How to show a random variable diverges from its mean? I am trying to prove the following statement which seem not to be very complicated but I cannot find a straightforward way to prove it (can it even be wrong?):
Suppose $X_n$ are a sequence of random variables with mean $\mu_n$ and variance $\sigma^2_n$. We know (by Chebyshev's inequality) that if $\lim_{n\to\infty} \dfrac{\sigma^2_n}{\mu_n^2} = 0$ then for any $\epsilon > 0$,
\begin{equation*}
\lim_{n\to\infty} \Pr\left[X_n \in [(1-\epsilon) \mu_n, (1+\epsilon) \mu_n] \right] = 1.
\end{equation*}
Now, I want to prove the following statement: If $\lim_{n \to \infty} \dfrac{\sigma_n^2}{\mu_n^2} = \infty$ then 
\begin{equation*}
\lim_{n\to\infty} \Pr\left[X_n \in [(1-\epsilon) \mu_n, (1+\epsilon) \mu_n] \right] = 0.
\end{equation*}
Does anyone have an idea how to show the above (or even if the above is correct)?

The statement is false.  Take 
$$X_n = \begin{cases} 0, & \text{ with probability } 1-1/n,\\
                      n^{3/4}, & \text{ with probability } 1/n. \end{cases}$$
Then $\mu_n=n^{-1/4}$ and $\sigma_n^2=\sqrt{n}-1/\sqrt{n}$, but 
$$\forall\epsilon>0: \lim_{n\to\infty} \Pr(|X_n-\mu_n|\leq \epsilon) =1.$$ 
