Suppose I have a real sequence $x_n\to 0$. Is it true that: $$ \left|\frac{x_{n+1}}{x_n}\right|\to r<1 $$ for some $r\in\mathbb{R}$? If not, is it true that: $$\exists N\in\mathbb{N}:\left|\frac{x_{n+1}}{x_n}\right|<1,\forall n>N$$
Thank you for your help and understanding! :D
On
Consider $u_n$ such that $u_{2n}=\frac 1 {2n}$ and $u_{2n+1}=\frac 1 {\sqrt{2n}}$.
$u_n\rightarrow 0$ but $\frac {u_{2n+1}} {u_{2n}}=\sqrt{2n}\rightarrow\infty$
On
Put $x_n = 1/n$ if $n$ is even and $x_n = 2/n$ if $n$ is odd. Then $$\left \lvert \frac{x_{n+1}}{x_n} \right \rvert = 2^{(-1)^n}\frac{n}{n+1}$$ which is less than $1$ for $n$ odd and greater than $1$ for $n$ even and has no limit as $n \to \infty$.
On
These are wildly false even for very simple examples.
Consider $x_n = \frac{1}{\ln n}$ for $n>1$, or, for any nonconstant polynomial $P(x)$, $x_n = \frac{1}{P(n)}$. We have $x_n \rightarrow 0$, $\left| \frac{x_{n+1}}{x_n} \right| \rightarrow 1$.
For $n \geq 0$, let $x_n = \prod_{j=1}^n \left( \frac{7}{6} + (-1)^j\frac{5}{6} \right)$ so our sequence is $(1,\frac{1}{3},\frac{2}{3}, \frac{2}{9}, \frac{4}{9}, \frac{4}{27}, \frac{8}{27}, \dots)$, where $\frac{x_{n+1}}{x_n} \in \{\frac{1}{3},2\}$. Then $x_n \rightarrow 0$ and $\frac{x_{n+1}}{x_n} = 2 > 1$ infinitely often.
For $n \geq 0$, let $x_n = \frac{\cos(\pi n)}{n}$, so our sequence is $(1,0,\frac{-1}{3}, 0, \frac{1}{5}, 0, \frac{-1}{7}, 0, \dots )$. We have $x_n \rightarrow 0$ and $\frac{x_{n+1}}{x_n}$ is undefined infinitely often. Replace the zeroes with $2^{-n}$ and $\frac{x_{n+1}}{x_n}$ will be unbounded (in both positive and negative directions). In fact any interleaving of $\text{polynomial}^{-1}$ and a sequence of successive negative powers (of an element of $(-1,1) \subset \Bbb{R}$), as long as both sequences appear infinitely often, will have this property.
Consider the sequence $$\frac{1}{1},-\frac{1}{1},\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\ldots$$
The terms go to zero, but consecutive terms have ratio that is $1$ in absolute value, half of the time.