Is it Cauchy?

Relevant wiki: Cauchy Sequences

Yes. Consider any $\epsilon>0$ , and take $N$ so large that $N>\sqrt{2/\epsilon}$ . Then, $\left|\frac{1}{n^2}-\frac{1}{m^2}\right|\leq \frac{1}{n^2}+\frac{1}{m^2}\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,$ so this sequence is Cauchy.

In a metric space, each convergent sequence is a Cauchy sequence. In this case, $\displaystyle \lim_{n\to\infty} \frac{1}{n^2} = 0$ . Furthemore, $\mathbb{R}$ with the usual metric $| \cdot|$ is a Banach space, i.e, is a complete metric space,i.e, a sequence in $\mathbb{R}$ is a convergent sequence $\iff$ it's a Cauchy sequence