\[\frac{I_{\nu}\left(z\right)}{I_{\nu-1}\left(z\right)}=\cfrac{\frac{1}{2}z/\nu}%
{1+}\cfrac{\frac{1}{4}z^{2}/(\nu(\nu+1))}{1+}\cfrac{\frac{1}{4}z^{2}/((\nu+1)(%
\nu+2))}{1+}\cdots,\]