Convergence of Cauchy sequence $(x_n)$, where $x_n=1/n^2$. I'm just looking at an example where we're asked to prove that the sequence $(x_n)$ is Cauchy, where $x_n=1/n^2$. (the example is from here at time 7:15 minutes; this requires registration but is free) Anyway, the proof goes exactly like this:

Let $\varepsilon>0$ and let $N=\lfloor 1/(2\varepsilon)\rfloor+1$. Then $q>p>N$ implies $q>p>1/(2\varepsilon)$ and
$$\left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{1}{q}+\frac{1}{p}\right)\leq 2\left(\frac{1}{q}-\frac{1}{p}\right)<2\frac{1}{2\varepsilon}.\tag{1}$$
Hence $$\frac{1}{q^2}-\frac{1}{p^2} < \varepsilon,\tag{2}$$ as required.

My questions are:
(i) the $\varepsilon$ in equation (2) does not follow from the $2\times 1/(2\varepsilon)$ in equation (1), so is this just a mistake in the given proof?
(ii) Shouldn't the proof be using absolute notation, e.g.
$$\left|\left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{1}{q}+\frac{1}{p}\right)\right|\leq 2\left|\frac{1}{q}-\frac{1}{p}\right|<2\frac{1}{2\varepsilon},$$ because $q>p$ implies that $$\frac{1}{q}-\frac{1}{p}<0$$ is negative, which is not something we should allow. In particular, equation (2) should not be negative because we are trying to show that $$0<\frac{1}{q^2}-\frac{1}{p^2}<\varepsilon,$$ i.e. goes to zero as $\varepsilon\to 0$.
 A: (i) Yes. This is a mistake.
(ii) The absolute value is necessary, OR they could have said $p > q$. However, they did define it wrong for their argument to work.
Here's how I believe that proof is supposed to go. (It is similar, but there appear to be typos in the above.
Let $\{x_n\}$ be a sequence defined by $x_n = \frac{1}{n^2}, \forall n \in \mathbb{N}$. Show $\{x_n\}$ is Cauchy.
Fix $\epsilon > 0$. Then, there exists an $N = \lceil\frac{4}{\epsilon} \rceil$ such that $\forall n,m > N$ we know that
$$ \left| \frac{1}{n^2} - \frac{1}{m^2} \right| = \left| (\frac{1}{n} - \frac{1}{m} ) (\frac{1}{n} + \frac{1}{m})\right| \le 2 \left(\frac{1}{n} + \frac{1}{m}\right) < 2 \left(\frac{\epsilon}{4} + \frac{\epsilon}{4}\right) = \epsilon,$$
where the second to last inequality follows since $\left(\frac{1}{n} + \frac{1}{m} \right) \ge \left| \frac{1}{n} - \frac{1}{m} \right|$, and the last inequality follows since $\frac{1}{m}$ and $\frac{1}{n}$ are both less than $\frac{1}{N} < \frac{\epsilon}{4}$.
Moreover, here is another method that is more straightforward.
Fix $\epsilon > 0$. Let $N > \left\lceil \sqrt{\frac{2}{\epsilon}} \right\rceil $, or equivalently, $\frac{1}{N^2} < \frac{\epsilon}{2}$. Then,$\forall n,m \ge N$
$$ \left| \frac{1}{n^2} - \frac{1}{m^2} \right| \le \frac{1}{n^2} + \frac{1}{m^2} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
