# Implications of zero limiting variance

Assume that I have a sequence of random variables $$X_1, X_2, \dots$$ with means $$\mu_1, \mu_2, \dots$$ such that $$\lim_{n \to \infty} \operatorname{Var}(X_n) = 0$$.

Can I claim that for large enough $$n$$ the respective $$X_n$$ will be close to $$\mu_n$$ with high probability? How can I can formalize such a statement?

Here is an example of a result that would make me happy:

For any $$\varepsilon > 0$$ and $$\delta > 0$$ there exists an $$N$$, such that for all $$n \ge N$$ it holds that $$\Pr( |X_n - \mu_n| \ge \delta) \le \varepsilon$$.

However, I'm not sure what tools I can use to prove it, or if there exists some similar known inequality.

Since $$\operatorname{Var}(X_n) \overset{n \to \infty}{\to} 0$$, we can pick $$N$$ such that for all $$n \ge N$$ we have $$\operatorname{Var}(X_n) \le \varepsilon \delta^2$$. Then by Chebyshev's inequality $$\Pr(|X_n - \mu_n| \ge \delta) \le \frac{\operatorname{Var}(X_n)}{\delta^2} \le \varepsilon,$$ which completes the proof.