\section{Decay after clearing denominators}\label{sec:decay} Clearing denominators multiplies each form by $\delta_n^S$. The prime number theorem gives $\delta_n^{1/n}\to\ep$, so it suffices to bound the $n$th-root limsup of $|F_{n,b}|$ strictly below $\ep^{-S}$. We estimate the alternating sums by their absolute values. The initial grid points are zeros, and the remaining terms lie to the right of $rn$, where $R_n$ is positive. The polynomial $W_n$ has fixed degree and will not change the exponential rate. \subsection{The exponential profile} We measure distance beyond the last zero in units of $n$, writing $t=nx$. The function $\Phi(x)$ below records the exponential size of $R_n(rn+t)$ at that scaled position. For $t>0$, the numerator and denominator products have the gamma representations \begin{align*} P_n(rn+t)&=t\prod_{\theta\in\Theta} \frac{\Gamma(t+Ln+\theta)}{\Gamma(t+\theta)},\\ \prod_{h=0}^n(rn+t+4h) &=4^{n+1}\frac{\Gamma(((r+4)n+t)/4+1)}{\Gamma((rn+t)/4)}. \end{align*} Together with the factorials in $C_n$, Stirling's formula gives \begin{align} \Phi(x)&=\lim_{\substack{n\to\infty\\n\text{ odd}}} \frac1n\log R_n(n(r+x))\notag\\ &=\log K+D\bigl((x+L)\log(x+L)-x\log x\bigr)\notag\\ &\quad-\frac S4\bigl((x+r+4)\log(x+r+4) -(x+r)\log(x+r)\bigr)+S\log4. \label{eq:Phi} \end{align} The coefficient of $n\log n$ from the factorials is $S-DL$; the product ratio contributes its negative. The linear Stirling terms also cancel. The remaining expression is \eqref{eq:Phi}. Let $f$ be the function in \eqref{eq:f-criterion}. Direct differentiation gives $4\Phi'(x)=\log f(x)$. The derivative of $\log f$ has positive denominator and numerator \[ 4\bigl((S-DL)x^2+L(S-DL)x-DLr(r+4)\bigr). \] This quadratic has exactly one positive zero. Thus $\log f$ first decreases and then increases. It starts at $+\infty$ and approaches $0$ from below, with leading term $-4(S-DL)/x$. Consequently it has exactly one zero $x_0>0$, where $\Phi$ attains its global maximum. The identity $\Phi'(x_0)=0$ simplifies that maximum to \begin{equation}\label{eq:G} G=\exp(\Phi(x_0)) =K(x_0+L)^{DL}4^S \left(\frac{(x_0+r)^r}{(x_0+r+4)^{r+4}}\right)^{S/4}. \end{equation} \subsection{Summing the profile and controlling the tail} \label{sec:profile-sum} Fix $A>0$. We must allow $t$ to stay as small as $1/B$ while $n$ grows. Only the gamma factors $\Gamma(t+\theta)$ can have bounded arguments. Stirling's formula for $t\ge1$ and continuity on the compact interval $[1/B,1]$ give, uniformly for $\theta\in\Theta$, \[ \log\Gamma(t+\theta)=t\log t-t+O_B(\log(2+t)), \qquad t\ge1/B. \] All other gamma arguments grow proportionally to $n$ on $1/B\le t\le An$. Their logarithmic errors are $O_A(\log n)$, as are the error above and the extra term $\log t$ from $P_n$. Substitution in the gamma formulas therefore gives the uniform estimate \[ \log R_n(rn+t)=n\Phi(t/n)+O_A(\log n), \qquad 1/B\le t\le An. \] The lower endpoint $1/B$ is sufficient: the first nonzero point on any of our grids lies at least that distance beyond $rn$. Homogeneity gives $|W_n(rn+t)|=O_A(n^E)$ on this range, which contains $O_A(n)$ grid points. Their absolute sum therefore has $n$th-root limsup at most $G$. We also need a bound uniform over the infinite tail. Set \begin{equation}\label{eq:tau} \tau=s-E>0, \end{equation} and take $A\ge\max(L,r+2)$. For $t\ge An$, each numerator factor is at most $2t$, each denominator factor is at least $t$, and $|W_n(rn+t)|\le C t^E$ for a fixed constant $C$. Thus \[ |T_n(rn+t)|\le C C_n2^{DLn+1} t^{-(S-DL)n-\tau}. \] For $q_n=(S-DL)n+\tau>1$, the tail on a grid of spacing $1/b$ is bounded by a first term and an integral: \[ \sum_{\substack{t\text{ on the grid}\\t\ge An}}t^{-q_n} \le(An)^{-q_n}+b\frac{(An)^{1-q_n}}{q_n-1}. \] Stirling's formula for $C_n$ then gives the tail $n$th-root bound \begin{equation}\label{eq:tail} 4^S K2^{DL}\ep^{-(S-DL)}A^{-(S-DL)}. \end{equation} The powers of $n^{S-DL}$ from the factorials cancel the corresponding powers from $(An)^{-(S-DL)n}$. Since $S>DL$, the bound in \eqref{eq:tail} tends to zero as $A$ grows. Combining the bounded range with the tail proves \begin{equation}\label{eq:uncleared-decay} \limsup_{\substack{n\to\infty\\n\text{ odd}}}|F_{n,b}|^{1/n}\le G. \end{equation} \subsection{The two rational inequalities imply decay} The estimate \eqref{eq:uncleared-decay} reduces the required decay to $\ep^S G<1$. If $f(u)<1$, the description of $f$ above implies $x_0