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.


(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 $|b| \leq 1.$


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.

  • 1
    $\begingroup$ Welcome to Math.SE. Please edit your question to include your solution using MathJax formatting, not as an image. $\endgroup$
    – aschepler
    Jun 15, 2022 at 21:19
  • $\begingroup$ For future reference, it's best not to attach crucial parts of a post as images. I'm working on an answer now but I can try to type that out in the post afterwards if you'd like (although I might need to check with you on some translation stuff - I can't read the language but I think I've understood your argument) $\endgroup$ Jun 15, 2022 at 21:19
  • $\begingroup$ If $a = 0$, then $\lim_{n \rightarrow \infty} \frac{x_{n+1}}{x_n}$ could be anything. For eg, for $x_n = b^{-n}$ where $b > 1$, we have $\lim_{n \rightarrow \infty} \frac{x_{n+1}}{x_n} = \frac{1}{b}$, so different values for different b. $\endgroup$
    – Anon
    Jun 15, 2022 at 21:20
  • 1
    $\begingroup$ @Kaind, First of all you probably meant $a = 0$, as otherwise the limit must be equal to $1$, and second of all the limit still can't literally be anything, for example it may not equal $2$. $\endgroup$
    – Adayah
    Jun 15, 2022 at 21:24
  • $\begingroup$ @Adayah So the OP is asking for an upper bound on the limit? The limit need not even be defined for $a = 0$ for eg: $x_n = 0$ $\endgroup$
    – Anon
    Jun 15, 2022 at 21:26

1 Answer 1


I think I've understood your attempt (as best as I can without being able to read the language) and here are my thoughts:

Your first point is correct: if $a \neq 0$ then we must have $\lim_{n \to \infty} \frac{x_{n+1}}{x_n} = \frac{\lim_{n \to \infty} x_{n + 1}}{\lim_{n \to \infty} x_n} = \frac{a}{a} = 1.$ If you'd like we could show this in the epsilon-delta form like this: if $|x_n - a| \leq \epsilon$ for all $n \geq n_0(\epsilon),$ then

$$|x_{n + 1} - x_n| \leq |x_{n + 1} - a| + |x_n - a| \leq 2\epsilon \ \text{for } n \geq n_0(\epsilon)$$

$$|x_n - a| \leq \epsilon \Rightarrow a - \epsilon \leq x_n \leq a + \epsilon \Rightarrow |x_n| \geq a - \epsilon \text{ for } n \geq n_0(\epsilon), \epsilon < a$$

$$\left|\frac{x_{n + 1}}{x_n} - 1\right| = \frac{|x_{n + 1} - x_n|}{|x_n|} \leq \frac{2\epsilon}{a - \epsilon} \text{ for } n \geq n_0(\epsilon)$$

and with a little algebraic manipulation we get that $\left|\frac{x_{n+1}}{x_n} - 1\right| \leq \epsilon_2$ for all $n \geq n_0\left(\frac{a\epsilon_2}{\epsilon_2 + 2}\right).$

For your second point, I agree with your counterexample for the case where $\frac{x_{n+1}}{x_n}$ need not have a limit when $a = 0$ (and I quite like the counterexample you came up with, good job) but I would justify the lack of existence of the limit a bit differently: rather than taking limits of the terms $\frac{x_{2n}}{x_{2n - 1}}$ and $\frac{x_{2n + 1}}{x_{2n}},$ I would just say that the sequence $b_n = \frac{x_{n + 1}}{x_n}$ oscillates between $1$ and $\frac{1}{p},$ and use that to say there is no limit. In any case you have the right idea, the limit does not exist.

For the third part of your argument, where you argue that if the limit exists in the $a = 0$ case then its magnitude must be no more than $1,$ I think you can avoid a little bit of annoyance with $\lambda$ relying on $n$ by tightening your imposed restriction on $\epsilon$: instead of saying there must be an $\epsilon > 0$ such that $|b| - \epsilon > 1$ when $|b| > 1,$ I would instead say there must be an $\epsilon > 0$ such that $|b| - \epsilon > \frac{|b| + 1}{2},$ which will happen when $0 < \epsilon < \frac{|b| - 1}{2}.$ Now you can proceed by proving that $x_n \geq x_{n_0} \cdot \left(\frac{|b| + 1}{2}\right)^{n - n_0}$ for $n \geq n_0,$ and from there your argument regarding comparing $x_n$ to a divergent exponential function follows.

Overall, good work. It's clear you know what you're doing, there are just a few spots I think things could be tightened up.

  • $\begingroup$ Thanks! Already love this forum. Never got answers to my questions so quickly before. I'll try what you suggested and post (tomorrow) in MathJax solution that includes dealing with $\lambda$ and another one with what you suggested. $\endgroup$
    – Cakeice
    Jun 15, 2022 at 22:14
  • $\begingroup$ @Cakeice Glad it helped - should I continue with my translation attempt of your original solution attempt? $\endgroup$ Jun 15, 2022 at 22:15
  • $\begingroup$ I'll add a completely translated version of my solution tomorrow. $\endgroup$
    – Cakeice
    Jun 15, 2022 at 22:19
  • $\begingroup$ @Cakeice Alright - for the convenience of other users I'm still going to add my attempt if you don't mind. Also it might save you some time having to rewrite some stuff $\endgroup$ Jun 15, 2022 at 22:20
  • $\begingroup$ I'm always happy to see other solutions here. $\endgroup$
    – Cakeice
    Jun 15, 2022 at 22:24

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