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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?

RobPratt
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Saikat
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  • $$\lim_{n\to \infty} a_n = \sum_{j\ge \min(0,J_{N_1})} (\lim_{n\to \infty} A_{n,j}) x^j\in R((x))$$ because each $a_n, n\ge J_{N_1}$ is in $a_{N_1}+R[[x]]$ – reuns Jan 11 '21 at 20:08
  • I am convinced with "each $a_n$, $n \ge J_{N_1}$ is in $a_{N_1}+R[[x]]$", now how can we tell that for any $n$ we cannot have $J_n<J_{N_1}$? – Saikat Jan 11 '21 at 20:20
  • And while summing up $j$ must run over all integers greater than or equal to $J_n$ for each $n \in \mathbb{N}$ Is this correct? Please help me understand these. – Saikat Jan 11 '21 at 20:22
  • $R[[x]]$ is the subring of formal power series, ie. Laurent series with $0$ negative coefficients – reuns Jan 11 '21 at 21:34
  • You don't necessarly need to use everything you did in your past inquiries. And you must use the fact that $(a_n){n \in \mathbb{N}}$ tends to $0$ (as opposed to just being Cauchy, i.e. just being convergent). So since it does, for all $N \in \mathbb{N}$, there is $k \in \mathbb{N}$ with $|a_n|<x^N$ for all $n\geq k$. Using that, can you prove that $(\sum \limits{k=0}^n a_k)_{n \in \mathbb{N}}$ is Cauchy, hence convergent? – nombre Jan 12 '21 at 07:48

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