The deterministic critical degree root stays above nine quarters

For $z\ge 1$, let $r=\sqrt{z}$ be the nonnegative root from [F6]. Then $r\ge 1$, since $0\le r<1$ would imply $r^2<1$. For every $M\ge 1$, $z\ge M^2$ implies $r\ge M$: otherwise $M^2-r^2=(M-r)(M+r)>0$ contradicts $r^2=z\ge M^2$. Thus $r\to \mathrm{\infty }$ as $z\to \mathrm{\infty }$. By [F2], $\log z=2\log r$, so [F3] gives $0\le \log z\mathrm{/}z\le 2(r-1)\mathrm{/}{r}^2\le 2\mathrm{/}r$. Given $\epsilon >0$, choosing $M>\max (1,2\mathrm{/}\epsilon )$ makes the last bound less than $\epsilon$ for $z\ge M^2$; hence $\log z\mathrm{/}z\to 0$. This yields $\log n\mathrm{/}n\to 0$ and $\log \log n\mathrm{/}\log n\to 0$. By [F2], $d_n\to \mathrm{\infty }$ and $\log (Kt{d}_n)\mathrm{/}\log n\to 0$ for each fixed $t>0$. Therefore $p_n\to 0$.

For fixed $n>1$, $d_n>0$ and $Kt{d}_n>0$ for $t\ge 1$. By [F5], the identity map and its scalar multiple $t\mapsto K{d}_{n}t$ are continuous on $[1,\mathrm{\infty })$; [F2] and composition give continuity of $t\mapsto \log (Kt{d}_n)$. The same continuity algebra applied to $I(t)=t\log t-t+1$, and then to its scalar multiple and sum with the logarithmic term, shows that $F_n(t)=d_{n}I(t)+(1\mathrm{/}2)\log (Kt{d}_n)$ is continuous on that interval, including at $1$. It is strictly increasing and tends to infinity as $t\to \mathrm{\infty }$ by [F1] and [F2]. Step 1.1 gives $F_n(1)\mathrm{/}\log n\to 0$ and $F_n(9\mathrm{/}4)\mathrm{/}\log n\to (3\mathrm{/}2)I(9\mathrm{/}4)\le 15\mathrm{/}16<1$. Hence for large $n$ these two values are below $\log n$. The intermediate value theorem gives a unique root above $9\mathrm{/}4$. The probability-parameter bounds follow from step 1.1.