Proving that a sequence is Cauchy 5

Question

We want to prove that the sequence $a_n = n^2$ is Cauchy in the metric space $(E, d)$, with $E = [0, \infty[$ and $d(x, y) = |\arctan(x) - \arctan(y)|$. I proceed in the following way:

$a_n$ is Cauchy $\Leftrightarrow$ $\forall \varepsilon >0$, $\exists N \in \mathbb{N}$ such that $\forall m, n \geq N$, $d(a_m, a_n) < \varepsilon$. In this case,

$$d(a_m, a_n) = |\arctan(m^2)-\arctan(n^2)| = |\arctan(m^2) + \arctan(-n^2)| \leq |\arctan(m^2)| + |\arctan(-n^2)| \leq \pi.$$

I doubt this is correct or sufficient for the demonstration.

Edit: Follow up: I've also tried by using integrals:

$$|\arctan(m^2)+\arctan(-n^2)| = \Big|\int_{-n^2}^{m^2} \frac{1}{t^2+1}dt\Big|$$

This seems to go to $0$ as $n, m$ go to $\infty$. I can't seem to prove it rigorously. I appreciate all feedback.

Answer 1

Hint: Your answer should look something like this:

Let $\varepsilon > 0$ be arbitrary. We may select an integer $N$ which is larger than [function of $\varepsilon$]. If $m$ and $n$ are taken to be bigger than $N$, then we have $$ d(a_m,a_n) = \cdots \leq \cdots \leq \cdots < \varepsilon $$ Thus, the sequence $a_n = n^2$ is indeed Cauchy, as desired.

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Hint 2: For example, we can say

We may select an integer $N$ so that $N > \sqrt{\tan(\varepsilon)}$

and fill in the rest of the proof from there.