Why does the sum 1/log(n) behave like 1/log(n)?

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Discussion Overview

The discussion revolves around the behavior of the sum \(\frac{1}{n} \sum_{k=3}^n \frac{1}{\log k}\) and its relationship to \(\frac{1}{\log n}\). Participants explore methods to establish equivalences and address convergence issues related to series and integrals.

Discussion Character

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

Main Points Raised

  • One participant notes that in their probability course, the sum \(\frac{1}{n} \sum_{k=3}^n \frac{1}{\log k} \propto \frac{1}{\log n}\) is used, seeking clarification on this relationship.
  • Another participant suggests using integrals to bound the sum as a method to find an equivalent expression.
  • Concerns are raised about the convergence of the series \(1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3}\), with references to factorial terms and divergent series.
  • There is a discussion about asymptotic series, where one participant emphasizes the importance of considering a limited number of terms to derive an equivalent, correcting a typographic error in their previous attachment.
  • A new question is introduced regarding the summation of terms like \(\frac{1}{\log(n - 2i)}\), which leads to a reminder about thread etiquette and the nature of the summation being discussed.

Areas of Agreement / Disagreement

Participants express differing views on the convergence of certain series and the methods used to establish equivalences. There is no consensus on the validity of the approaches or the convergence of the series mentioned.

Contextual Notes

Some participants mention the need for careful consideration of convergence and the limitations of their approaches, particularly regarding the use of asymptotic series and the treatment of infinite sums.

nonequilibrium
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Hello,

In my probability course it is used that \frac{1}{n} \sum_{k=3}^n \frac{1}{\log k} \propto \frac{1}{\log n}.

I hope I'm not being daft: can somebody enlighten me on why this is so?
 
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Hello,

a method to find a equivalent is to border the sum by integrals (in attachment)
 

Attachments

  • Equivalent.JPG
    Equivalent.JPG
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Are you sure jjacquelin that this series 1+1/x+2/x^2+6/x^3 converges? I take it to be sum_k k!/x^k.

I took a similar approach by integrating by parts, but sadly also ended up with a divergent series. I think this is a non-trivial question to prove from scratch.
 
Are you sure jjacquelin that this series 1+1/x+2/x^2+6/x^3 converges? I take it to be sum_k k!/x^k.
We don't consider the infinite series which doesn't converge. We consider a limited number of the first terms in order to derive an equivalent. This is the usal way for asymptotic series.
In fact, in my previous attachment, where a typographic character which is not exactly the good one. In the asymptotic formula the "equal" should be the "tilde". (see attachment)
 

Attachments

  • Series.jpg
    Series.jpg
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Thank you
 
I have not seen that before, thank you for the explanation.
 
Hi,

what is the summation of 1/log(n) + 1/log(n-2) + 1/log(n-4) ..

Regards
Vishy
 
Please do not "hijack" someone elses thread to ask a different questions- start your own thread.

Also, for fixed n, n- i will eventually be 0 so if you mean an infinite sum, that will not exist. I presume you mean
\sum_{i= 0}^{n-1} \frac{1}{ln(n- i)}
 
sorry abt that ..
actually i wanted to know the summation
n/2−1
∑ 1/lg(n−2i)
i=0
 

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