Continuation beyond the natural boundary with $f_+(z)=\lim_{x \to 1-} f(z,x)=\lim_{x \to 1-}\sum_{n=1}^{\infty}\frac{z^{n^2}}{n^n}n^{-n^2 (1-x)} x^n$

Question

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.

And this paper :

It's a natural boundary. The singularities on the boundary are too dense to try to analytically extend through it. — Brevan Ellefsen, Dec 15 '23 at 03:10
@BrevanEllefsen the boundary is a natural boundary but it is bounded and $C^{\infty}$. The continuation is not analytic continuation. That would ofcourse be inpossible. Not every continuation is analytic continuation. We might have an analytic function beyond the boundary. Such functions exist. — mick, Dec 15 '23 at 03:33
@BrevanEllefsen quote : The theory of generalized analytic continuation studies continuations of meromorphic functions in situations where traditional theory says there is a natural boundary. — mick, Dec 15 '23 at 03:36

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