Let $a_n$ be a Cauchy sequence in Q, the rationals. For n an element of the Natural Numbers, Let $b_n$ = $a_{n^2}$ Prove that ($a_n$) ~ ($b_n$).
I'm not really sure where to start.
I believe that if ($a_n$) ~ ($b_n$), then $a_n->c$ and $b_n->k$, where c=k, then given some value of $\epsilon$, there should be some value of N>0 such that $-\epsilon < a_N - c < \epsilon$. And then because N>0 and an integer, $N^2 >= N$, it should also converge to the same value.
On
It all depends on how much you already know. This theorem is rather trivial if you know the right theorems. Firstly, $a_n$ and $b_n$ are both Cauchy sequences, so therefore they converge. Furthermore, $b_n$ is a subsequence of $a_n$, and since $a_n$ converges to a value, all of its subsequences must also converge to that same value.
Something tells me that the last statement is one that you haven't learned yet, but in any case, your work on the topic is almost there. Noticing that n^2>n is a crucial step... Now how can we use that in the definition of convergent/Cauchy sequences?
This assertion is false in the special case where you take on $\mathbb{Q}$ the metric $d(p,q)=|\arctan(p)-\arctan(q)|.$ Indeed for $n\in\mathbb{N}^*, $take $a_n=n\in\mathbb{Q}.$ The sequence $(a_n)_{n\in\mathbb{N}^*}$ is Cauchy for $d,$ you get $b_n=a_{n^2}=n^2$ and $$\frac{a_n}{b_n}=\frac{n}{n^2}=\frac{1}{n}\underset{n\to+\infty}{\longrightarrow}0\neq1.$$