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How come this sequence does not approach any limit?

$\{\max((-1)^n,0)\}^\infty _{n=1} : {0,1,0,1,0,1,0,1,...}$

I read that since this alternates between 0 and 1 this does not approach any limit. Hence not convergence. Is it safe to say that it does not diverge either?

Since:

A sequence can be divergent by having terms that increase (decrease) without limit. Example:
2,4,8,16,32,64,...

Does all sequences that alternates not approach a limit? For example this one too:

$3,1,3,1,3,1,3,1...$

So what name does these sequences have and what does it mean? It's neither convergent nor divergent. How do you prove that it is neither?

For example in the Collatz problem you "always" run into cycles (sub-sequences) that are similar to this.

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    $\begingroup$ Any sequence which does not converge is said to diverge. There are different types of divergence, including but not limited to alternating sequences as well as diverging to infinity (i.e. increasing without limit). $\endgroup$ – JMoravitz Jul 10 '18 at 17:55
  • $\begingroup$ Ok, I see. So it is diverging.. Thanks. $\endgroup$ – Natural Number Guy Jul 10 '18 at 17:56
  • $\begingroup$ You might do well to read the formal definition of a limit of a sequence. $\endgroup$ – JMoravitz Jul 10 '18 at 17:58
  • $\begingroup$ Another example of a divergent sequence would be $3,1,4,1,5,9,2,6,5,3,5,8,9,7,9,\dots$, the sequence of the digits of pi in base 10. This can be shown to never reach a point where it stops on a number indefinitely and thus never converges (else $\pi$ would have been a rational number), though this sequence does not simply alternate between values nor does it increase without bound. We can give a name to the "doesn't increase without bound" property though, we call such sequences "Bounded sequences." $\endgroup$ – JMoravitz Jul 10 '18 at 18:05
  • $\begingroup$ As an aside, in the Collatz problem, it is conjectured that you always run into cycles, but the conjecture hasn't yet been proven to be true or false. $\endgroup$ – JMoravitz Jul 10 '18 at 18:06
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Every sequence of real numbers either converges or diverges. This is trivial, since divergence means the opposite of convergence.

And the sequences that you mentioned diverge. That is, there is no real number $L$ such that$$(\forall\varepsilon>0)(\exists p\in\mathbb{N})(\forall n\in\mathbb{N}):n\geqslant p\implies|L-a_n|<\varepsilon.$$

You are wrong when when assert that that's what happens in the Collatz problem. If we knew that, it wouldn't be a problem anymore.

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  • $\begingroup$ On the Collatz-sequences, what I meant is that it has been shown (computer-checked) by that of initial-numbers up to $87\times2^{60}$, but not proven that it ends in a $4\rightarrow2\rightarrow1$-cycle or $2\rightarrow1$-cycle. I wasn't clear enough about that. $\endgroup$ – Natural Number Guy Jul 10 '18 at 21:03
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You have missed the definition of a divergent sequence.

A divergent sequence does not have to be unbounded, it simply does not have a limit.

$$ 1,0,1,0,1,0,... $$ does not converge so it is divergent.

Simply put, if a sequence is not convergent we call it divergent regardless of its other properties.

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  • $\begingroup$ Yes, I was confused by the definition. So the Collatz iterating function: $(3x+1)/2$ when odd and $x/2$ when even is most certainly bounded and divergent? Atleast if it always ends in the $2\rightarrow1$-cycle? $\endgroup$ – Natural Number Guy Jul 10 '18 at 21:22
  • $\begingroup$ The Collatz function is still a conjecture and so far the sequences of iterations seem to merge to the cycle $2\to1 \to 2\..$ $\endgroup$ – Mohammad Riazi-Kermani Jul 10 '18 at 23:41
  • $\begingroup$ I tried to make this as an acceptable answer as well. But it worked with only one. Thanks for all the feedback. It made everything clearer. $\endgroup$ – Natural Number Guy Jul 13 '18 at 1:23

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