What is the relation between convergence and a quotient of consecutive terms of it?

Question

Let $\{x_n\}_{n=0}^\infty$ be a convergence to zero sequence . Define $r_n=\frac{x_n}{x_{n-1}}$.

1) What is the relation between the convergence of a sequence $\{x_n\}_{n=0}^\infty$ and $\lim_{n\to\infty}r_n$?.

2) Is $\lim_{n\to\infty}r_n$ is the rate of convergence of the sequence $\{x_n\}_{n=0}^\infty$?

$x_n$ does not equal zero.

Answer 1

There is no clear-cut answer, many things can happen, and it's not clear to me whether one should write an exhaustive account of what could or could not be the relationship between convergence of $(r_n)_n$ and rate of convergence of $x_n)_n$. Therefore, writing this as Community Wiki -- feel free to improve or edit this answer.

1. $(r_n)_n$ might not be defined. For instance, if infinitely many of the $x_n$'s are zero, you're in trouble.

$$ x_{2n} = 0, \qquad x_{2n+1} = \frac{1}{2n+1} $$ for $n\geq 0$, for instance.

Assuming now that all $x_n$'s are positive.

2. Then $(r_n)_n$ is not necessarily convergent. $$ x_{2n} = x_{2n+1} = \frac{1}{2^n} $$ for $n\geq 0$, for instance.

3. Worse, it is not even guaranteed to be bounded. $$ x_{2n} = \frac{1}{n+1}, \qquad x_{2n+1} = \frac{1}{(n+1)^2} $$ for $n\geq 0$, for instance.

4. Assuming $(r_n)_n$ does converge to some limit $\ell\in \mathbb{R}$:

   • One must have $\ell\in[-1,1]$ since $(x_n)_n$ converges to $0$. Indeed, if $\lvert \ell\rvert > 1$, then for $\varepsilon \stackrel{\rm def}{=} \frac{\lvert \ell\rvert-1}{2}$ there exists $N\geq 0$ such that, for every $n\geq N$, $\lvert r_n\rvert \geq \lvert \ell\rvert - \varepsilon = \frac{\lvert \ell\rvert+1}{2} > 1$. But then, for $n\geq N$ we have $$\lvert x_n\rvert \geq \lvert x_{n-1}\rvert \cdot \frac{\lvert \ell\rvert+1}{2} \geq \dots \geq \lvert x_N\rvert \cdot \left(\frac{\lvert \ell\rvert+1}{2}\right)^{n-N} \xrightarrow[n\to\infty]{}\infty $$ which is a contradiction.

   • If $\ell\in(-1,1)$, then $(x_n)_n$ asymptotically converges very fast (exponentially) to $0$, by comparison to a geometric series. Same argument as before: since $\lvert \ell\rvert < 1$, setting $\varepsilon \stackrel{\rm def}{=} \frac{1-\lvert \ell\rvert}{2}$ there exists $N\geq 0$ such that, for every $n\geq N$, $\lvert r_n\rvert \geq \lvert \ell\rvert + \varepsilon = \frac{\lvert \ell\rvert+1}{2} < 1$. But then, for $n\geq N$ we have $$\lvert x_n\rvert \leq \lvert x_{n-1}\rvert \cdot \frac{\lvert \ell\rvert+1}{2} \leq \dots \leq \lvert x_N\rvert \cdot \left(\frac{\lvert \ell\rvert+1}{2}\right)^{n-N} $$ and the RHS converges to $0$ at an exponential (geometric) rate.

   • $\ell\in\{-1,1\}$ is possible. For instance, consider $(x_n)_n$ defined by $x_n =\frac{1}{n+1}$.

   • Note that inspecting the proof of the (second bullet) above (where we did assume convergence of $(x_n)_n$ to $0$ to conclude that $\ell\in[-1,1]$) also immediately proves that, if we did not assume convergence of $(x_n)_n$, we still have $$ r_n \xrightarrow[n\to\infty]{} \ell \in(-1,1) \Rightarrow x_n \xrightarrow[n\to\infty]{} 0 $$ and further that the convergence to $0$ is exponentially (geometrically) fast. In the case where $\lvert \ell \rvert > 1$, the proof in the first bullet above also implies that $(x_n)_n$ diverges ($\lvert x_n\rvert \xrightarrow[n\to\infty]{} \infty$ exponentially fast). But if $\ell \in\{-1,1\}$, one cannot conclude anything. There are examples (as the one above, with $\frac{1}{n+1}$) where $x_n \xrightarrow[n\to\infty]{} 0$. There are others (take $x_n = \ell$) where $x_n \xrightarrow[n\to\infty]{} \ell \neq 0$. There are others (take $x_n = n$) where $x_n \xrightarrow[n\to\infty]{} \infty$. And there are where $(x_n)_n$ is bounded but divergent (and even, say, dense in $[-1,1]$).