Why Does Summation Not Converge?

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Discussion Overview

The discussion revolves around the convergence of various infinite series, specifically comparing series such as summation from n=1 to infinity of (-1)^n, i^n, 1/n, -1/n, and 1/n^2. Participants explore the conditions under which these series converge or diverge, examining the implications of terms tending towards zero.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions why certain series do not converge despite their terms tending to zero, specifically citing (-1)^n, i^n, 1/n, and -1/n.
  • Another participant clarifies that while terms tending towards zero is necessary for convergence, it does not ensure it, using the harmonic series as an example of divergence.
  • A participant notes that (-1)^n and i^n do not tend to zero, suggesting that this affects the convergence of their associated series.
  • One participant introduces the alternating series summation from k=1 to infinity of (-1)^(k+1)/k, claiming it converges to ln(2), prompting further exploration of this series.
  • Another participant suggests that the Taylor series for the logarithm may be relevant to understanding the convergence of the series mentioned.
  • There is a proposal to start from the series 1/(1+x) and integrate it term by term to derive a related formula, with a specific interest in evaluating it at x=1.
  • A participant expresses understanding of the previous points but seeks clarification on another series involving the digamma function and the Euler-Mascheroni constant.

Areas of Agreement / Disagreement

Participants express differing views on the convergence of specific series, with some agreeing on the divergence of certain series while others propose methods to demonstrate convergence for others. The discussion remains unresolved regarding the broader implications of convergence criteria.

Contextual Notes

Limitations include varying interpretations of convergence criteria, dependence on specific series definitions, and unresolved mathematical steps in the proposed methods for demonstrating convergence.

hola
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Hey. This has been bugging me for a long time:
why does summation from n=1 to infinity of (-1)^n or i^n or 1/n or -1/n not converge, because summation from n=1 to infinity of 1/n^2 conveges. Don't the terms in (-1)^n or i^n or 1/n or -1/n(-1)^n tend to 0?
 
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While tending towards 0 is a requirement for convergence, it does not guarantee convergence. The most classic example, which you cited, would be the harmonic series, 1/n which does have a limit of 0 but still diverges. Though a limit of anything other than 0 would guarantee divergence.
 
hola said:
Hey. This has been bugging me for a long time:
why does summation from n=1 to infinity of (-1)^n or i^n or 1/n or -1/n not converge, because summation from n=1 to infinity of 1/n^2 conveges. Don't the terms in (-1)^n or i^n or 1/n or -1/n(-1)^n tend to 0?
[itex](-1)^n[/itex] and [itex]\quad i^n[/itex] do not tend to zero, so the associate series is not convergent.
 
Well,the trick is with the series
[tex]\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}[/tex].
IIRC it converges to
[tex]\ln 2[/tex].

Daniel.
 
dextercioby said:
Well,the trick is with the series
[tex]\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}[/tex].
IIRC it converges to
[tex]\ln 2[/tex].

Daniel.

Hum, interesting. Think I might try to show that. From your post the other day, seems Euler Gamma functions may be involved.
 
I think it's just a matter of using the Taylor series for the logarithm.
 
You can start from this series
[tex]\frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}-...[/tex]

and integrate it term by term and you'll have to put x=1 in the new series to find the formula which I've hinted.

Daniel.
 
dextercioby said:
You can start from this series
[tex]\frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}-...[/tex]

and integrate it term by term and you'll have to put x=1 in the new series to find the formula which I've hinted.

Daniel.

Thanks, I understand now (should know that I know). Actually I'd like to understand this one:

[tex]\sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\psi(n+1)+\frac{1}{2}\gamma[/tex]

I'll do some work on it and if I can't figure it out, I'll make a separate post in the general math and well, ask for help.
 

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