I've found (as have others), that for some analytic functions, a Padé approximant of it has an infinite convergence radius, whereas its associated Taylor series has a finite convergence radius. $f(x)=\sqrt{1+x^2}$ appears to be one such function. My questions are:

1) Is there any function where the Taylor series has the largest convergence radius of all associated Padé approximants? If so, is the Taylor series radius strictly larger, or only equal to the convergence radius of other Padé approximants (i.e. excluding the Taylor series itself)?

2) If not, is there any function that is analytic everywhere, and yet for which there is no (limit of) Padé approximant(s) that has an infinite convergence radius?

It would be both very cool and very useful if there is always a (limit of) Padé approximant(s) that has an infinite convergence radius for any function that is analytic everywhere, though I haven't the slightest how one checks/analyzes convergence of Padé approximants if the degrees of numerator and denominator both approach infinity. :)

One extra question, if there is always such a Padé approximant:

3) Is there always a numerically stable method of computing this approximant up to a finite order?

toosurprised, since they come from recursively applying a real, non-negative perturbation to rational functions. However, I was expecting that, like with the Taylor series of them, the rational perturbation would only be valid up to a finite value of the parameter. Anyway, I'm just rambling now. – Neil Dickson Sep 9 '10 at 7:23