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My attempt to understand the proof is as follows:
If $R=\lim\sup\limits_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|<1$ implies that for $\epsilon$ such that $0<\epsilon<1-R$ we can find an $N\in\mathbb{N}$ such that: $$|{a_{n+1}}|\leq (R+\epsilon)|{a_n}|\Rightarrow |a_{n+1}|<|a_n| , \forall n\geq N$$ That is, the sequence $|a_n|_{n\in\mathbb{N}}$ is going to be decreasing strictly for n large enough. Also redefining:
$x_n=|a_{N+n}|$ and $y_n=(R+\epsilon)^n|a_{N}|$. Then the sequence $x_n$ is dominated by the geometric progression $y_n$.

I.e, the series $\sum_{n=0}^{\infty}a_n$ is absolutely convergent if from a certain moment the sequence $|a_n|$ it is strictly decreasing and there is a geometric progression of a ratio less than 1 that dominates it(and that begins at that moment). Is this reasoning correct?

Now if $R=1$, I still can't understand what it means.

Toby Bartels
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David Morante
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    That the series is eventually decreasing, while true, isn't necessary. It's non-negative and bounded above by a convergent geometric series; that's enough for the Direct Convergence Test. But the reasoning is correct. When $R=1$, you would need $0 < \epsilon < 0$, which is impossible, so the proof doesn't go through in that case. (And of course, $R > 1$ is only worse.) – Toby Bartels Jul 07 '20 at 23:12

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This is exactly the right idea yes. If the limit of the ratio is less than $1$, it essentially tells you at some definable point - the remaining infinite number of terms in the series becomes bounded by some constant multiplied by a geometric series with ratio less than $1$ (which we know to be convergent) and hence the series overall has to converge:

$${\sum_{n=1}^{N}|a_n| + \sum_{n=N}^{\infty}|a_n|\leq \text{sum of finite terms} + \text{constant multiple of geometric series with ratio ${<1}$}}$$

And since the series has been shown to be absolutely convergent, then ${\sum_{n=1}^{\infty}a_n}$ must also be convergent (that is, it's also conditionally convergent).

If the limit does not exist, or the ratio ${=1}$ then in fact we cannot say anything. Easy examples:

(1) If you apply this test to the harmonic series, ${\sum_{n=1}^{\infty}\frac{1}{n}}$ - you get a ratio of ${1}$. And the harmonic series diverges.

(2) If you apply this test to the basal problem - ${\sum_{n=1}^{\infty}\frac{1}{n^2}}$ - then you get a ratio of ${1}$, yet the series converges (and to a nice result as well! - ${\frac{\pi^2}{6}}$).

Riemann'sPointyNose
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  • We know that $\dfrac{1}{n^2}$ goes to $0$ faster than $\dfrac{1}{n}$. Thinking like this, in the case of $R= 1$, could we find a sequence $(a_n)$ whose trend speed to $0$ is considered as a parameter ?, that is, $\dfrac{1}{n}$ goes to $0$ slower than $(a_n)$, then it diverges. And $\dfrac{1}{n^2}$ goes faster than $(a_n)$, so it converges. – David Morante Jul 07 '20 at 23:30
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    @DavidMorante, that's probably best asked as a separate question. I think it's unlikely, but it depends on how you define "goes to $0$ faster than". One problematic example I have in mind is $x_n = 1/n$ if $n$ is a power of $2$, and $1/n^2$ otherwise. Maybe the question is more interesting if you only consider monotone sequences? This is probably also relevant. – Izaak van Dongen Jul 08 '20 at 12:14
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    @DavidMorante, I've just remembered another useful example. Let $x_n = 2^{-n} + 1 / a_n \log a_n$, where $a_n$ is the next number at least as big as $n$ of the form $2^{k^2}$ for some $k$. Then $x_n$ is monotone decreasing, $\sum_n x_n$ converges, but $x_n$ is at least as large as $1 / n \log n$ infinitely often, the sum of which diverges. – Izaak van Dongen Jul 08 '20 at 12:50