Consider a real function $f(x)$. Can the function series of $f(x)$ be convergent where $f(x)$ is not defined "naturally"? More specifically, can the series converge where the function is not defined naturally and not continous expandable?
Take for example
$$1+\frac{1-x}{x} \mathrm{log}(1-x)=\sum_{n \geq 0} \frac{x^n}{n(n+1)}$$
For $x=0$ and $x=1$ the function $f(x)=1+\frac{1}{1-x} \mathrm{log}(1-x)$ is not defined naturally, but $\sum_{n \geq 0} \frac{x^n}{n(n+1)}$ converges for that values of $x$.
Nevertheless $$\mathrm{lim}_{x \to 0} f(x)=0=\sum_{n \geq 0} \frac{0^n}{n(n+1)}$$ And
$$\mathrm{lim}_{x \to 1} f(x)=1=\sum_{n \geq 0} \frac{1^n}{n(n+1)}$$
This means that the function is continuous expandable in $x=0$ and $x=1$.
So can I have that a series (of functions) converges where the sum function is not defined, but not even continous expandable?
The answer, in general, is yes. Please note that the function $1+\frac{1}{1-x} \mathrm{log}(1-x)$ is well defined in $0$.
The thing is, the series $\sum_{n \in \Bbb N}^\ \frac{x^n}{n(n+1)}$ is a power series with convergence radius $1$: $$\limsup_{n\to\infty}{\frac{1}{\sqrt[n]{\,a_n}}} = \limsup_{n\to\infty}{\sqrt[n]{n(n+1)}} = 1,$$ where $a_n = \frac{1}{n(n+1)}$.
Therefore convergence to the function is not guaranteed for $x=1$.