I also stumbled upon this very interesting paper:
http://arxiv.org/pdf/hep-th/9206074
It mentions methods that sum power series in the Mittag-Leffler star. One formula it gives, is this: given a principal branch of an analytic function, represented by its power series = \sum_{n=0}^{\infty} a_n z^n)
at z = 0,
 = \int_{0}^{\infty} \exp(-\exp(t)) \sum_{n=0}^{\infty} a_n \frac{(tz)^n}{\mu(n)} dt)
with
 = \int_{0}^{\infty} \exp(-\exp(t)) t^n dt)
.
But it the formulas don't seem to work when tested numerically. Try it with the reciprocal series for = \frac{1}{1 - z})
, which they mention in the paper, i.e.  = \sum_{n=0}^{\infty} z^n)
. Then try evaluating using these formulas at 
, which outside the convergence radius for the series, but inside the Mittag-Leffler star. It seems to give huge values. Of course (and I highly suspect this is the case), I've missed something here... what might it be?
ADDENDUM: I see now, this does work... it's just that the terms "hump up" for reasonably large z-values before they get smaller and the thing converges... it needed 768 terms to converge to a few places for z = -2, but got ~0.3333 like I'd expect for
. I think it gets better after that point since once you're over the "hump" the terms get small fairly quick, I suppose 1024 terms would get much more accuracy but calculating )
takes a long time.
http://arxiv.org/pdf/hep-th/9206074
It mentions methods that sum power series in the Mittag-Leffler star. One formula it gives, is this: given a principal branch of an analytic function, represented by its power series
with
But it the formulas don't seem to work when tested numerically. Try it with the reciprocal series for
ADDENDUM: I see now, this does work... it's just that the terms "hump up" for reasonably large z-values before they get smaller and the thing converges... it needed 768 terms to converge to a few places for z = -2, but got ~0.3333 like I'd expect for
