\[\frac{B^{2n+2}_{2m}(q)}{B^{2n+2}_{2}(q)}=\frac{(-1)^{m}}{(m!)^{2}}\left(\frac{%
q}{4}\right)^{m}\frac{q\pi\left(1+O\left(m^{-1}\right)\right)}{w^{\prime}_{%
\mbox{\tiny I}}(\frac{1}{2}\pi;b_{2n+2}\left(q\right),q)}.\]