So C is the set of all Cauchy Sequences. So where ~ is an equivalence relation such that where $\{a_n\},\{b_n\} \subset C$ then $\{a_n\} \sim \{b_n\}$ is true if the difference $\{a_n\} - \{b_n\}$ is a null sequence.
Show that $$\{\frac{1}{n^2}\} \sim \{\frac{1}{2^n}\}$$
I'm inclined to start with something like $\{\frac{1}{n^2} \sim \frac{1}{2^n}\}$ and then $|\frac{1}{n^2} - \frac{1}{2^n}| < \epsilon$ but I'm not sure where to go from there.
Show by induction that $\forall n \in \mathbb{N}_{>0}: 2^n\geq n$. It follows that $\frac{1}{2^n}\leq \frac{1}{n}$. Now, let $\varepsilon>0$. Choose $N\in \mathbb{N}$ such that $N>\max\left(\frac{\sqrt{2}}{\sqrt{\varepsilon}},\frac{2}{\varepsilon}\right)$.
Now, $\forall n\geq N$, we get: $\left|\frac{1}{n^2}-\frac{1}{2^n}\right|\leq \left|\frac{1}{n^2}\right|+\left| \frac{1}{2^n}\right| \leq \left|\frac{1}{n^2}\right|+\left| \frac{1}{n}\right|<\frac{1}{2}\varepsilon+ \frac{1}{2}\varepsilon=\varepsilon$. Here, the first inequality is triangular, the second uses $\forall n \in \mathbb{N}_{> 0}: 2^n\geq n$, and the last inequality uses the fact that $n\geq N > \max\left(\frac{\sqrt{2}}{\sqrt{\varepsilon}},\frac{2}{\varepsilon}\right)$.
It follows that $\left(\frac{1}{n^2}\right)_{n\in\mathbb{N}_{> 0}}\sim\left(\frac{1}{2^n}\right)_{n\in\mathbb{N}_{>0}}$.