Consider $z$ is complex and
$$f(z) = \sum_{n=1}^{\infty} \frac{z^{n^2}}{n^n}$$
This function has a natural boundary at $|z| = 1$.
Now I was thinking about summability methods or continuations beyond the natural boundary.
Define $$f(z,x) = \sum_{n=1}^{\infty} \frac{z^{n^2}}{n^n} n^{-n^2 (1-x)}x^n$$
Notice $f(z,1) = f(z)$.
Now let $f_+(z)$ be the proposed continuation of $f(z)$ that converges within some annulus $1 < |z| < r$.
Can we say
$$f_+(z) = \lim_{x \to 1-} f(z,x) = \lim_{x \to 1-} \sum_{n=1}^{\infty} \frac{z^{n^2}}{n^n}n^{-n^2 (1-x)} x^n $$
If so, what is the value of $r$ ?
And if not, why not ?
How do we even do such limits ?
edit
Some are confused and think this is about analytic continuation, but that is not the case !
Analytic continuation through a natural boundary does not exist.
But it might be definable on the other side of the boundary.
Such as for instance
$$t(z) = \sum \frac{z^n}{(z^n - 1) n^3}$$
Is defined within and outside of the unit circle and it has a taylor series that only works within the unit circle.
See also : generalized analytic continuation and Continuation of functions beyond natural boundaries.
And this paper :