How to prove mathematically a sequence as a Cauchy sequence in general?

Question

How to take epsilon and what exactly do one mean by the existence of natural number. Definition is itself a vague to me.

Can anyone help me with an mathematical algorithm for proving a Cauchy sequence which would be generla to every problem I consider?

Thank you well in advance.

Answer 1

I definitely understand. It can take awhile to get used to the reasoning involved in Real Analysis. A Cauchy sequence is a sequence whose terms eventually start clustering together. That is they eventually start getting arbitrarily close together. For real numbers, a sequence is Cauchy if and only if it converges. If you think about it, a convergent sequence eventually starts clustering around its limit, and so all the terms that are clustering near the limit point are therefore also clustered near each other. The way to prove a sequence is Cauchy is to show that given any positive number, no matter how small, eventually the terms of that sequence are within that distance of each other, and stay within that distance of each other. Consider the following definition of a Cauchy sequence

$$\text{For all } \xi > 0, \text{ there exists an } N\in\mathbb{N} \text{ such that } n,m \geq N \text{ implies } \left|x_n - x_m\right| < \xi$$

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The absolute value is a good way of expressing distance, that is, $|x-y|$ can be thought of as the distance between the points $x$ and $y$

Think of the $\xi$ as a challenge. Somebody gives you $\xi$, say $\xi = .001$ and says "show me the terms of the sequence are eventually always within $.001$ of each other. The $n,m \geq N$ corresponds to the "eventually" part, and the number $N$ is the cutoff point. That is, if you pick any two terms of the sequence that are higher than $N$ they will be within $.001$ of each other. Your goal is to find the cut-off point $N$. You have to show that you can find such an $N$ for any $\xi > 0$. I realize it can be hard thinking in this way but I can promise you with practice it will get easier. I'm afraid there is no algorithm that one can use, but as a general rule the triangle inequality can frequently be helpful.

Answer 2

Well, let's give it a try to explain the definition of a Cauchy sequence with a simple example like $a_n=\frac1n$: we need to construct for any given $\epsilon>0$ a number $N(\epsilon)$ so that $|a_n-a_m|<\epsilon$, as soon as both $n$ and $m$ are $>N(\epsilon)$. But if $n>N(\epsilon)$ and $m>N(\epsilon)$, $\frac1n<\frac1{N(\epsilon)}$ and $\frac1m<\frac1{N(\epsilon)}$, so $$|a_n-a_m|=\left|\frac1n-\frac1m\right|\le\frac1n+\frac1m<\frac2{N(\epsilon)},$$ and that's certainly $<\epsilon$, if only $N(\epsilon)>\frac2\epsilon$.