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My question is,

Is it true that $x_k\rightarrow x$ iff. $\exists N \in \Bbb{N}$ st. $k>N$ implies $|x_k-x|<a_k$ for some $a_k$ where $a_k>0$ and $a_k \rightarrow 0$ as $k \rightarrow 0$.

Necessity is obvious by squeeze theorem, i.e.

$|{x_k} - x| < {a_k} \Rightarrow x - {a_k} < {x_k} < x + {a_k} \Rightarrow \mathop {\lim }\limits_{k \to \infty } (x - {a_k}) \le \mathop {\lim }\limits_{k \to \infty } {x_k} \le \mathop {\lim }\limits_{k \to \infty } (x + {a_k}) \Rightarrow {x_k} \to x$

However, I was not successful in proving the sufficiency, that is, to prove the existence of such $a_k$ given ${x_k} \to x$. Hope someone can help. Thank you!

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Define $a_k=2|x-x_k|$, and it works automatically.

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Let $a_k = |x_k-x| + 1/k$, for example.

NB: $a_k = c*|x_k-x|$, for $c>1$ won't work for the cases where $x_k = x$ for some k.

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