View source for E (0,1)(z)=1/(1-z) for abs(z) less than 1

==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==
* {{PaperReference|Mittag-Leffler Functions and Their Applications|2011|H.J. Haubold|author2=A.M. Mathai|author3=R.K. Saxena|prev=Mittag-Leffler|next=E(1,1)(z)=exp(z)}}: $(2.1)$ (uses notation $E_0$ instead of $E_{0,1}$)

[[Category:Theorem]]
[[Category:Unproven]]