Is this a valid convergence test (for sequences)?

Question

Let $(x_n)$ be a bounded sequence (from above by $M$) such that $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=1$. Does $(x_n)$ converge?

My first idea was to prove the sequence is Cauchy: Given $\epsilon\gt0$ we need to find $N\ge1$ such that for every $m,n\ge N$ $$|\frac{x_m}{x_n}-1|\lt\frac{\epsilon}{M}$$ $$|x_m-x_n|\lt\frac{\epsilon |x_n|}{M}\le\epsilon$$ Now $$\frac{x_m}{x_n}=\frac{x_m}{x_{m-1}}\cdots\frac{x_{n+1}}{x_n}$$ and each factor can be arbitrarily close to $1$, say in $(1-\delta,1+\delta)$ for some $\delta$, so $$\frac{x_m}{x_n}\in ((1-\delta)^{m-n},(1+\delta)^{m-n})$$ however it is impossible to choose $\delta$ such that $$(1-\epsilon,1+\epsilon)\subseteq((1-\delta)^{m-n},(1+\delta)^{m-n})$$ for every $m,n$ greater than something: if $m\to\infty$ then the interval with the $\delta$ becomes $(0,\infty)$

What is missing? I've also tried the $|x_m-x_n|\le|x_m-x_{m-1}|+\cdots+|x_{n+1}-x_n|$ trick but arrive at the same problem (the product {sum} of a lot of things close to one {zero} is far away from it)

Maybe there is some very simple counterexample I'm not seeing

Answer 1

Idea of counterexample: Construct an oscillating function whose "rate of oscillation" decreases as $n$ increases. The idea is to make the differences between successive values smaller due to the decreasing "rate of oscillation", while the fact that that it oscillates ensures that it does not converge.

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Counterexample: Consider the sequence $x_n = \sin \sqrt{n} + 2$. This is obviously bounded above by $M = 3$ and does not converge.

Finding the limit is not straightforward, though. $$ \lim_{n \to \infty} \frac{x_{n+1}}{x_n} = \lim_{n \to \infty} \frac{\sin \sqrt{n+1} + 2}{\sin \sqrt{n} + 2} \\ =1 + \lim_{n \to \infty} \left[ \frac{\sin{\sqrt{n+1}} - \sin \sqrt{n}}{\sin \sqrt{n} + 2}\right] $$ But we also have $$ \lim_{n\to \infty} (\sin{\sqrt{n+1}} - \sin \sqrt{n}) = 2 \lim_{n\to \infty} \left[ \cos \frac{\sqrt{n+1} + \sqrt{n}}{2} \sin \frac{\sqrt{n+1} - \sqrt{n}}{2} \right] $$ Since $\lim_{n \to \infty} (\sqrt{n + 1} - \sqrt{n}) = 0$, then by a couple of applications of the squeeze theorem we have $$ \lim_{n \to \infty} \left[ \frac{\sin{\sqrt{n+1}} - \sin \sqrt{n}}{\sin \sqrt{n} + 2}\right] = 0, $$ and thus $$ \lim_{n \to \infty} \frac{x_{n+1}}{x_n} = 1, $$ as desired.

This is not a terribly elegant counterexample, and I wouldn't be surprised if there are other ones out there.

EDIT: A slightly less cumbersome counterexample is the sequence $x_n = e^{\sin \sqrt{n}}$. The proof proceeds in much the same way, and relies on the fact that $$ \lim_{n \to \infty} (\sin \sqrt{n + 1} - \sin \sqrt{n}) = 0.$$ As noted in by @DanielFischer in the comments, the mean value theorem implies that $$ \frac{\sin \sqrt{n+1} - \sin \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \sin x $$ for some $x \in [\sqrt{n}, \sqrt{n+1}]$; and since $\sin x$ is bounded between -1 and +1, this means that $$ |\sin \sqrt{n+1} - \sin \sqrt{n}| \leq \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} $$ for all $n$. The convergence of both of the counterexamples follows from there.

Answer 2

This is not a solution, but I think you can solve it if you can fill all gaps.

Since $(x_n)_n$ is a bounded sequence, there exists a Cauchy subsequence $(x_{a(n)})_n$ (hence convergent). Let $r=\lim_{n\to\infty} x_{a(n)}$. Since $\frac{x_{n+1}}{x_n}\to 1$, for $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that $$x_n(1-\varepsilon)<x_{n+1}<x_n(1+\varepsilon)\qquad\qquad\forall n\geq N.$$ This implies $$x_n(1-\varepsilon)^{m-n}<x_{m}<x_n(1+\varepsilon)^{m-n}\qquad\qquad\forall m\geq n\geq N.$$ The upshot is to estimate the difference between $r$ and $x_n(1\pm \varepsilon)^{n-a(k)}$, where $a(k)\leq n\leq a(k+1)$.

Answer 3

Let $(a_n)_{n\in\Bbb N}$ be the sequence$$1,2,\frac32,1,\frac43,\frac53,2,\frac74,\frac64,\frac54,1,\frac65,\ldots$$It is bounded and it fails to converge (you have $a_n=1$ infinitely many times and you also have $a_n=2$ infinitely many times). It is not hard to prove that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$.

Answer 4

Assume we have a strictly increasing sequence of natural numbers $a_1=1,a_2,\dots$ with $a_{m+1}-a_m\to\infty.$ Say $a_m=m^2.$ Then we can define, for $a_m\leq n<a_{m+1}:$ $$x_n=\begin{cases}1+\frac{n-a_m}{a_{m+1}-a_m}&m\text{ odd}\\ 2-\frac{n-a_m}{a_{m+1}-a_m}&m\text{ even}\end{cases}$$

Then we get $|x_{n+1}-x_n|=\frac1{a_{m+1}-a_m}.$ The only time this is not obvious is when $n+1=a_{m+1},$ but it isn’t hard to check.

Since $x_{n}\geq 1$ for all $n,$ we get:

$$\left|\frac{x_{n+1}}{x_n}-1\right|\leq\frac1{a_{m+1}-a_m}\to\infty$$

We also get $1\leq x_n\leq 2,$ so $x_n$ is bounded.

But $x_{a_m}$ is $1$ when $m$ is odd and $2$ when $m$ is even. So $x_n$ diverges.

To remove $m$ from the equation, we can replace $a_k$ with $k^2$ and $m$ with $\lfloor \sqrt{n}\rfloor.$. Or $a_k=2^{k}$ and $m=\lfloor \log_2n\rfloor.$

Essentially, $x_n$ is bouncing between $1$ and $2,$ but at slower and slower intervals. So this isn’t much different than the $2+\sin(\sqrt n)$ solution, but maybe requiring less trigonometry, and actually reaching the supremes and infimum.