At what point do the sums of the reciprocals converge?

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Discussion Overview

The discussion centers around the convergence of series formed by the sums of reciprocals, specifically exploring the transition point at which these sums switch from divergence to convergence based on the power of the denominators. The scope includes theoretical aspects of series convergence and mathematical reasoning.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant notes that the Harmonic Series diverges while the series of reciprocals of squares converges, questioning the power at which this transition occurs.
  • Another participant references the integral test for convergence, stating that the series converges for powers greater than one, as indicated by the integral of the function.
  • A later reply expresses appreciation for the explanation provided regarding the convergence criteria.
  • One participant shares a link to an external resource, suggesting it may be of interest to the discussion.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the specific power at which the sums switch from divergence to convergence, though there is acknowledgment of the integral test's implications.

Contextual Notes

The discussion does not clarify the assumptions behind the convergence criteria or the specific conditions under which the integral test applies.

thegreatjared
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It is well known that the Harmonic Series diverges (1/1+1/2+1/3+1/4+...), but that the series [1/1+1/4+1/9+1/16+...] converges. In the first series, the denominators are the integers, whereas in the second example, the denominators are the integers to the power of 2. My question is, at what power do the sums of the reciprocals "switch" from divergence to convergence?
 
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The integral test for convergence tells us that the infinite sum:
\sum_1^\infty f(n)
and the integral:
\int_1^\infty f(n)
either both converge or both diverge. Since:
\int_1^\infty \frac{1}{x^{1+\delta}} = \frac{1}{\delta}
This tells us that the infinite series:
\sum_1^\infty \frac{1}{x^{1+\delta}}
will converge as long as delta is greater than zero, no matter how small it is.
 
Wow, simple as that. Thanks!
 

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