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Can somone help me find an assymptotic formula for n, for fixed x , for this sum , perhaps an inequality would be even better, or some bound on the error. $$\sum_{k=1}^n \frac{1}{\log(kx)}$$

I need somthing better then the integral from 1 to n of ln(kx) with respect to k. Its also okay if you use special functions, like the logarithmic integral.

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In what limit do you want an approximation? $x \to \infty$ for fixed $n$? $n \to \infty$ for fixed $x$? – Robert Israel Nov 16 '12 at 1:50
n->infinity, for fixed x, that is an asymptotic formula in terms of n and x – boby Nov 16 '12 at 1:59
@boby Why can't you use one account to post your questions? I have already told this to you at-least twice (…) and (…) – user17762 Nov 16 '12 at 2:18
Use the Euler-Maclaurin series as suggested below. The expression doesn't in and of itself have a closed form that simplifies to elementary functions, nor have I found any special function that neatly fits your requirement. – GregRos Nov 17 '12 at 1:43

The Euler-Maclaurin series for $\sum_k 1/\log(kx)$ starts $$ F(k) = \frac{\text{Li}(kx)}{x} - \frac{1}{2 \ln(kx)} - \frac{1}{12 \ln(kx)^2 k} + \frac{3 + 3 \ln(kx) + \ln(kx)^2}{360 \ln(kx)^4 k^3} + \ldots $$

That is, we should have $$\sum_{k=1}^n \frac{1}{\log(kx)} \approx F(n+1) + C$$ where $C$ is a constant.

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