Why does every oscillating sequence diverge? Is this statement true?
$$\bf\text{Every oscillating sequence diverges.}$$
My thoughts: $\bf{False}$. $s_n = (-1)^n$ does not converge. But it's bounded, therefore, not divergent either. Divergent means diverging to $-\infty$ or $+ \infty$, yes?
Solution key: $\bf{True}$. If a sequence oscillates, then its limit inferior and limit superior are unequal. If follows that it cannot converge, for if it converged all its subsequences would converge to the same limit.
Three other places discussing oscillating convergence:

*

*This website says: "Oscillating sequences are not convergent or
divergent. Such as 1, 0, 3, 0, 5, 0, 7,..." I agree.


*This SE post says: "Diverge means doesn't converge."  But, I
think it can be neither?


*This SE post says: "$\sin xe^{-x}$ is oscillating and
convergent." I agree.
So, is the solution key correct? Who's right here?

In the usual definitions that I have picked up from various instructors and textbooks, "this sequence diverges" just means "this sequence doesn't converge". You could self-consistently define things the other way, but this is unusual in my experience.
By contrast, "this sequence oscillates" is usually not rigorously defined. Thus when we somewhat casually say "this sequence doesn't converge because it oscillates forever" we really mean "this sequence doesn't converge because it oscillates with an amplitude bounded away from zero forever". With no formal definition setting the context, I would probably say that $\frac{(-1)^n}{n}$ both oscillates and converges.
