    At what point would you be adding zero to the sum? 1/k is never zero.
     
  4. Dec 4, 2007 #3
    Here's one reasoning:

    1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

    Can be grouped as follows:

    1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + ... + 1/15 + 1/16) + ...

    Notice that

    1/3 + 1/4 > 1/4 + 1/4 = 2(1/4) = 1/2

    1/5 + 1/6 + 1/7 + 1/8 > 1/8 + 1/8 + 1/8 + 1/8 = 4(1/8) = 1/2

    1/9 + 1/10 + ... + 1/15 + 1/16 > 1/16 + 1/16 + ... + 1/16 + 1/16 = 8(1/16) = 1/2

    You can do this indefinitely, so that the harmonic series is always bigger than

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

    which clearly diverges.
     
  5. Dec 4, 2007 #4
    That is my problem, I can't say when, just like you can't. But if the upper "bound" is infinity that means that at some point you have the "term 1/infinty"
    and that equals zero.

    I know that my reasoning is flawed in that it is impossile to determine where that occurs.....but it seems that it does.

    Casey
     
  6. Dec 4, 2007 #5
    No, infinity is not a number. It's a concept. The fraction 1/k is NEVER zero, because no value of k will make that fraction 0. 1/k approaches 0, or we say that

    [tex]\lim_{k\to \infty} \frac{1}{k} = 0 [/tex]

    but again, no value of k will produce 1/k = 0. So "intuitively," you are indeed adding on terms that are extremely small. However, they are still "big enough" so as to contribute a "large amount" when you have "enough" of them. Just look at the explanation I gave earlier on how "enough" of these small values can be combined so as to contribute a significant amount (1/2).
     
  7. Dec 4, 2007 #6

    Dick

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    Well, it just doesn't. Zero is the best number to describe the behavior of 1/k for large k. But 1/k is never exactly zero. You KNOW that. Now try to digest it.
     
  8. Dec 5, 2007 #7

    dynamicsolo

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    Have you dealt with integrals as the infinite limit of Riemann sums? There is more that could be said on this topic because the following question is left begging. The general term 1/k is never zero, so you are always adding something to the series. Why is it then that the infinite sum of terms 1/k diverges, and the infinite sum of terms, say, 1/(k^0.5) diverges, but the infinite sum of terms, say, 1/(k^2) converges? Something called the Integral Test for infinite series helps deal with this. [As it turns out, the harmonic series is the "threshold case" for infinite sums of terms (1/k^p)... The point is that there is a sort of "rate" at which additional terms approaching zero can be added to still lead to a finite sum; excedding that "rate" produces a sum which will not be finite!]
     
    Last edited: Dec 5, 2007
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