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?

This depends on your definition of oscillation of sequences.According to the Wikipedia page here: https://en.wikipedia.org/wiki/Oscillation_(mathematics)#Oscillation_of_a_sequence, the oscillation is zero if and only if the sequence converges. So the answer key is correct because the difference between limit superior and the limit inferior is non-zero and hence the oscillation is non-zero, and so the sequence does not converge.
