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

  • Thread starter nonequilibrium
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In summary, there is a question about why the sum \frac{1}{n} \sum_{k=3}^n \frac{1}{\log k} \propto \frac{1}{\log n} in a probability course. Someone suggests using integrals to find an equivalent and another person asks for clarification on whether the series 1+1/x+2/x^2+6/x^3 converges. It is explained that only a limited number of terms are being considered for an asymptotic series. There is a discussion about the notation of an asymptotic formula and a new question is asked about the summation of 1/log(n) + 1/log(n-2) + 1
  • #1
nonequilibrium
1,439
2
Hello,

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

I hope I'm not being daft: can somebody enlighten me on why this is so?
 
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  • #2
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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  • #3
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.
 
  • #4
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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  • #5
Thank you
 
  • #6
I have not seen that before, thank you for the explanation.
 
  • #7
Hi,

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

Regards
Vishy
 
  • #8
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
[tex]\sum_{i= 0}^{n-1} \frac{1}{ln(n- i)}[/tex]
 
  • #9
sorry abt that ..
actually i wanted to know the summation
n/2−1
∑ 1/lg(n−2i)
i=0
 

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

What is the significance of the sum 1/log(n)?

The sum 1/log(n) is a mathematical series that represents the harmonic series with natural logarithmic terms. It is commonly used in mathematical analysis and has applications in fields such as number theory and physics.

What does it mean for \sum 1/log(n) to "go like" something?

When we say that \sum 1/log(n) goes like something, we mean that it has a similar behavior or trend as that particular value or function. In other words, as n increases, the sum 1/log(n) will approach the same limit or follow a similar pattern as the specified value or function.

What is the asymptotic behavior of \sum 1/log(n)?

The asymptotic behavior of \sum 1/log(n) is logarithmic, meaning that as n increases, the sum will approach infinity at a slower rate than other types of functions such as polynomial or exponential functions.

How is \sum 1/log(n) related to the natural logarithm function?

The sum 1/log(n) is closely related to the natural logarithm function, ln(n). In fact, as n approaches infinity, the sum 1/log(n) approaches the same limit as ln(n), which is infinity. This means that the growth rate of both functions is equivalent.

Can \sum 1/log(n) ever reach a finite value?

No, the sum 1/log(n) will always approach infinity as n increases. This is because the terms in the series decrease very slowly, and as n increases, there are an infinite number of terms in the series, making the sum infinite. However, for certain values of n, the sum may appear to approach a finite value, but it will never actually reach it.

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