\[\frac{A^{2n+1}_{2m+1}(q)}{A^{2n+1}_{1}(q)}=\frac{(-1)^{m+1}}{\left({\left(%
\tfrac{1}{2}\right)_{m+1}}\right)^{2}}\left(\frac{q}{4}\right)^{m+1}\frac{2%
\left(1+O\left(m^{-1}\right)\right)}{w^{\prime}_{\mbox{\tiny II}}(\frac{1}{2}%
\pi;a_{2n+1}\left(q\right),q)},\]