How do I determine $\lim_{n\to \infty}P(|\sum_{i=1}^{n^2}X_i - n^2|\leq n^{\alpha})$ where $(x_i)$ is a sequence of random variables?

Question

Let $x_1, x_2, \dotsb$ be a sequence of independent random variables distributed uniformly on the interval $[0, 2]$. Find all possible values of the following limit, where $\alpha$ is a positive number:

$$\lim_{n\to \infty}P\left(\left|\sum_{i=1}^{n^2}X_i - n^2\right|\leq n^{\alpha}\right).$$

Answer 1

The $X_i$ have mean $\mu=1$ and variance $\sigma^2=\frac13$. By the central limit theorem, the distribution of

$$ S_n=\frac n\sigma\frac1{n^2}\left(\sum_{i=1}^{n^2}(X_i-\mu)\right) $$

converges to the standard normal distribution for $n\to\infty$. As

$$ \left|\,\sum_{i=1}^{n^2}X_i-n^2\,\right|\le n^\alpha $$

is equivalent to

$$ |S_n|\le\frac{n^{\alpha-1}}\sigma\;, $$

the limit is $0$ if $\alpha\lt1$, $1$ if $\alpha\gt1$, and

$$ \operatorname{erf}\left({\sqrt\frac32}\right)\approx0.917 $$

if $\alpha=1$.