Convergence of sample mean using CLT 
Assume $X_i$s are i.i.d. random variables with mean $\mu$ and variance $\sigma^2$. Prove:
  $$\lim_{n\to\infty}n^2\mathbb{P}\left(\left|\frac{\sum_{i=1}^{n} X_i}{n}-\mu\right|>n^{-1/4}\right)=0.   \,\,\,\,\,\,\  (1)$$

My effort: I used Chebyshev inequality but did not work. Then, I thought about using the central limit theorem (CLT). By CLT:
$$\lim_{n\to\infty}\mathbb{P}\left(\sqrt{n}\left|\frac{\sum_{i=1}^{n} X_i}{n}-\mu\right|>\gamma\right)=1-\mathrm{erf}\left(\frac{\gamma}{\sigma}\right).$$
The problem is that $\gamma$ cannot be a function of $n$. Otherwise, I could let $\gamma=n^{1/4}$ and prove what I need. Is there a trick I can use here? How can I prove (1)? If (1) only holds for certain conditions, please let me know.
Lastly, if we know $X_i=A_i^2 B_i^2$ where $A_i$ and $B_i$ are Gaussian random variables with non-zero means, can we prove (1)?
 A: Replacing $X_i$ by $X_i-\mu$, we can assume that $\mu=0$. Let 
$$
p_n:=n^2\Pr\left\{\left\lvert \sum_{i=1}^nX_i\right\rvert>n^{3/4}\right\}.
$$
Then 
$$
n^2\Pr\{\left\lvert  X_1\right\rvert>2n^{3/4}\}
\leqslant p_n+n^2\Pr\left\{\left\lvert \sum_{i=1}^{n-1}X_i\right\rvert>n^{3/4}\right\}\leqslant p_n+\frac{n^2}{(n-1)^2}p_{n-1}.
$$
hence if (1) holds, then necessarily, 
$$
\tag{(C)}\lim_{t\to +\infty}t^{8/3}\Pr\{\left\lvert  X_1\right\rvert>t\}=0.
$$
To do the opposite direction, we use the following inequality (Theorem B.3 p. 172 in these notes by Emmanuel Rio): for each independent sequence of  random variables $(Y_i)_{i=1}^N$, each $V\geqslant \sum_{j=1}^N\mathbb E\left[Y_j^2\right]$ and each $\lambda$, $x\gt 0$, 
$$
\Pr\left\{\max_{1\leqslant n\leqslant N}\left\lvert\sum_{i=1}^nY_i \right\rvert   >\lambda\right\}\leqslant 2\exp\left(-\frac V{x^2}h\left(\frac{\lambda x}V\right)\right)+2
\sum_{i=1}^N\Pr\left\{  \left\lvert Y_i \right\rvert  >x \right\},
$$
where $h(u)=(1+u)\log(1+u)-u$.
Then apply this with $N=2^n$, $\lambda =2^{3n/4}$, $x=\sigma^2 2^{3n/8}$ and $V=\sigma^22^n$ to get that 
$$\max_{2^{n-1}\leqslant k\leqslant 2^n}p_k\leqslant 
2\exp\left(-\sigma^{-2}2^{n/4}h(2^{n/8}) \right)+2^{n+1}\Pr\{\left\lvert Y_1 \right\rvert>\sigma^2 2^{3n/8}  \}$$
hence $p_n\to 0$. 
