Is it true that any non-convergent function or series must eventually "terminate" to some infintitely repeating sequence or series?

Question

I don't know really any real analysis aside from the mindset of heavily proving things in a rigorous manner. I'm just trying to see if something I think is true is true or not. It might relate to something I might think about further down the line.

With that out of the way... let's say we have some sequence or function f(x). It doesn't really matter since $f$ is either discrete or continuous.

Suppose that $\lim_{x \to \infty} f(x)$ diverges and doesn't increase or decrease without bound. I wish to show that ultimately it must diverge in the sense that it cycles through the same discrete or continuous set of values.

I don't know the proper term for this or any rigorous way to define it. I hope the statement makes sense.

Ultimately this relates to me potentially trying to construct a way of making undefined limits elements of some non-complex number set and I wanted to make sure whether or not the periodicity could at all be a property I choose or choose not to use in my definitions. Either way I am still curious.

This is not the primary question (the highlighted conjecture is), but I believe this is primarily useful as if I understand correctly all limits can be reconstructed as infinite limits of functions/sequences. Please confirm this as well. I assume this is something most people will know from the basics.

Answer 1

What you can show is that if $f(x)$ is bounded then its values have an accumulation point. This is the Bolzano-Weierstrass theorem. It certainly does not have to have any regular cycle. For a sequence, you can have $f(n)=\{10^{n-1}\pi\}$, basically taking the decimal expansion of $\pi$ starting from the $n^{th}$ place. It is nicely bounded between $0$ and $1$ but bounces around between those limits erratically. On the assumption that $\pi$ is normal, its values are dense in the interval, so every point in $(0,1)$ is an accumulation point. I described a function $(0,1)\to(0,1)$ that takes all values in any interval here. It is very messy. You can make it go from $\Bbb R \to (0,1)$ with your favorite bijection and it is also bounded.

Functions can be much messier than the nice curves we draw on the whiteboard.

Answer 2

Ross's answer already provides a counterexample, but there is a little more to be said.

First I will take it that by saying the function doesn't increase or decrease without bound you mean it is bounded, that is from some point on $f(x)$ will be into some interval $[a,b]$. We can do better and choose the $a$ and $b$ such that they are the $\lim\;\mathrm{inf}$ and $\lim\;\mathrm{sup}$.

Now if the function is continuous you're lucky and you get what you wanted. All the numbers (strictly) between $a$ and $b$ get hit infinitely often. This is not hard to prove by using the intermediate value theorem: "Take any value $c$ in $(a,b)$ since $a$ is $\lim\;\mathrm{inf}$ take $x_1$ such that $f(x_1)<c$ and since $b$ is $\lim\;\mathrm{sup}$ take $x_2>x_1$ such that $f(x_2)>c$ than there is a $x_1<x_{1,2}<x_2$ such that $f(x_{1,2}=c$." Rinse and repeat countably many times.

Notice here we use the non-convergence to get $a\neq b$.

Answer 3

No, the sequence $s_n=\sin n$ is non converging and aperiodic, by the irrationality of $\pi$. The values will be non-repeating. Even if the values are discrete, the sequence may be non-repeating. Take $\lfloor\sin n\rfloor+1$:

$1 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 ,\cdots$

For an aperiodic, bounded function of a real variable, you can take $\sin x+\sin\sqrt2x$ (by irrationality of $\sqrt2$). Anyway, as it is continuous, it will "cycle" inside $(-2,2)$ and come back infinitely many times to all values.

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You will get periodic sequences when considering finite-state automata, i.e. sequences such that the next value is computed as a function of the current value only. If the number of possible values is finite, you will necessarily pass twice by the same value, and this will repeat forever in the same order.

Answer 4

Another example of a sequence that doesn't have this property would be the sequence $ a_n = \sum_{i = 1}^\infty 2^{-i}h_a(i)$ where $h_a(i)$ is a function that is 1 if the Turing machine a haults on input i and returns 0 otherwise.

A sequence that cycled though the same values would be computable however this sequence isn't computable.