# Alternate proof that a sequence is Cauchy

Is it sufficient to show that for any $\epsilon > 0$ that there exists $N$ such that if $n$ is greater than or equal to $N$, then $d(s_n, s_{n+1})$ is less than $\epsilon$ to prove that a sequence ${s_n}$ is Cauchy?

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– David Mitra Jan 3 '12 at 1:37

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FYI, this is actually true in the p-adic world. ;) – Mark Schwarzmann Jan 3 '12 at 1:42
– David Mitra Jan 3 '12 at 1:59

## 4 Answers

In general, no, common counterexamples being $s_n = \sqrt{n}$ or $s_n = \log n$ or $s_n = \sum_1^n \frac{1}{i}$ as mentioned by Chris.

But it's not completely unfeasible that there exist metric spaces $(X,d)$ in which this is true. For example if $d$ is the discrete metric on $X$ then it is certainly true; I wonder if there are others.

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It's true, for example, in any ultrametric space. This includes the $p$-adics, as Mark notes in the comments. – Chris Eagle Jan 3 '12 at 1:55

No. For example, take $s_n=\sum_{i=1}^n \frac{1}{i}$.

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Intuition and or process when looking for such a series: Firstly, it's going to be easier to look for an unbounded solution - it will require ingenuity to find a bounded one (though simple sequences can be found). Secondly, we don't want it to grow very quickly - we can tell if something grows quickly by taking it's derivitve. Canonical functions with such things are log(x).

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No, this is false in general. A standard counter-example is the harmonic series: $H_n = \sum \limits_{i=1}^n \frac{1}{i}$. It is well-known that $H_n$ diverges while $H_{n} - H_{n-1} = \frac{1}{n} \to 0$.

This situation is the sequence-analogue of the “$n$-th term test” for series:

If $\sum a_n$ converges, then $a_n \to 0$. Equivalently, if $a_n$ does not converge to $0$, then $\sum a_n$ diverges.

You might know that the $n$-term test is a divergence test and cannot be used to assert convergence of a series. Similarly, we cannot conclude that a sequence $s_n$ converges given only that $s_{n} - s_{n-1} \to 0$.

In fact, this connection is not a coincidence. Let $\sum a_n$ be a series, and define $s_n$ be its $n$-th partial sum. Then $s_{n} - s_{n-1} = a_n$, so asking if $s_n - s_{n-1} \to 0$ is equivalent to asking whether the $n$-th term of the series goes to $0$.

Finally, let me also touch upon Mark Schwarzmann's comment. In any ultrametric space (with an addition operation satisfying the usual properties), a sequence $s_n$ converges if and only if $s_{n+1} - s_n \to 0$. By the preceding discussion, it follows that any series $\sum a_n$ in such a space converges if and only if it passes the $n$-term test!

Examples of such spaces include
(i) any discrete metric,
(ii) the $p$-adic numbers, and
(iii) the ring of formal power series $\mathbf C[[X]]$ (with the metric $d(a, b) = 2^{-i}$ where $i$ is the smallest index where $a$ and $b$ differ).

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