How fast does convergence have to be

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Discussion Overview

The discussion centers on the convergence of series, specifically comparing the harmonic series \(\sum 1/k\) and the series \(\sum 1/(k^{1+\epsilon})\) for \(\epsilon > 0\). Participants explore the mathematical concepts that determine the rate of convergence and question the reasoning behind these distinctions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants express confusion over why \(\sum 1/k\) diverges while \(\sum 1/(k^{1+\epsilon})\) converges, questioning the arbitrariness of the required rate for convergence.
  • One participant suggests the Cauchy condensation test as a relevant mathematical concept to consider in this context.
  • Another participant argues that the reasoning that terms becoming small implies convergence is flawed, emphasizing that the relative magnitudes of terms matter significantly.
  • Discussion includes the concept of limits of sequences, noting that the sequence of partial sums for the harmonic series grows without bound, indicating divergence.
  • Participants mention the integral test as a method to prove convergence or divergence, though opinions vary on its role as a definitive measure.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the reasoning behind the convergence rates, with multiple competing views and interpretations of the concepts involved.

Contextual Notes

There are unresolved assumptions regarding the definitions of convergence and divergence, as well as the implications of the integral test and other mathematical concepts discussed.

gottfried
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Reading through a real analysis textbook 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.
 
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gottfried said:
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.

See Cauchy condensation test.

Your specific example is related to the Riemann zeta function, which has a pole a s=1.
 
gottfried said:
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.
Using that faulty argument, any infinite sum having terms approaching zero, will converge.
Essentially what I'm asking is who or what mathematical concept determined the required rate for convergence because it seems arbitrary to me.
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.
 
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.
 
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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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