Is this a valid convergence test (for sequences)?

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.

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.


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 $a(k)\leq n\leq a(k+1)$.

Tags:

Convergence Divergence

Cauchy Sequences

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