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Li(x) function

  1. Mar 28, 2005 #1
    How can i compute : [tex] S(n)=\sum_{k=2}^n\frac{1}{log(k)}[/tex]...I tried with the Li(x) function, but couldn't manage till the end...I just have to know the asypmtotic behaviour when n->infty.

  2. jcsd
  3. Mar 29, 2005 #2


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    Did you have an asymptotic for Li(x)? If not integrating by parts might help:

    [tex]Li(x)=\int_2^x\frac{dt}{\log t}=\frac{x}{\log x}-\frac{2}{\log 2}+\int_2^x\frac{dt}{(\log t)^2}[/tex]

    If you already had this, please give some more details on what you've tried and where you got stuck.
    Last edited: Mar 29, 2005
  4. Mar 30, 2005 #3
    Ok...so more precisely my question is : what is [tex] \lim_{x\rightarrow\infty} \frac{Li(x)}{x} [/tex] ? (I suppose this should give the same as [tex] \lim_{n\rightarrow\infty}\frac{S(n)}{n}[/tex] ?)
    My hope would be that this limit is not 0...but I think it is.
  5. Mar 30, 2005 #4


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    You might expect the x/log(x) term to be the 'main' term of Li(x), so you should consider the limit

    [tex] \lim_{x\rightarrow\infty} \frac{Li(x)}{x/\log(x)} [/tex]

    You can use l'hopital to find that limit if you like or make use of the more satisfying inequality (both using Li(x) after you've integrated by parts once-or more if you want to adapt this to a more accurate asymptotic):

    [tex]\int_2^x\frac{dt}{(\log t)^2} =\int_2^{x^{1/2}}\frac{dt}{(\log t)^2}+\int_{x^{1/2}}^x\frac{dt}{(\log t)^2}=\frac{x^{1/2}-2}{(\log 2)^2}+\frac{x-x^{1/2}}{(\log x)^2/4}[/tex]

    Where the integrals were estimated trivially (max of the integrand times width of the interval).
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