Exponential bound for Euler's zeta function?

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Discussion Overview

The discussion centers around the behavior of Euler's zeta function, specifically examining the convergence of the series when s=1 and the implications of applying an exponent L to the divergent series.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant poses a question about the existence of an exponent L that could limit the finiteness of the expression (\sum_{n=1}^{\infty}1/n^s)^L when s=1.
  • Another participant states that when s=1, the series diverges, indicating that no value of L can render it finite.
  • A third participant reflects on the realization that applying an exponential to a diverging series is ineffective.
  • A later reply offers reassurance, suggesting that such misunderstandings are common.

Areas of Agreement / Disagreement

Participants generally agree that the series diverges when s=1, but the initial question about the exponent L introduces a point of contention regarding its implications.

Contextual Notes

The discussion does not resolve the broader implications of applying exponents to divergent series, nor does it explore the limits of convergence for other values of s.

Who May Find This Useful

This discussion may be of interest to those studying series convergence, the properties of the zeta function, or mathematical analysis involving divergent sequences.

Loren Booda
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Let Euler's zeta function be given by

<br /> \sum_{n=1}^{\infty}1/n^s<br />

Is there an exponent L which limits the finiteness of

<br /> (\sum_{n=1}^{\infty}1/n^s)^L<br />

for the case where s=1?
 
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When s=1, that series diverges, so there is no value of L that would make it finite.
 
Gib Z,

The ineffectiveness of an exponential on a diverging sequence should have been obvious to me.
 
Don't worry, we all have brain farts once in a while =]
 

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