$\forall \epsilon>0, \lim \sup_{n \rightarrow \infty} |a_n - a| < \epsilon \iff (a_n) \rightarrow a$?

Question

Is it true that $\forall \epsilon>0, \lim \sup_{n \rightarrow \infty} |a_n - a| < \epsilon \iff (a_n) \rightarrow a$?

Intuitively, I think it is true, but I am unsure of how to prove it formally. My "intuitive" reasoning is as follows:

$(\impliedby)$ If $(a_n) \rightarrow a$, then $\forall \epsilon >0$, $\exists N$ such that $|a_n - a| < \epsilon$ for all $n \ge N$. So, intuitively, this means that after a specific point, the sequence $|a_n-a|$ is less than $\epsilon$. Since $\lim \sup_{n\rightarrow \infty}|a_n-a|$ can be thought of as the limit of the "upper envelope" of the sequence $|a_n-a|$, then we must have $\lim \sup_{n\rightarrow \infty}|a_n-a| < \epsilon$ as well.

$(\implies)$ Fix $\epsilon>0$, $\lim\sup_{n \rightarrow \infty}|a_n-a|<\epsilon$ means that the limit of the "upper envelope" of the sequence $|a_n-a|$ is smaller than $\epsilon$, so this means that somewhere along the sequence we must have a point after which $|a_n-a|<\epsilon$. Hence $(a_n) \rightarrow a$.

How can I make my arguments more rigorous using formal definitions of $\lim\sup$ rather than relying on intuitive understanding?

Answer 1

Since $\limsup_n|a_n-a|<\varepsilon$, for any $\varepsilon>0$, $\limsup_n|a_n-a|\leqslant0$. And since $(\forall n\in\mathbb{N}):|a_n-a|\geqslant0$ and $\liminf_n|a_n-a|\leqslant\limsup_n|a_n-a|$,$$\liminf_n|a_n-a|=\limsup_n|a_n-a|=0.$$But this is equivalent to the assertion that $\lim_n|a_n-a|=0$, which, in turn, is equivalent to the assertion that $\lim_na_n=a$.

On the other hand, $\lim_n|a_n-a|=0\implies\limsup_n|a_n-a|=0$. So, for every $\varepsilon>0$, $\limsup_n|a_n-a|<\varepsilon$.

Answer 2

For example the direction $(\implies)$ : Let $\epsilon >0$ since

$\limsup_{n\to \infty}|\alpha_n-\alpha| =\lim_{n \to \infty} \sup_{k\geq n}|\alpha_k-\alpha| <\epsilon $. There exists $n_{0}$ such that for every $n\geq n_{0}$

$\sup_{k\geq n}|\alpha_k -\alpha| <\epsilon$. So for every $n\geq n_{0}$ you have $|\alpha_n-\alpha| \leq \sup_{k\geq n_{0}}|\alpha_k-\alpha|<\epsilon.$Which

means that $\lim_{n \to \infty} |\alpha_n -\alpha|=0 \implies \lim_{n\to\infty} \alpha_n=\alpha$.