The asymptotic order of the series

I want to know the order of $$\sum\limits_{n = 2}^\infty {{{{{\left( {\ln n} \right)}^{-n}}}}{x^n}}$$ when $ x \to + \infty $, or a fine upper bound. I appreciate any good ideas or solutions. I asked a similar question earlier, but I was wrong. I want to get a lower order than $\exp(\exp(x))$, as small as possible.

Solution 1:

A rough argument is as follows. For fixed $x$, the maximum of $$ n \mapsto {\frac{{x^n }}{{(\log n)^n }}} $$ occurs at $$ n = \exp \left( { - \frac{1}{{W( - 1/x)}}} \right) \sim e^{x - 1} , $$ where $W$ is the principal branch of the Lambert $W$-function. The approximation $e^{x-1}$ is quite accurate for large positive $x$. Thus, \begin{align*} \sum\limits_{n = 2}^\infty {\frac{{x^n }}{{(\log n)^n }}} & = \sum\limits_{n = 2}^{\left\lfloor {2e^x } \right\rfloor } {\frac{{x^n }}{{(\log n)^n }}} + \sum\limits_{n = \left\lfloor {2e^x } \right\rfloor + 1}^\infty {\frac{{x^n }}{{(\log n)^n }}} \\ & \ll e^x \left( {\frac{x}{{x - 1}}} \right)^{e^{x - 1} } + \sum\limits_{n = \left\lfloor {2e^x } \right\rfloor + 1}^\infty {\frac{{x^n }}{{(\log n)^n }}} . \end{align*} But \begin{align*} \sum\limits_{n = \left\lfloor {2e^x } \right\rfloor + 1}^\infty {\frac{{x^n }}{{(\log n)^n }}} & \ll \int_{2e^x }^{ + \infty } {\frac{{x^t }}{{(\log t)^t }}dt} = 2e^x \int_1^{ + \infty } {\left[ {\frac{x}{{x + \log (2s)}}} \right]^{2se^x } ds} \\ & = 2e^x \int_1^{e^x } {\left[ {\frac{x}{{x + \log (2s)}}} \right]^{2se^x } ds} + 2e^x \int_{e^x }^{ + \infty } {\left[ {\frac{x}{{x + \log (2s)}}} \right]^{2se^x } ds} \\ & < 2e^{2x} + 2e^x \int_{e^x }^{ + \infty } {4^{ - se^x } ds} \ll e^{2x} \ll e^x e^{e^{x - 1} /x} \le e^x \left( {\frac{x}{{x - 1}}} \right)^{e^{x - 1} } . \end{align*} Therefore, $$ \sum\limits_{n = 2}^\infty {\frac{{x^n }}{{(\log n)^n }}} \ll e^x \left( {\frac{x}{{x - 1}}} \right)^{e^{x - 1} } , \quad x\to +\infty. $$ Note that $$ e^x \left( {\frac{x}{{x - 1}}} \right)^{e^{x - 1} } \ll \exp \left( {(1 + \varepsilon )\frac{{e^{x - 1} }}{x} + x} \right) $$ for any $\varepsilon>0$, so the estimate is better than $\exp(\exp(x))$.