Let's say we have two real sequences $(a_n)_{n\in\mathbb{N}}$ and $(c_n)_{n\in\mathbb{N}}$ with $c_n\in o(\frac1n)$ (i.e. $c_n(\frac1n)^{-1}\xrightarrow{n\rightarrow\infty}0$). And for all $\epsilon>0$ holds that $$\frac{1-\epsilon}n-\frac1\epsilon |c_n|\leq a_n\leq \frac{1+\epsilon}n+\frac1\epsilon |c_n|.$$Is it true that $a_n\sim \frac1n$ i.e. $a_n(\frac1n)^{-1}\xrightarrow{n\rightarrow\infty}1$? I think it should hold but I'm really confused by the $\epsilon$'s because I can't just consider the limit for $\epsilon\rightarrow0$.
Thanks and I'm sorry if it's a really stupid question!
Fix $\epsilon\gt0$. Then $n|c_n|\to0$ and, for every $n$, $$ 1-\epsilon-\epsilon^{-1}n|c_n|\leqslant na_n\leqslant1+\epsilon+\epsilon^{-1}n|c_n|, $$ hence $$ 1-\epsilon\leqslant\liminf_{n\to\infty}\,na_n\leqslant\limsup_{n\to\infty}\,na_n\leqslant1+\epsilon. $$ This holds for every $\epsilon\gt0$ hence $$ 1=\liminf_{n\to\infty}\,na_n\leqslant\limsup_{n\to\infty}\,na_n=1, $$ which implies $$ \liminf_{n\to\infty}\,na_n=\limsup_{n\to\infty}\,na_n=1, $$ that is $$ \lim_{n\to\infty}na_n=1. $$