# Terms that get closer and closer together [duplicate]

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?

## marked as duplicate by Henning Makholm, user21436, J. M. is a poor mathematician, Rahul, t.b.Apr 25 '12 at 13:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Morally, because lots and lots of really small distances can add up to a really big distance. – Arturo Magidin Apr 25 '12 at 5:15
• Here's a fun example - in a non-archimedean field like p-adics, all Cauchy-ish sequences do converge! Unfortunately, this is because in those fields, all Cauchy-ish sequences are Cauchy. – user29743 Apr 25 '12 at 5:40
• "For distances, which are non-negative, to be decreasing means to approach zero, correct?" No, they could approach a non-zero limit from above. For instance, 1.1, 1.01, 1.001,... You have to say that the distance between successive terms tends to zero if that is what you mean. – TonyK Apr 25 '12 at 9:53
• @TonyK noted and changed. Thanks – chharvey Apr 25 '12 at 11:45

## 4 Answers

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.

• Great answer. I can definitely see the connection between Cauchy-ish sequences and sequences of partial sums of Cauchy sequences. This may have been the reason I was looking for. – chharvey Apr 25 '12 at 12:15

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.

• Great example of a Cauchy-ish sequence. For some reason, this example satisfies my curiosity, and maybe that's because it's not monotone. When I think of a monotone Cauchy-ish sequence, I can't help but visualize that it will converge, even though I know I'm wrong. – chharvey Apr 25 '12 at 12:05

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.

• (also, correct this if I'm wrong because I'm only currently reading through the sequences chapter of baby rudin) – user18862 Apr 25 '12 at 6:20

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 $x_n \rightarrow \infty$.

• And this is basically the same example as $\ln$ (as $\ln(n+1)-\ln{n}$ is rougly equal to $\frac{1}{n}$) :) – penartur Apr 25 '12 at 5:30
• Euler and Mascheroni would differ :-). – copper.hat Apr 25 '12 at 5:34
• I mean that the sequences chosen by you and the OP are basically taken in the same place (if we will go from $\mathbb N$ to $\mathbb R^+$, the $\ln$ would be exactly the result of integrating $1/x$). – penartur Apr 25 '12 at 5:42
• It was a joke, the difference between the Harmonic series and $\log$ converges to $\gamma$, the Euler-Mascheroni constant. – copper.hat Apr 25 '12 at 5:52