Let $x_n$ be a real number sequence.
I have managed to prove that,for a sequence $x_n$ of positive terms :
if lim sup $(x_{n+1}/x_{n}) < 1$ then $x_n$ is not only eventually monotonicly non-increasing ( which implies it converges ) but it converges to 0.
I was seeking for some intuition ( maybe ilustrated by examples ) into why positive and eventually monotonicly non-increasing sequences $x_n$ that satisfy $(x_{n+1}/x_{n}) < 1$ must converge to 0 while the others that don't , don't need to converge to 0.
Suppose $x_n$ is a sequence of positive real numbers such that
$$\limsup_{n \rightarrow +\infty} \frac{x_{n+1}}{x_n} = a < 1$$
Fix any $b \in (a,1)$ (for example $b=a+ \frac{1-a}{2}$). Then eventually you have $\frac{x_{n+1}}{x_n} <b$. We can suppose without loss of generality (since we are wondering if $x_n$ is convergent) that for all $n$
$$x_{n+1}<bx_n$$ holds.
Then you can easily prove by induction that for all $n \geq 1$
$$x_n < x_0b^n $$ Hence $$\lim_n x_n \leq \lim_n x_0b^n = x_0 \cdot 0 = 0$$
So $x_n$ converges to $0$ by the sandwich theorem.
Note that the converse does not hold. For example you can consider the following sequence $x_n$: $$1, 2, \frac{1}{2}, \frac{2}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \dots, \frac{1}{m}, \frac{2}{m}, \dots$$ for which you have $\limsup_{n \rightarrow +\infty} \frac{x_{n+1}}{x_n} = 2$
Moreover, if you consider the sequence $\frac{1}{n}$, you easily see that it converges to $0$, but the limsup of the quotients is $1$. So you really need the hypothesis $a <1$.