Terms that get closer and closer together [duplicate]

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.

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.

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.