# Existence of a Bound Related to Erratically Converging Sequences?

I have a question related to the way an erratically converging sequence $(x_{k})$ in $\mathbb{R}$ approaches its limit.

The definition of the convergence of $(x_{k})$ implies that for all $\varepsilon$ > 0 there is a $k_0(\varepsilon)$ such that when k > $k_0(\varepsilon)$ then $x_{k}$ is in an $\varepsilon$ -neighbourhood of its limit.

An erratically converging sequence may enter and exit an $\varepsilon$ -neighbourhood of its limit, possibly multiple times, before it settles down in the $\varepsilon$ -neighbourhood. I think there are naturally occurring examples in Number Theory of this behaviour.

Let E($\varepsilon$) be the set of k such that $x_{k-1}$ is in an $\varepsilon$ -neighbourhood of the limit of $(x_{k})$ but $x_{k}$ is not. E($\varepsilon$) will be finite for an erratically converging sequence and empty for a monotonically converging sequence.

My question is, are there theorems that by placing conditions on $(x_{k})$ (or otherwise) derive an upper bound on the set E($\varepsilon$) for an erratically converging sequence?

My own search has not turned up anything so if someone could point me in the right direction, it would be appreciated.

REPLY TO COMMENTS/ANSWERS (@ 8 JAN 1340)

Thanks for your comments/answers.

Brian's comments appear to be about possible upper bounds on the number of members of E($\varepsilon$), i.e. on |E($\varepsilon$)|, and not on the members of E($\varepsilon$), i.e. on max E($\varepsilon)$. Is this correct ? I am interested in upper bounds on the latter.

If I have understood correctly, Betty starts with a convergent sequence, regarded as a subsequence, and wants to construct a convergent parent sequence, by inserting into the subsequence, members that are outside the $\varepsilon$ -neighbourhood of the limit, L, of the resulting parent sequence.

I can see how a suitable choice of the inserted members can produce a parent sequence that is erratically converging but ultimately all members of the parent sequence, including the inserted members, must be within the $\varepsilon$ -neighbourhood of limit L, otherwise the parent sequence will fail to meet the definition of being convergent. This means that the set E($\varepsilon$), as I have defined it, will be a non-empty and finite subset of the Naturals in the case of an erratically converging sequence.

BEHAVIOUR OF E($\varepsilon$) AS $\varepsilon$ GETS SMALLER (@ 23 JAN)

Betty, this is further to my reply to your comment (this reply is here because it was rejected as too long for a comment).

I do not know how |E($\varepsilon$)| changes as $\varepsilon$ gets smaller. It seems to me it could go either way. |E($\varepsilon$)| could get smaller because as $\varepsilon$ gets smaller the $\varepsilon$-neighbourhood becomes more exclusive so some segments of the sequence that previously entered then exited the $\varepsilon$-neighbourhood may now be entirely outside the smaller $\varepsilon$-neighbourhood. On the other hand, segments of the sequence that were previously entirely inside the $\varepsilon$-neighbourhood will now be nearer to the border of the neighbourhood so they may enter and exit it thereby increasing |E($\varepsilon$)|.

-
I'm not quite convinced that the parent sequence can't a)converge and b) provide an unbounded E. As I am describing it, the parent sequence has $2^n$ "bad" points between $a_n$ and $a_{n+1}$ of the original sequence, where each jump is of height $2^{-n}$ . I think that means E is becoming unbounded as $\epsilon$ gets smaller. That is why I suggested we look at bounded variation. If you think this is not correct, could you explain again in more detail. – Betty Mock Jan 11 '14 at 1:40
Apologies for the delay in replying. My understanding of the definition of a "convergent" sequence is that for any pre-selected $\varepsilon$ > 0, after some finite value of k= $k_0(\varepsilon)$ all members of the sequence $(x_{k})$ must belong to an $\varepsilon$-neighbourhood of its limit, however the sequence may be constructed. Hence once k= $k_0(\varepsilon)$ is reached the sequence never exits the $\varepsilon$-neighbourhood again. The exits of the sequence prior to k= $k_0(\varepsilon)$ is all there are so the set E($\varepsilon$) is finite. – gjh Jan 22 '14 at 8:52
Certainly once you pick $\epsilon$ you must be able to find an N so that for n > N the elements are within $\epsilon$ of the limit. For every sequence you have to pick N big enough to achieve that. I believe my construction meets that requirement, because the jumps are getting smaller with each $2^n$. So for any given $\epsilon$ the jumps of $2^{-n}$ will eventually (for large enough n) put the entire tail of the sequence in the epsilon neighborhood of the limit; but you will still get all those erratic elements -- more and more as $n \rightarrow \infty$. – Betty Mock Jan 23 '14 at 3:37

## 2 Answers

In an earlier comment I suggested thinking about a sequence {$a_n$} which has a subsequence {$a_{4n+1}$} converging monotonically to the limit L. I was thinking of this as decreasing. Then I inserted 3 $a_n$'s between each {$a_{4n+1}$,} each of which increases by $2^{-n}$.

This clearly is an erratically converging sequence, since for any neighborhood where $a_{4n+1}$ is close to L, the subsequent 3 entries will be out of the neighborhood.

This brings us to clarifying the definition of E. You have said that if $a_{k-1}$ is in the neighborhood and $a_k$ is not, then E picks up a +1. As I constructed it, I was thinking that E would pick up +4 since $a_{k-1}$ is in the neighborhood and then next 3 $a_k$ are not -- which is not the same thing. However, bear with me for a moment.

One can construct the first set of increasing values with 4 elements, the next set with 8 elements, etc., which by the second definition of E would get us $E(\epsilon) \rightarrow \infty$ as $\epsilon \rightarrow 0$.

With this idea in mind, going back to your definition that E picks up a +1 only for each pair, I think half my 4, 8, 16... values have to drop back to equal $a_{k-1}$, while the others can stay where they are.

This still gives you an E which is unbounded. However, it also gives an idea of what conditions might force a bound on E.

Brian Rushton says that constraining the sequence to be convex would force a bound, which I certainly believe.

I am wondering whether it would be true that if the sequence is of bounded variation E is bounded. I don't think the sequence I describe above is of bounded variation.

This is not a full answer, but since you asked for pointers, I think it is worth considering.

-

Here is one condition that would place a bound on $E$: If a sequence in the real numbers is convex, we can bound $E(\epsilon)$ by 3 (and probably 2).

-
why would the sequence be convex? – Betty Mock Jan 3 '14 at 4:51
If it is the restriction of a convex function on the reals – Brian Rushton Jan 3 '14 at 5:01
sorry I am still lost -- I'm sure I can invent a sequence that fulfills the requirements and is not convex. Since we were in the reals to begin with, what function is it we are restricting, and from where are we restricting it. I do not mean to be a pain here -- I'm really trying to understand. Thanks. – Betty Mock Jan 3 '14 at 21:41
What about requiring $a_n-a_{n-1}\leq a_{n+1}-a_n$? This is equivalent to $a_n\leq(a_{n-1}+a_{n+1})/2$. – Brian Rushton Jan 3 '14 at 22:17
Suppose $a_1, a_5, a_9 (i.e. a_{4n+1})$ converge monotonically to L. Suppose $a_{4n+2} = a_{4n+1} + 2^{-n}; a_{4n+3} = a_{4n+2} + 2^{-n}; a_{4n+3} = a_{4n+2} + 2^{-n}$. Doesn't $a_n$ converge to L? Is the sequence {a_n} convex? What have I misunderstood? – Betty Mock Jan 4 '14 at 3:46