# Divergent real sequence for which $\forall \epsilon >0 \exists a \in \mathbb{R}: |x_n-a|<\epsilon$ for almost all n

Looking at the definition of convergence:

$$\exists a \in \mathbb{R}\forall \epsilon >0: |x_n-a|<\epsilon$$ for almost all n

I switched the order of the first two quantifiers. Is there a sequence for which

$$\forall \epsilon >0 \exists b \in \mathbb{R}: |x_n-b|<\epsilon$$ for almost all n

holds but it's not convergent?

Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $$\mathbb{R}$$. To see this, suppose $$\epsilon > 0$$. Then there exists $$b$$ such that $$|x_n - b| < \frac{\epsilon}{2}$$ for almost all $$n$$. In order words, for such a $$b$$, there exists $$N$$ such that $$|x_n - b| < \frac{\epsilon}{2}$$ whenever $$n \ge N$$. But then whenever $$m, n \ge N$$, we have $$|x_m - x_n| = |(x_m - b) - (x_n - b)| \le |x_m - b| + |x_n - b| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$