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How fast does convergence have to be

  1. Oct 24, 2012 #1
    Reading through a real analysis text book I noticed that [itex]\sum[/itex] 1/K
    diverges but [itex]\sum[/itex] 1/K1+[itex]\epsilon[/itex] converges for all [itex]\epsilon[/itex] > 0.

    This is confusing because 1/k will eventually be equally as small as the terms in 1/K1+[itex]\epsilon[/itex] and therefore it should also converge. It may take much longer but surely the numbers eventually become equally small and adding more on becomes pointless.

    I also saw that one can prove this using the integral test but the integral test isn't the 'decider' there must a more rigorous reason.

    Essentially what I'm asking is who or what mathematical concept determined the required rate for convergence because it seems arbitrary to me.
     
    Last edited: Oct 24, 2012
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  3. Oct 24, 2012 #2

    pwsnafu

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    See Cauchy condensation test.

    Your specific example is related to the Riemann zeta function, which has a pole a s=1.
     
  4. Oct 24, 2012 #3
    Using that faulty argument, any infinite sum having terms approaching zero, will converge.
    You can start with the mathematical concept of a finite sum. For example, you can easily verify that finite partial sums of the harmonic series can exceed any given number, so there should be no talk of convergence in that case.
     
  5. Oct 24, 2012 #4
    The concept is limits of sequences. In the case of series, the sequences are the partial sums. If you have that epsilon, then the sequence of partial sums converges to a number. Without the epsilon, the sequence does not converge because it grows without bound. And the integral test is the "decider" because it can be used to prove those statements.

    See http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)) for a few good explanations (including a picture that explains the integral test) for why it diverges.
     
    Last edited: Oct 24, 2012
  6. Oct 24, 2012 #5

    arildno

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    "This is confusing because 1/k will eventually be equally as small as the terms in 1/K1+ϵ and therefore it should also converge."

    This is conletely wrong!

    If you look at the RELATIVE magnitudes of like terms, this will go as (1/K)ê, i.e, go to zero.
    Thus, as K increases, the terms in the convergent sequence converges much faster to zero than for the harmonic series.
     
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