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:

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

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

  3. This SE post says: "$\sin xe^{-x}$ is oscillating and convergent." I agree.

So, is the solution key correct? Who's right here?

  • 1
    $\begingroup$ (2) and (3) are right; The example in (1) only if you allow the limit to be $\pm\infty$. $\endgroup$ Jul 22, 2020 at 18:28
  • 4
    $\begingroup$ It's a question of terminology, and the standard terminology (as far as I'm aware), is that divergence covers anything that doesn't converge. $\endgroup$ Jul 22, 2020 at 18:29
  • $\begingroup$ Is the solution key correct? $\endgroup$
    – user13985
    Jul 22, 2020 at 18:30
  • $\begingroup$ @ElchananSolomon Could you point to some authority sources? $\endgroup$
    – user13985
    Jul 22, 2020 at 18:35
  • 1
    $\begingroup$ @user13985 : mathworld.wolfram.com/DivergentSequence.html or merriam-webster.com/dictionary/divergent. I'm afraid that if you're expecting a Biblical or Constitutional authority on math definitions, you may be disappointed. $\endgroup$ Jul 22, 2020 at 19:23

3 Answers 3


You have to be very precise with your definitions. Generally, we call a sequence divergent if it does not converge. This means that convergent and divergent are each other's opposite.

As far as I know, there is no accepted definition for oscillating sequence.

The sequence $(-1)^n$ diverges, because it does not converge, while the sequence $\frac{(-1)^n}{n}$ converges to zero. Depending on the precise definition, you may or may not call the latter sequence oscillating, so you have to consult your textbook there.

  • $\begingroup$ Oscillation is a particular case of a more general one being a sequence with multiple "adherence" values (limit point in English ?) which is called divergent too. $\endgroup$
    – zwim
    Jul 22, 2020 at 19:09

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.