If $x_n \to a$, what can be said about $\displaystyle{\lim_{n\to \infty}\frac{x_{n+1}}{x_n}}$ ?
I do have a solution but I wonder whether there is a more simple way to find the answer.
Solution.
(1) If $a \neq 0,$ then $$\lim_{n \to \infty} \frac{x_{n + 1}}{x_n} = \frac{\displaystyle\lim_{n \to \infty} x_{n+1}}{\displaystyle\lim_{n \to \infty} x_n} = \frac{a}{a} = 1$$
(2) If $a = 0,$ then $\displaystyle \lim_{n \to \infty} \frac{x_{n + 1}}{x_n}$ may not exist.
Consider the sequence $$x_n=\underbrace{\frac{1}{p}}_{\color{grey}{x_1}}, \underbrace{\frac{1}{p}}_{\color{grey}{x_2}}, \underbrace{\frac{1}{p^2}}_{\color{grey}{x_3}}, \underbrace{\frac{1}{p^2}}_{\color{grey}{x_4}}, \cdots, \underbrace{\frac{1}{p^k}}_{\color{grey}{x_{2k-1}}}, \underbrace{\frac{1}{p^k}}_{\color{grey}{x_{2k}}}, \cdots \space (p>1)$$
We have $\displaystyle \lim_{n\to\infty}x_n=0$. But
$$\lim_{n \to \infty} \frac{x_{2n}}{x_{2n - 1}} = \lim_{n \to \infty} \frac{\frac1{p^n}}{\frac1{p^n}} = 1$$
$$\lim_{n \to \infty} \frac{x_{2n+1}}{x_{2n}} = \lim_{n \to \infty} \frac{\frac1{p^{n+1}}}{\frac1{p^n}} = \frac{1}{p}$$
so the limit of $\displaystyle \frac{x_{n+1}}{x_n}$ does not exist.
Another example is a stationary sequence $\{x_n\}=0$ which does have a limit $$\displaystyle \lim_{n\to\infty}x_n=\lim_{n\to\infty}0=0$$ but the limit of $\displaystyle \frac{x_{n+1}}{x_n}$ does not exist: $$ \lim_{n\to\infty}\frac{x_{n+1}}{x_n}=\lim_{n\to\infty}\frac{0}{0}=\frac{0}{0} - undefined$$
(3) Suppose $\displaystyle \frac{x_{n + 1}}{x_n}$ does have a limit and its value is $b.$ Let's prove that $|b|$ must be $\leq 1$. For contradiction, let's suppose that $|b| > 1$. By limit property $\displaystyle \lim_{n \to \infty} \left| \frac{x_{n+1}}{x_n} \right| =|b|>1$. $$\left|\left|\frac{x_{n + 1}}{x_n}\right| - |b|\right| < \epsilon$$
which implies that starting from some $n_{\varepsilon}$ for all $n>n_{\varepsilon}$
$$|b| - \varepsilon < \left|\frac{x_{n + 1}}{x_n} \right|< |b| + \varepsilon$$
Because this must hold for all $\varepsilon$, we can consider those $\varepsilon$ such that $|b| - \varepsilon > 1$. Now we have $$1<\left|\frac{x_{n+1}}{x_n} \right|<|b|+\varepsilon$$ $$\left|\frac{x_{n+1}}{x_n}\right| = \frac{|x_{n+1}|}{|x_n|} > 1$$ for $n > n_{\varepsilon}.$
Now, noting that
$$|x_n| = |x_{n_\varepsilon + 1}| \cdot \frac{|x_{n_\varepsilon + 2}|}{|x_{n_\varepsilon + 1}|} \cdot \frac{|x_{n_\varepsilon + 3}|}{|x_{n_\varepsilon + 2}|} \cdot \ldots \cdot \frac{|x_n|}{|x_{n-1}|}$$
let $\displaystyle \lambda = \min\left\{\frac{|x_{n_\varepsilon + 2}|}{|x_{n_\varepsilon + 1}|}, \frac{|x_{n_\varepsilon + 3}|}{|x_{n_\varepsilon + 2}|}, \ldots, \frac{|x_n|}{|x_{n-1}|}\right\}.$ Because $\displaystyle \frac{x_{n+1}}{x_n} > 1$ for all $n > n_\varepsilon,$ this means that $\lambda > 1,$ so $$|x_n| > |x_{n_\varepsilon + 1}| \cdot \lambda^{n - (n_\varepsilon + 1)}$$ implies that
$$\lim_{n \to \infty} |x_n| \geq \lim_{n \to \infty} \left[ |x_{n_\varepsilon + 1}| \cdot \lambda^{n - (n_\varepsilon + 1)} \space \right]= \frac{|x_{n_\varepsilon + 1}|}{\lambda^{n_\varepsilon + 1}} \cdot \lim_{n \to \infty} \lambda^n = +\infty$$ $$\lim_{n\to\infty}|x_n|=+\infty$$
However, this contradicts our definition of $\displaystyle a = \lim_{n \to \infty} x_n = 0,$ so by contradiction we must have that $|b| \leq 1.$
Summary:
if $\displaystyle \lim_{n\to\infty}x_n=a \not = 0$, then $\displaystyle \lim_{n\to\infty}\frac{x_{n+1}}{x_n}=1 $;
if $\displaystyle \lim_{n\to\infty}x_n = a=0$, then either $\displaystyle \lim_{n\to\infty}\frac{x_{n+1}}{x_n}=b\in [-1,1] $ or the limit doesn't exist.
This is my first time here, btw.
