# \sum 1/log(n) goes like ?

## Main Question or Discussion Point

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?

Hello,

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

#### Attachments

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disregardthat
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

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Thank you

disregardthat
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

HallsofIvy
Homework Helper

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