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($\varepsilon$)|.
 A: 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. 
A: 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).
