In a proof, I encountered the following situation:
For every $n \in \mathbb{N}$ between $k^2 \leq n \leq (k+1)^2$, we have:
$$a_k \leq b_n \le c_{k+1}$$
where $a_k, c_k \to 0$ if $k \to \infty$.
They then conclude that $\lim_{n \to \infty} b_n = 0$. I can see this intuitively but can't write it out rigorously.
On
Let $\varepsilon > 0$. Since $a_k, c_k \xrightarrow{k\to\infty} 0$, there exists $k_0 \in \mathbb{N}$ such that $k \ge k_0 \implies |a_k|, |c_k| < \varepsilon$.
Let $n \ge k_0^2$. Clearly $$n \in \left[k_0^2, +\infty\right\rangle \cap \mathbb{N} = \bigcup_{k \ge k_0} \left[k^2, (k+1)^2\right] \cap \mathbb{N}$$
so there exists $k \in \mathbb{N}, k \ge k_0$ such that $k^2 \le n \le (k+1)^2$.
We have
$$-\varepsilon \le -|a_k| \le a_k \le b_n \le c_{k+1} \le |c_{k+1}| \le \varepsilon$$
so $|b_n| < \varepsilon$.
We conclude $b_n \xrightarrow{n\to\infty} 0$.
Define $A_n$ and $C_n$ as follows: if $k^2\leq n<(k+1)^2$, let $A_n=a_k$ and $C_n=c_{k+1}$. Then, $A_n\leq b_n\leq C_n$ so it suffices to show $A_n\to 0$ and $C_n\to 0$ and use the Sandwich Theorem.
For example, with $A_n$: let $e>0$ be given. As $a_k\to 0$, there is $K$ such that $|a_k|<e$ for all $k\geq K$. Then, for all $n\geq K^2$, we have $A_n=a_{k}$ for some $k\geq K$ and so $|A_n|<e$.