Ratio Test intuitive Idea 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.
 A: 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! - ${\frac{\pi^2}{6}}$).
