This is a continuation of this question. I want to check if $(a_n) \to 0$, then $\sum\limits_{n=1}^\infty a_n$ is convergent.
As @reuns stated in his answer to the question in his above link. If $(a_n)$ is convergent in $R((x))$ then $(a_n)$ is Cauchy. So as @reuns stated:
Writing $a_n = \sum\limits_{j\ge J_n} A_{n,j} x^j$ we get that $(A_{n,j})_{n\ge 1}$ is constant for $n$ large enough and $$\lim_{n\to \infty} a_n = \sum_{j\ge J_{N_1}} (\lim_{n\to \infty} A_{n,j}) x^j\in R((x))$$
So in $\sum\limits_{n=1}^ \infty a_n = \sum\limits_{n=1}^\infty \left(\sum\limits_{j\ge J_n} A_{n,j}x^j\right) = \sum\limits_{j}\left(\sum\limits_{n=1}^\infty A_{n,j}\right)x^j \underset{?}\in R((x))$ As I think if $\min\{J_n: n \in \mathbb{N}\}$ does not exists. Then I think belongingness is not true.
Also if min exists then $\sum\limits_{n=1}^\infty A_{n,j}$ as almost finitely many nonzero terms. Hence the summation terminates after finitely many steps. So we can tell the series is summable.
Is my line of thinking ok? If yes then how do I write it rigorously? and do the belongingness hold?