Cut the Cards

[Gerson W. Barbosa]

For the inverse digamma function I’ve been using a continued fraction with increasing number of terms. I am sure about only the first half of them, though. It’s difficult to be sure about the other half when one is limited to only 12 digits.

\(\rm{e}^{{x}}+\frac{1}{2+\frac{1}{6\rm{e}^{{x}}-\frac{1}{2+\frac{1}{\rm{e}^{{x}-1}-\frac{1}{2+\frac{1}{\rm{e}^{{x}}+\frac{1}{2}}}}}}}\)