For which values of parameter $a$, the series $Z_{n} = n^{\frac{1}{3}}(S_{n}-a)$ is convergent in $L^2$?
Where $ S_{n} = \frac{1}{n}(X_{1} + \cdots + X_{n}) $ with density $f(x) = 3x^{2}\mathbb{1}_{[0,1]}$
Edit: $X_{1} \ldots X_{n}$ are i.i.d. from, they have the same density function .
Converges in the p-th mean $X_{n}$ to $X $ in $L^p$ if the r-th absolute moments $E(|X_{n}|^{p})$ and $E(|X|^{p})$ of $X_{n}$ and $X$ exist, and $\lim _{n\to \infty }\operatorname {E} \left(|X_{n}-X|^{2}\right)=0$.
From Strong law of large number, we have that
$S_{n}$ converges almost surely to $E(X_{1}) = \frac{3}{4}$
If $ a \neq E(X_{1}) $ then $Z_{n}$ not converges.
IF $ a = E(X_{1}) $ then $Z_{n}$ converges to $0$.
This is my result.
This is indeed correct, but some justifications are needed. For example, when $a\neq 3/4$ we know that we do not have the almost sure convergence to $0$ but this would not a priori exclude that we have convergence in $L^2$. But convergence in $L^2$ implies almost sure convergence of a subsequence hence it $Z_n$ was convergent in $L^2$ to $0$, we would have that $Z_{n_k}\to 0$ almost surely for a subsequence $(Z_{n_k})$, a contradiction.
When $a=3/4$, $Z_n=\frac 1{n^{4/3}}\sum_{i=1}^n\left(X_i-3/4\right)$ and the variance can be computed by noticing that the random variables $X_i-3/4$, $1\leqslant i\leqslant n$, are uncorrelated.