E (0,1)(z)=1/(1-z) for abs(z) less than 1
From specialfunctionswiki
Theorem
The following formula holds for $|z|<1$: $$E_{0,1}(z)=\dfrac{1}{1-z},$$ where $E_{0,1}$ denotes the Mittag-Leffler function.
Proof
References
- H.J. Haubold, A.M. Mathai and R.K. Saxena: Mittag-Leffler Functions and Their Applications (2011)... (previous)... (next): $(2.1)$ (uses notation $E_0$ instead of $E_{0,1}$)
