2
$\begingroup$

My teacher said there are two main ways a sequence can diverge, it can increase in magnitude without bound, or it can fail to resolve to any one limit.

But maybe that second kind of divergence is too diverse? There is a big difference between the divergent sequence 1, -1, 1, -1 . . . And the sequence formed by taking a digit from pi, g, then adding up the next g digits of pi and dividing that by g. (6/3, 25/5, 36/5, 11/2, 18/4, . . . )

Yet both of the above are more orderly than a sequence of random numbers. From what little I understand of randomness.

So maybe we should say that we have:

  1. Sequences that increase in magnitude without bound.
  2. Sequences the can be decomposed in to convergent sub sequences, or in to sequences as in #1
  3. Sequences based on a rule.
  4. Random sequences.

Yet, a random sequence, with even distribution will have convergent sub sequences to every number in it's range...suddenly randomness seems orderly.

What do professionals say about types of divergence?

  • 3
    You are awfully close to http://www.multicians.org/thvv/borges-animals.html ! :)2011-07-18

2 Answers 2

1

Every sequence that doesn't increase in magnitude without bound can be decomposed into convergent subsequences.

EDIT: Maybe a useful way of classifying divergent sequences (of real numbers) would be by the set of accumulation points, that is, by the set of limit points of convergent subsequences. One could ask

  1. Is $+\infty$ an accumulation point?

  2. Is $-\infty$ an accumulation point?

  3. Are there any finite accumulation points?

  4. Is there more than one finite accumulation point?

  5. Are there infinitely many accumulation points?

  6. Are there uncountably many accumulation points?

  7. Is every real number an accumulation point?

  • 4
    @Gerry: It’s worth pointing out that the limits of the subsequences produced in this way don’t necessarily exhaust the cluster pts. of the sequence. They *must* do so if the set $C$ of cluster pts. is discrete; *may* do so if $C$ is countably infinite but not discrete; and *cannot* do so if $C$ is uncountable (and hence of power $2^\omega$, since $C$ is closed).2011-07-18
1

As Mariano aptly suggests in his (borgian) comment, your classification tends to mix different concepts, and hence can hardly be useful (useful classifications, in math, are usually muttually exclusive and complete). When speaking about convergence, the traditional and most useful classification is simply

  • convergent sequences: have a (finite) limit
  • divergent sequences = non convergent

Another classification (in principle "orthogonal" to the above, i.e. not necessarily related) is

  • bounded sequences
  • unbounded sequences

It results that the set of convergent sequences is a proper subset of the bounded sequences.

Further, inside the set of the unbounded sequences, (all divergent), we can put apart the sequences that "tend to (plus/minus) infinity" ( We say a sequence tends to infinity if, however large a number we choose, the sequence becomes greater than that number, and stays greater - see eg here). One can loosely say that these sequences "have a limit" (plus/minus infinity), but they are not properly convergent sequences.

Other non convergent sequences (bounded or not), simply "have no limit". Bounded sequences always include at least a convergent subsequence, and some can be decomposed into a finite set of convergeent subsequences, but this is a less important classification.

Regarding or classification about "random" sequences and "following a rule", that's a very interesting thing but also quite difficult, I would not advise to mix those concepts with the divergent/convergent classification (for example, we could even have convergent random sequences).