Analyze the convergence or divergence of $\{1/n^2\}$ 
Analyze the convergence or divergence of $\{1/n^2\}$

$$\left|\frac{1}{m^2} - \frac{1}{n^2}\right| < \left|\frac{1}{m^2}\right| < \frac{1}{N^2} < \varepsilon$$
whenever $N > \dfrac{1}{\sqrt{\varepsilon}}$ (first step because $n^2 > 0$ and hence $\dfrac{1}{n^2} > 0$ and $\dfrac{1}{m^2} - \dfrac{1}{n^2} < \dfrac{1}{m^2}$)
So the sequence is Cauchy and it certainly converges to zero.
Is my approach correct?

The reasoning is not quite correct, but it is in fact a Cauchy sequence that converges to $0$.
$\frac{1}{n^2}>0$ is true when $n$ is a positive integer, and this does imply that $\frac{1}{m^2}-\frac{1}{n^2}<\frac{1}{m^2}$.  However, that does not mean that $\left|\frac{1}{m^2}-\frac{1}{n^2}\right|<\frac{1}{m^2}$.  For example, consider $n=1$, $m=4$.
Instead, you could note that $\left|\frac{1}{m^2}-\frac{1}{n^2}\right|$ is equal to either $\frac{1}{m^2}-\frac{1}{n^2}$ or $\frac{1}{n^2}-\frac{1}{m^2}$, and find a number that is an upper bound for both of these.
The reasoning could also be made more complete by specifying how $m$ and $n$ are related to $N$.
