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Let $f^{k+1}(x)=f^k(x)\log(f^k(x))$ and $f^1(x)=\log x$.

What is $g(x)=\sum_{k=1}^\infty \frac{1}{xf^k(x)}$ for $x\gg0$?

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    I did some TeX improvements, writing \log with a backslash, which not only de-italicizes it but also provides proper spacing between "log" and "x". And I changed x>>0 to $x\gg 0$ (the TeX code is "x \gg 0"). But I stopped short of changing the superscripts to subscripts. But subscripts would be more conventional and less likely to be confused with either of two other uses of superscripts.2011-11-16

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I will follow Michael's suggestion, and use lower indexes. Let's inspect few terms in the sequence $f_k(x)$: $ f_1(x) = \log(x) \qquad f_2(x) = \log(x) \cdot \log(\log(x)) \qquad f_3(x)= \log(x) \cdot \log(\log(x)) \cdot \log(\log(\log(x))) $ Therefore $f_k(x) = \prod_{n=1}^k \log^{(\circ n)}(x)$. The sum, thus, has a form: $ g(x) = \frac{1}{h_1} + \frac{1}{h_1 h_2} + \frac{1}{h_1 h_2 h_3} + \ldots = \frac{1}{h_1} \left( 1 + \frac{1}{h_2} \left( 1 + \frac{1}{h_3} \left( 1+ \ldots \right)\right) \right) $ with $h_1(x) = x \log(x)$, $h_2(x) = \log(\log(x))$, $h_3(x) = \log(\log(\log(x)))$, and so on.

Thus $ x \log(x) g(x) = h_1(x) g(x) = 1 + o(1) $ and thus for $x \gg 0$, $g(x) = O \left( \frac{1}{x \log(x)} \right)$.