Let $\{X_n\}$ be a pairwise uncorrelated sequence of random variables such that there exists a fixed constant $c>0$ such that $E(X_n^2)\leq c$ for all $n\geq1$. Does it imply that for any $\alpha>0.5$, $Y_n=\dfrac{1}{n^\alpha}\sum_{k=1}^nX_k\xrightarrow{L^2}0?$
Well, for $\alpha>1$ I have found the result is true. My working is as follows:
$$E(\dfrac{1}{n^\alpha}\sum_{k=1}^nX_k)^2$$$$=\dfrac{1}{n^{2\alpha}}E(\sum_{k=1}^nX_k^2+\sum_{i\neq j}X_iX_j)$$
Now $$E(\sum_{k=1}^nX_k^2)\leq nc$$
Also $$E(X_iX_j)\leq\sqrt{E(X_i^2)E(X_j^2)}\leq c$$
So $$\dfrac{1}{n^{2\alpha}}E(\sum_{k=1}^nX_k^2+\sum_{i\neq j}X_iX_j)^2\leq\dfrac{1}{n^{2\alpha}}(nc+n(n-1)c)=n^{2-2\alpha}c$$
The above converges to $0$ iff $2-2\alpha<0\iff\alpha>1$.
What happens when $\alpha\in(\dfrac{1}{2},1]$?
Please note that I have not used the zero correlation condition. I cannot figure out how we can apply it.
The general answer to your question is that it doesn't converge to zero for $\alpha \leq 1$. Consider the case $\alpha = 1$. This is the Law of Large Numbers. If your random variables were iid then that converges to the mean which may not be zero. For $\alpha > 1/2$ we have the sum being even larger. And in the case $\alpha = 1/2$ (which you didn't ask about) we have the central limit theorem.