Show the sample mean $\mathfrak{T}_t$ converges to the population mean faster than $n^{1/3}$. 
Let $\mathfrak{T}_{t}$ be an iid random variable with support $\mathfrak{T}_{t} \in [0,1]$.
Prove $n^{1/3}\frac{1}{n} \sum\limits_{t=1}^{n} (\mathfrak{T}_{t} - \mathbb{E}[\mathfrak{T}_{t}] ) \xrightarrow{p} 0$.

Progress
We have a version of the CLT that supposes iid, finite variance, and the shows the sample mean converges to $N(0, \sigma^2)$. What is throwing me off is the $n^{1/3}$. The normal CLT uses $n^{1/2}$, and I'm not sure why this new variable makes it any different.

The expectation of the square of $\sum\limits_{t=1}^{n} (\mathfrak{T}_{t} - \mathbb{E}[\mathfrak{T}_{t}] )$ is $n\mathbb E\left[(\mathfrak{T}_{1} - \mathbb{E}[\mathfrak{T}_{1}])^2\right]$, hence 
$$\mathbb E\left(n^{-2/3}\sum\limits_{t=1}^{n} (\mathfrak{T}_{t} - \mathbb{E}[\mathfrak{T}_{t}] )\right)^2=n^{-4/3}n\mathbb E\left[(\mathfrak{T}_{1} - \mathbb{E}[\mathfrak{T}_{1}])^2\right].$$
To conclude the convergence in probability, use Markov's inequality.
