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)?
 A: 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 
$$\forall\epsilon>0: \lim_{n\to\infty} \Pr(|X_n-\mu_n|\leq \epsilon) =1.$$ 
A: Similar to JPi's, here's another attempt at a counter example, with the multiplicative concentration. Define $X_n$ to be the random variable supported on $\{-n,\frac{1}{1-1/n},n\}$ by
$$
X_n \stackrel{\rm def}{=} \begin{cases}
n &\text{ w.p. } \frac{1}{2n} \\
\frac{1}{1-1/n} &\text{ w.p. } 1-\frac{1}{n} \\
-n &\text{ w.p. } \frac{1}{2n}
\end{cases}
$$
You have $\mu_n\stackrel{\rm def}{=}\mathbb{E}X_n = 1$, and the variance is defined and equal to 
$$
\sigma^2_n\stackrel{\rm def}{=}\operatorname{Var}X_n = \mathbb{E}[X^2_n] - 1 = n + \frac{1}{1-\frac{1}{n}} - 1 = n+o(1)
$$
so that indeed $\frac{\sigma^2_n}{\mu^2_n} = n+o(1) \xrightarrow[n\to\infty]{} \infty$. But on the other hand, fixing any $\epsilon > 0$, you have that for any $n\geq 1/\epsilon+1$
$$
\mathbb{P}\{\lvert X_n - \mu_n\rvert \leq \epsilon \mu_n \} = \mathbb{P}\{\lvert X_n - 1\rvert \leq \epsilon \} = \mathbb{P}\left\{X_n = \frac{1}{1-\frac{1}{n} }\right\} = 1-\frac{1}{n} \xrightarrow[n\to\infty]{} 1.
$$
