Show that $\lim_{x\to \infty} \frac{f(x)}{g(x)}=1$

Question

Consider $f(x)=\ln(\ln(\zeta(\exp(\exp(-x)))))$ and $g(x)=\ln(x),$ where $\zeta(x)=\sum n^{-x}$ for $\Re(x)>1.$

Show that $$\lim_{x\to \infty} \frac{f(x)}{g(x)}=1$$

I obtained $f(x)$ from $\zeta(x)$ after performing two consecutive log-log coordinate transforms on $\zeta(x).$ $f(x)$ appears to converge, quite quickly, to $\ln(x)$ as $x$ increases. I've numerically verified several values such as: $$f(2) \approx 0.69978 $$ $$g(2)\approx 0.69315 $$ $$ f(8)\approx 2.07944 $$ $$ g(8)\approx 2.07944$$

Answer 1

Let $x=-\log(\log(t))$ so that

$$\log\left(\log\left(\zeta\left(e^{e^{-x}}\right)\right)\right)=\log\left(\log\left(\zeta\left(t\right)\right)\right)$$

and as $x\to \infty$, $t\to 1^+$.

Then, we wish to evaluate the limit

$$\lim_{t\to 1^+}\frac{\log\left(\log\left(\zeta\left(t\right)\right)\right)}{\log(-\log(\log(t)))}$$ Recalling that near $t=1^+$, we have $\zeta(t)=\frac1{t-1}+O(1)$ and $\frac1{\log(t)}=\frac1{t-1}+O(1)$, it is straightforward to see that the value of the limit of interest is $1$.

Answer 2

As $x\to\infty$, $\exp-x\to0^+$. Define $s:=\exp\exp -x$ so $\zeta(s)\sim\frac{1}{s-1}\sim\exp x$. So $\ln\zeta=x+o(1),\,\ln\ln\zeta\sim\ln x$ as required.