Series

[kleinwolf]

How can i compute : S(n)=\sum_{k=2}^n\frac{1}{log(k)}...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.

Thanx.

[shmoe]

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

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}

If you already had this, please give some more details on what you've tried and where you got stuck.

[kleinwolf]

Ok...so more precisely my question is : what is \lim_{x\rightarrow\infty} \frac{Li(x)}{x} ? (I suppose this should give the same as \lim_{n\rightarrow\infty}\frac{S(n)}{n} ?)
My hope would be that this limit is not 0...but I think it is.

[shmoe]

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

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

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):

\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}

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