- #1

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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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- Thread starter nonequilibrium
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- #1

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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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- #3

disregardthat

Science Advisor

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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

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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.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.

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)

- #5

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Thank you

- #6

disregardthat

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I have not seen that before, thank you for the explanation.

- #7

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Hi,

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

Regards

Vishy

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

Regards

Vishy

- #8

HallsofIvy

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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

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sorry abt that ..

actually i wanted to know the summation

n/2−1

∑ 1/lg(n−2i)

i=0

actually i wanted to know the summation

n/2−1

∑ 1/lg(n−2i)

i=0

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