Consider next sum \begin{eqnarray} \label{PF_spindef} Z = \sum_{r=0}^{N N_f} h^{2r} \ Q(r) . \end{eqnarray} and \begin{equation} Q(r) \ = \ \sum_{\sigma \vdash r} s_{\sigma}(1^{N_f}) \ s_{\sigma}(1^{N_f}) \ . \label{QSUN_def} \end{equation} where $s_{\sigma}(1^{N_f})$ is Schur function and $\sigma \vdash r$ run over partition.

Expansion around $h=0$ in the Veneziano limit $N \to \infty$ (where $\frac{N_f}{N }= \kappa $) explicitly gives $$ Z= 1 + h^2 N^2 \kappa^2 + \frac{h^4}{2} N^2 \kappa^2( N^2 \kappa^2 + 1) + \frac{h^6 }{6} N^2 \kappa^2 (N^2 \kappa^2 +1)(N^2 \kappa^2 +2) + ... = e^{ N^2 \kappa^2 \log \frac{1}{1- h^2} } $$ $$ \lim_{N \to \infty} \frac{1}{N^2} Z = \kappa^2 \log \frac{1}{1- h^2} $$

How to re-expand initial $Z$ (in terms of Schur functions) around $h=1$ and takes the same Veneziano limit ?