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Reportedly, the last problem of this contest is:

Given a positive decreasing sequence $\displaystyle{\{a_n\}}_{n\geqslant1}$ satisfying $\displaystyle\lim_{n\to\infty}a_n=0$, show that if the series $\displaystyle\sum_{n=1}^\infty{a_n}$ diverges, then the integral $\displaystyle\!\int_1^{+\infty}{\!\dfrac{\ln{\operatorname{\it{f}}{(x)}}}{x^2}\operatorname{d}{\!x}}$ diverges as well, where $\displaystyle\operatorname{\it{f}}{(x)}≔\sum_{n=1}^\infty{a_n^nx^n}$.

Its suggested solution (in S̲i̲m̲p̲l̲i̲f̲i̲e̲d̲ Chinese) can be found in the link above. However, I attempt to solve this via another approach, which is as follows:

  • It is well known that $\displaystyle\sum_{n=1}^\infty{\mspace{-1.5mu}n^{-n\mspace{-0.5mu}}x^n}=\int_0^1{\mspace{-1mu}{xt^{-xt}\mspace{1.5mu}}\mathrm{d}t}$ (see below) and then $\displaystyle{(\sum_{n=1}^\infty{n^{-n}x^n\!})^{1/x}}\sim{\mspace{-1.5mu}\exp{\!\left(\mspace{-0.75mu}\frac1{\mathrm{e}}\mspace{-0.75mu}\right)}}$ (This is AMM11982.) as $x\to+\infty$ and hence $\displaystyle\int_1^{+\infty}{\mspace{-3.5mu}\dfrac{1}{x^2}\mspace{0.5mu}\ln{\mspace{-5mu}\left(\sum_{n=1}^\infty{\mspace{-1.5mu}\frac{x^n}{n^n}}\mspace{-2.5mu}\right)\mspace{-4.5mu}}\operatorname{d}{\!x}}$ diverges. The main method is: For all sufficiently large $x\ge1$, $\operatorname{\it{f}}{(x)}$ should be positive, so, if one can prove that $\sqrt[x]{\operatorname{\it{f}}{(x)}}=\operatorname{\mathcal{O\!}}{\left(\mathrm{e}^{x^{\varepsilon}}\right)}$ (which means $\dfrac{\ln{\operatorname{\it{f}}{(x)}}}{x^2}=\dfrac1x\ln\!\left(f(x)\right)^{\mspace{-1mu}1/x}=\operatorname{\mathcal{O\!}}{\left(\mspace{-1.5mu}\dfrac1{x^{1-\varepsilon}}\!\right)}$) as $x\to+\infty$ for some non-negative real number $\varepsilon$ (depending on $\{a_n\}$), the respective integral must diverge.
  • Now, since $$\begin{align}\operatorname{\it{f}}{(x)}&=\sum_{n=1}^\infty{a_n^nx^n}=\sum_{n=0}^\infty{a_{n+1}^{n+1}\frac{x^{n+1}}{n!}\operatorname{\!\Gamma\!}{\left(n+1\right)}}=\sum_{n=0}^\infty{a_{n+1}^{n+1}\frac{x^{n+1}}{n!}\mspace{-4.5mu}\int_0^1{\mspace{-4.5mu}\ln^n\!\!\left(\mspace{-2.25mu}\frac1y\mspace{-2.25mu}\right)\mspace{-4.5mu}\operatorname{d}{\!y}}}\text{, }\\&=x\sum_{n=0}^\infty{\frac{x^n}{n!}\mspace{-3.5mu}\int_0^{+\infty}{\mspace{-4mu}u^n\mathrm{e}^{\mspace{-1.5mu}-\frac{\mspace{-4.5mu}u}{a_{\mspace{2.5mu}n+1\mspace{7mu}}}\mspace{-5mu}}\operatorname{d}{\!u}}}=x\sum_{n=0}^\infty{\frac{x^n}{n!}\mspace{-3.5mu}\int_{0^+}^1{\left(-t\ln{t}\right)^{n}t^{a_{n+1}^{-1}\mspace{2mu}-\left(n+1\right)}\operatorname{d}{\!t}}}\text{,}\end{align}$$ provided that the termwise integration is allowed, we have $$\operatorname{\it{f}}{(x)}=x\mspace{-0.5mu}\int_0^1{\sum_{n=0}^\infty{\mspace{-1mu}\frac{\left(-x\mspace{1mu}t\ln{t}\right)^{n}}{n!}\mspace{-2.5mu}\cdot\underline{t^{a_{n+1\mspace{5.5mu}}^{-1}-\left(n+1\right)}}\operatorname{d}{\!t}}}\text{.}$$ When $a_n=n^{-1}$, direct evaluation leads to the above equation, and using one theorem (see below), its divergence can be proven. But for other $\{a_n\}$ (where we cannot ignore the underline part), I am not able to obtain similar result as a smallest divergent series is nonexistent (cf. Nonexistence of boundary between convergent and divergent series? and Is there a slowest rate of divergence of a series?)! What can we say for any positive decreasing sequence $\{a_n\}$ such that $\lim_{n\to\infty}a_n=0$ and $\sum_{n=1}^\infty{a_n}=+\infty$?
  • Why I write $f(x)$ in that form? A theorem is: $${{g,h\in\operatorname{\mathscr{C\!}}{\left(\left[a,b\right]\right)}}\land{h>0}}\implies{\displaystyle\lim_{n\to\infty}\sqrt[n]{\int_a^b{\left\lvert\operatorname{\it{g}}{(x)}\right\rvert^{n}h(x)\operatorname*{d\!}{x}}}=\max_{{a}\leqslant{x}\leqslant{b}}\left\lvert\operatorname{\it{g}}{(x)}\right\rvert}\tt.$$ And ignoring the underline part, $\displaystyle\left(x\int_0^1\sum_{n=0}^\infty{\mspace{-1mu}\frac{\left(-x\mspace{1mu}t\ln{t}\right)^{n}}{n!}}\times\left(\textrm{?}\right)dx\right)^{\frac1x}\overset{?}\approx{x^{1/x}\left(\int_0^1\left(\frac1{t^t}\right)^{x}\times\left(\textrm{¿}\right)dx\right)^{1/x}}$ is just similar to the left hand side of this theorem. Unfortunately, I do not know how to use those condition of $\displaystyle\{a_n\}$ in next steps (based on my “another” approach). Is it really unworkable in general? I have no idea.

Besides, Although I reckon that the problem at the begining is not new at least in MSE (or other forums like AoPS), I can't find such old posts after searching. So it can be still suitable for posting these questions… I'd be grateful for any help one may provide.

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