Nowhere analytic superexponential convergence

[mike3]

well im sorry too Bo.

but it seems you missed the point.

lim n -> oo log^[n] ( exp^[n] (x) ) also doesnt have a somewhere non-zero convergence radius.

but clearly on the real line f(x) = lim n -> oo log^[n] ( exp^[n] (x) ) is real-analytic because it simply reduces to f(x) = x for real x , which is clearly real-analytic.

No, it does have a non-zero radius, at every step of the way, since on the real line for \( x > 0 \), it equals \( x \).