Terms that get closer and closer together 
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Why doesn't $d(x_n,x_{n+1})\rightarrow 0$ as $n\rightarrow\infty$ imply ${x_n}$ is Cauchy? 

I was thinking about sequences where it appears the terms get closer and closer together, and wondered if they converge.
Now let's first define a few things. When I say "the terms get closer and closer together", I mean "the distance between any two consecutive terms approaches zero." In other words, for a sequence $\left(x_n\right)$,
$$|x_n-x_{n-1}| \to 0$$
consecutive terms become closer and closer together.
Let's look at an example: $\left(\ln n\right)$. Clearly,
$$\bigl|\ln (n) - \ln (n-1)\bigr|\to 0$$
and this can be verified by looking at a graph. At first, I saw this and thought $\left(x_n\right)$ and $\left(\ln n\right)$ looked like Cauchy sequences, and this was bugging me for the longest time, because I knew $\left(\ln n\right)$ was not supposed to be Cauchy! But I realize now that there is a subtle difference: for a Cauchy sequence $\left(y_n\right)$,
$$|y_n-y_{m}| \to 0$$
So in the cases of $\left(x_n\right)$ and $\left(\ln n\right)$, it may be true that consecutive terms are closer together, but two arbitrary terms aren't necessarily close together. So $\left(\ln n\right)$ is definitely not Cauchy.
What can we call sequences such as $\left(x_n\right)$ and $\left(\ln n\right)$, where consecutive terms become closer together? I'd like to propose a name: let's call them Cauchy-ish.
An intuitive geometric view of a Cauchy-ish sequence $\left(x_n\right)$ might be that you have a bunch of points on a line, and as you move forward in the sequence, the points get closer and closer together. It seems to me that this sequence would converge, no? Obviously my intuitive side and my analytical side disagree, because $\left(\ln n\right)$ is Cauchy-ish but is not Cauchy.
My question, finally, is, why don't "Cauchy-ish" sequences necessarily converge?
 A: Suppose that you are given $(x_n)_{n \geq 1}$ and define a sequence $(a_n)_{n \geq 1}$ for all $n$ by
$$
a_n = \begin{cases} x_1, & n = 1, \\ x_n - x_{n-1}, & n > 1. \end{cases}
$$
Or, alternatively, suppose that you are given $(a_n)_{n \geq 1}$ and define a sequence $(x_n)_{n \geq 1}$ for all $n$ by
$$
x_n = \sum_{j=1}^n a_j.
$$
In either setting, note:


*

*$x_n - x_{n-1} \to 0$ as $n \to \infty$ if and only if $a_n \to 0$ as $n \to \infty$, and

*The sequence $(x_n)_{n \geq 1}$ converges if and only if the series $\sum_{n=1}^{\infty} a_n$ converges.  


So your question is some sense equivalent to asking why a series can diverge even if its $n$th term goes to $0$.
I'm not sure that I have a good, short, "intuitive" answer to either question, but at least this fact gives you a large stock of examples.  And whatever understanding you might have of series can now be sort of "imported" to an understanding of this phenomenon.
A: My answer would be: Because they aren't Cauchy.
I think that the logical reason is quite clear. Every convergent sequence is Cauchy. Suppose you had some sequence that wasn't a Cauchy sequence. Then there exists an $\varepsilon$ such that for every $N$ there is always some $n,m>N$ for which the distance between $x_n$ and $x_m$ is greater than $\varepsilon$. So that means if the sequence were to converge to some value $y$, it would have to compete with some other number, because the gap between the sequence and $y$ can NOT be made arbitrarily small. The fact that there is always a difference between two far (meaning large $n,m$) numbers in the sequence isn't special in itself, its the fact that this difference does not vanish. If the difference doesn't vanish as $n$ and $m$ get larger, what would YOU suggest the limiting value be?
If being cauchy-ish was a sufficient criterion for convergence, then $\log(n)$ would converge even though intuitively, it gets infinitely large! Cauchy sequences really help define convergence in the first place.
A: Consider the sequence
$$0, 1, \frac{1}{2},0,\frac{1}{3},\frac{2}{3},1,\frac{3}{4}, \frac{2}{4},\frac{1}{4},0 ,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}, 1,\frac{5}{6},\frac{4}{6},\frac{3}{6},\frac{2}{6},\frac{1}{6},0, \frac{1}{7},\frac{2}{7},\dots.$$
Successive terms get close to each other, but the sequence travels back and forth between $0$ and $1$ forever, so does not converge. In fact every real number between $0$ and $1$ is the limit of a subsequence of our sequence.
A: An example is the Harmonic series, let $x_n = 1+\frac{1}{2}+...+\frac{1}{n}$, then $x_n-x_{n-1} = \frac{1}{n}$, but $x_n \rightarrow \infty$.
