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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?

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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 $$|x_m - x_n| = |(x_m - b) - (x_n - b)| \le |x_m - b| + |x_n - b| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

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