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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 $x$?2012-11-16
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    n->infinity, for fixed x, that is an asymptotic formula in terms of n and x2012-11-16
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    @boby Why can't you use one account to post your questions? I have already told this to you at-least twice (http://math.stackexchange.com/questions/233268/simplifying-very-large-euler-product#comment517819_233268) and (http://math.stackexchange.com/questions/229399/proving-a-simple-inequality#comment510818_229399)2012-11-16
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    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.2012-11-17

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