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?
 A: UPDATE:
To answer straight from the definition, take $N$ such that $1/N < \sqrt{\epsilon/2}$
Then using the triangle inequality for $m,n\geq N$
$$|\frac{1}{n^{2}}-\frac{1}{m^{2}}| \leq |1/m^{2}|+|1/n^{2}|< \epsilon/2 + \epsilon/2 = \epsilon$$

Your approach is flawed because your first inequality is incorrect. The easiest way to show this sequence is Cauchy is to show it converges to 0. (A convergence sequence is Cauchy)
Fix $\epsilon>0$.
There exists an $N$ (by the Archimedean property) such that for all $n\geq N$,
$1/n \leq 1/N < \sqrt{\epsilon}$
Then 
$$|1/n^{2} - 0| = 1/n^{2} <\epsilon$$
A: 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 $N$.
A: I assume that you want to prove that the sequence $\{a_n \}$ given by $a_n = \frac{1}{n^2}$ is convergent. Indeed as you see from the other answer, you can first prove that the sequence is Cauchy, and then conclude that it is convergent. So that approach is correct. I just wanted to point out that you actually can prove "directly" that the sequence is convergent.
You all ready have guessed that the limit of the sequence is zero. So let $\epsilon > 0$ be given. We want to find $N>0$ such that if $n \geq N$, then $\lvert a_n \lvert < \epsilon$. So we need
$$\frac{1}{n^2} < \epsilon.
$$
Note that this equivalent to saying that 
$$
\frac{1}{\sqrt{\epsilon}} < n.
$$
Hence we can pick any $N$ such that $\frac{1}{\sqrt{\epsilon}} < N$. Then indeed if $n \geq N$ then you have
$$\begin{align}
\frac{1}{\sqrt{\epsilon}} < N \leq n  &&\Rightarrow \\
\lvert a_n \lvert = \frac{1}{n^2} < \epsilon.
\end{align}
$$
And we are done.
