\section{Decay of the differentiated sums}\label{sec:decay} Clearing denominators multiplies a form by $\delta_n^{S+k}$. Since $\delta_n^{1/n}\to\ep$, it suffices to bound its $n$th-root limsup strictly below $\ep^{-(S+k)}$. The initial grid terms through $rn$ vanish. We estimate the remaining terms by their absolute values, first on a bounded scaled interval and then on the infinite tail. \subsection{The exponential profile} For $t>0$ the products in \eqref{eq:P}--\eqref{eq:R} give \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*} Set $x=t/n$. Stirling's formula, including the normalization $C_n$, gives the exponential profile \begin{align} \Phi(x)&=m\log K +mD\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 $n\log n$ in the factorial normalization is $S-mDL$; the products contribute its negative. The linear Stirling terms cancel as well. For fixed $A>0$, the estimate \begin{equation}\label{eq:uniform-Stirling} \log R_n(rn+t)=n\Phi(t/n)+O_A(\log n), \qquad 1/B\le t\le An, \end{equation} is uniform. Only the factors $\Gamma(t+\theta)$ can have bounded arguments. Stirling's formula for $t\ge1$ and continuity on $[1/B,1]$ give, uniformly for $\theta\in\Theta$, \[ \log\Gamma(t+\theta)=t\log t-t+O_B(\log(2+t)), \qquad t\ge1/B. \] The other gamma arguments are proportional to $n$, and the extra $m\log t$ term is also $O_A(\log n)$. This proves \eqref{eq:uniform-Stirling} down to the first possible nonzero grid point beyond $rn$. \subsection{Why differentiation preserves the exponential rate} \label{sec:derivative-decay} Use Cauchy's estimate on the circle $|z|=\varepsilon=1/(2B)$ about $rn+t$, where $t\ge1/B$. Every numerator factor is positive at the real center. If its value is $c$, its absolute-value ratio on the circle is at most $1+\varepsilon/c$. The sum of these reciprocals is $O(\log n)$ on $1/B\le t\le An$, including the extra factor $t$. The denominator factors are at least $rn+t$, and their reciprocal sum is $O(1)$. Thus, for a fixed constant $C$, \[ \sup_{|z|=\varepsilon}|R_n(rn+t+z)| \le n^C R_n(rn+t). \] Homogeneity gives $|W_n(rn+t+z)|=O_A(n^E)$ on this range. Consequently \begin{equation}\label{eq:Cauchy} |V_n(rn+t)|\le\varepsilon^{-k} \sup_{|z|=\varepsilon}|T_n(rn+t+z)|. \end{equation} There are $O_A(n)$ grid points in the bounded scaled interval. By \eqref{eq:uniform-Stirling}, their absolute sum has $n$th-root limsup at most $\exp(\max\Phi)$. Here $m$, and hence the derivative order $k$, is fixed while $n\to\infty$. We also need a bound over the infinite tail. Take $A\ge\max(L,r+2,1)$ and $t\ge An$. On the same circle each numerator factor is at most $3t$ in absolute value, each denominator factor is at least $t$, and $|W_n(rn+t+z)|\le C't^E$. Equation~\eqref{eq:Cauchy} gives \[ |V_n(rn+t)|\le C''C_n3^{mDLn+m} t^{-(S-mDL)n-\tau_0}. \] For $q_n=(S-mDL)n+\tau_0>1$, a first term and an integral bound the sum on any grid of spacing $1/b$: \[ \sum_{\substack{t\text{ on the grid}\\t\ge An}}t^{-q_n} \le(An)^{-q_n}+b\frac{(An)^{1-q_n}}{q_n-1}. \] Stirling's formula for $C_n$ gives the tail $n$th-root bound \begin{equation}\label{eq:tail} 4^S K^m3^{mDL}\ep^{-(S-mDL)}A^{-(S-mDL)}. \end{equation} The powers of $n^{S-mDL}$ from the factorials cancel those from the lower tail endpoint. Since $S>mDL$, \eqref{eq:tail} tends to zero as $A$ grows. Together with the bounded range, this proves \begin{equation}\label{eq:uncleared-decay} \limsup_{\substack{n\to\infty\\n\text{ odd}}}|F_{n,b}|^{1/n} \le G:=\exp\!\left(\max_{x\ge0}\Phi(x)\right). \end{equation} \subsection{The two inequalities imply cleared decay} The function $f$ in \eqref{eq:f} satisfies $4\Phi'(x)=\log f(x)$. The derivative of $\log f$ has a positive denominator and numerator \[ 4\bigl((S-mDL)x^2+L(S-mDL)x-mDLr(r+4)\bigr). \] This quadratic has exactly one positive zero. Thus $\log f$ first decreases and then increases. It starts at $+\infty$ and tends to $0$ from below, with leading term $-4(S-mDL)/x$. It has exactly one zero $x_0>0$, where $\Phi$ reaches its global maximum. Substituting $\Phi'(x_0)=0$ into \eqref{eq:Phi} gives \begin{equation}\label{eq:G} G=K^m(x_0+L)^{mDL}4^S \left(\frac{(x_0+r)^r}{(x_0+r+4)^{r+4}}\right)^{S/4}. \end{equation} If $f(u)<1$, then $x_0