In this appendix we show that, given the choice of
$z = \left(\frac{e}{k}\left( \frac{1}{k^4} \right) ^ \frac{1}{k}\right)^d$ in Section~\ref{sec:gog}, $\sum_{{k'}} H_{d, k'} z^{k'}  = O(k^{1-3d})$.
\begin{align*}
\sum_{k'=3}^k H_{d, k'} z^{k'}  &=  \sum_{k'=3}^k (k'!)^d \left( \frac{e}{k}\left( \frac{1}{k^4} \right) ^ \frac{1}{k}\right)^{d \cdot k'} \\
  &=  \sum_{{k'}} \left(\sqrt{2\pi k'} \left(\frac{k'}{e}\right)^{k'}    \left(     \frac{e\phantom{'}}{k\phantom{'}} \right)^{k'}        \left(  \frac{1}{k^4} \right) ^{\frac{k'}{k}} \right)^d \\
\intertext{(The error of Stirling's approximation for $k'!$ can be bounded by a factor of $1+O(k'^{-1})$, which we omit.)}
%  &=  \frac{1}{e^2} \sum_{{k'}} \sqrt{2\pi k'} \left(\frac{k'}{e}\right)^{k'}    \left(   \frac{1}{\sqrt{2\pi k}} \right)^\frac{k'}{k}   \left( \frac{e}{k} \right)^{k'}      \left(  \frac{1}{k^4} \right) ^{\frac{k'}{k}}  \\
%  &\le  \frac{1}{e^2} \sum_{{k'}} \sqrt{2\pi k'} \left(\frac{k'}{e}\right)^{k'}   \left( \frac{e}{k} \right)^{k'}      \left(  \frac{1}{k^4} \right) ^{\frac{k'}{k}}  \\
 &\le  \sum_{{k'}} \left (\sqrt{2\pi } \left(\frac{k'}{k}\right)^{k'}  k^{1/2} \left(  \frac{1}{k^4} \right) ^{\frac{k'}{k}} \right)^d \\
 &=   \sum_{{k'}} \left( \sqrt{2\pi }  \left(\frac{k'}{k} \right) ^{k'}  k^{1/2-\frac{4k'}{k}} \right)^d \\
 &=  \left(\frac{\sqrt{2\pi }}{ k^{3}}\right)^d \sum_{{k'}} \left( \left(\frac{k'}{k} \right) ^{k'}  k^{3.5-\frac{4k'}{k}} \right)^d
\end{align*}
It can be shown that each term within the sum is less than one \cite{Zhang-gildea-tr06}
\[ \left( \frac{k'}{k} \right) ^{k'}   k^{3.5-\frac{4k'}{k}}  < 1               .\]
\begin{align}
% \intertext{in order to show that the entire expression is $O(k^{-2})$.
% First note that for all $k'$,}  
%  \left(\frac{k'}{k} \right) ^{k'} &< 1 \notag\\
% \intertext{which implies that eq.~\ref{eq:ltone} holds if} 
%    k^{3.5-\frac{4k'}{k}}   &< 1 \notag\\
%  3.5-\frac{4k'}{k} &< 0 \notag\\
%  k' &> \frac{7}{8}k \notag\\
% \intertext{Thus eq.~\ref{eq:ltone} always holds for higher values of $k'$; now we must show that 
% it also holds for the lower terms where $  k' < \frac{7}{8}k$.  First note that:}
%  k^{3.5-\frac{4k'}{k}} &< k^{3.5} \notag\\
% \intertext{so it is sufficient to show that:}
%  \left(\frac{k'}{k} \right) ^{k'}  k^{3.5}  &< 1 \notag\\
% %%%\intertext{Taking the logarithm of both sides, we need:}
%  k' ( \log {k'} - \log {k} )   + 3.5 \log k  &< 0 \notag\\
%  k' \log k' &< ( k' - 3.5 ) \log k   \label{eq:k3}\\
% \intertext{Since we are looking at the range where $k' <  \frac{7}{8}k$,  we know that:}
%  \log k' &< \log k + \log\frac{7}{8} \notag\\
% \intertext{so we need:}
%  k' ( \log k + \log \frac{7}{8} )  &< ( k' - 3.5 ) \log k   \notag\\
%  k' ( \log \frac{7}{8} )  &<  - 3.5  \log k   \notag\\
%  k' &>    \frac{- 3.5}  { \log \frac{7}{8} } \log k   \notag\\
%  k' &>     \frac{3.5}{ \log \frac{8}{7} } \log k   \notag\\
% \intertext{Another way of satisfying eq.~\ref{eq:k3} is:}
%  \frac{ k'} { k' - 3.5 } \log k' &< \log k   \notag\\
% \intertext{Since $k' \ge 4$, we know that $\frac{ k'} { k' - 3.5 } \le 8$ so we just need:}
%  8 \log k' &< \log k\notag\\
%  k' &< k^{\frac{1}{8}} \notag\\
\intertext{%As $k$ grows, the bounds  $k^{\frac{1}{8}}$ and $\frac{3.5}{\log \frac{8}{7} } \log k $ will meet.
Therefore, for sufficiently large $k$,
}
\sum_{{k'}} H_{d, k'} z^{k'}  &\le  \left(\frac{\sqrt{2\pi }}{ k^{3}}\right)^d \sum_{{k'}} \left( \left(\frac{k'}{k} \right) ^{k'}  k^{3.5-\frac{4k'}{k}} \right)^d \notag\\
%\frac{1}{e^2k^3} \sum_{{k'}}  \left(\frac{k'}{k} \right) ^{k'}  k^{3.5-\frac{4k'}{k}} \notag\\
 &\le \left(\frac{\sqrt{2\pi }}{ k^{3}}\right)^d \sum_{{k'}}  1 \notag\\
 &\le \left(\frac{\sqrt{2\pi }}{ k^{3}}\right)^d k\notag\\
% &\le \frac{1}{e^2k^{3}} k \notag\\
% &= \frac{1}{e^2k^{2}}  \notag \\
 &= O(k^{1-3d})  \notag
\end{align}