Prove $\{a_n\}^r$ is a Cauchy sequence.

Question

For any $r \in \mathbb{Q}$, consider the sequence $\{a_n\}^r$ given by $a_n = r$ for all $n \in \mathbb{N}$. Prove $\{a_n\}^r$ is a Cauchy sequence.

I don't have any idea how to prove $\{a_n\}^r$ yet, but I do have an idea to show that $\{a_n\}$ is Cauchy sequence. So far, I know that the difference between any two terms is equal to zero. Can we take $\epsilon = 0$ because the difference is zero? Does this mean that the sequence converges to zero? any idea or hints would be really appreciated.

EDIT: I don't know if this makes sense, but I will write down what I have in my mind.

Since the difference between any two terms is equal to zero, we have $|a_{i+1} - a_i| = 0 < \epsilon$. Thus, we can take $r = 0 \in \mathbb{Q}$. Then, there exists $k \in \mathbb{N}$ such that $|a_{i+1} - a_i| < \epsilon$ for $i > k$. Therefore, $\{a_n\}$is a Cauchy sequence. Since $r = 0$, $\{a_n\}^r$ is a Cauchy sequence that converges to zero.

Answer 1

A Cauchy sequence is just a sequence where the values eventually get arbitrarily close to each other. Formally, we say that for an arbitrarily small value $\varepsilon > 0$, there is a “threshold” $N \in \mathbb{N}$ such that for all indices $n, m \geq N$, the values $a_n$ and $a_m$ differ by no more than $\varepsilon$. This definition may seem daunting, but if you break it down, it should make sense. Following from this definition, here's how you would solve your problem:

1. Start by picking some arbitrary value of $\varepsilon$. Here, the word “arbitrary” indicates that $\varepsilon$ can be anything—notice that we aren't picking a specific value for $\varepsilon$ (as you did when you said “Can we take $\varepsilon = 0$ because the difference is zero?”).

2. Next, identify your “threshold” $N$. All values of your sequence past this threshold must be within a distance of $\varepsilon$ of each other. Usually, the tricky part is choosing the right value for $N$, but it shouldn't be too difficult for your problem.

3. Prove that for all indices $n, m \geq N$, the distance $|a_n - a_m| < \varepsilon$. This is to just verify that the threshold you picked in (2) is indeed valid.

And that's it! We have shown that for arbitrarily small values of $\varepsilon$ (1), there is a threshold $N$ (2) such that all values past this threshold are $\varepsilon$-close to each other (3). Almost every time you want to prove a sequence is Cauchy, you will follow these steps.