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 $x\gg0$?
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 $x\gg0$?
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 $x \gg 0$, $g(x) = O \left( \frac{1}{x \log(x)} \right)$.