Suppose $\{X_n\}$ is uncorrelated sequence,meaning $\mathrm{Cov}(X_i,X_j)=0, i\not= j$

Question

Suppose $\{X_n\}$ is uncorrelated sequence,meaning $$\mathrm{Cov}(X_i,X_j)=0, i\not= j$$ If there exists a constat $c>0$ such that $\mathrm{Var}(X_n)\leq c$ for all $n\geq 1$, then for any $\alpha > \frac{1}{2}$ we have $$\frac{\sum_{i=1}^{n}X_i}{n^\alpha}\stackrel{L_2}{\longrightarrow}0.$$

Proof

I want to show that $E[|\frac{\sum_{i=1}^{n}X_i}{n^\alpha}-0|^2]\rightarrow 0$ as $n\rightarrow \infty$. Observe $$E[|\frac{\sum_{i=1}^{n}X_i}{n^\alpha}-0|^2]=E[(\frac{\sum_{i=1}^{n}X_i}{n^\alpha})^2]=E[\frac{(\sum_{i=1}^{n}X_i)^2}{n^{2\alpha}}]=E[(\sum_{i=1}^{n}X_i)^2)]\frac{1}{n^{2\alpha}}\rightarrow 0$$

Answer 1

The conclusion is not correct without further assumptions. For example, you might have $X_i = 2^i$ almost surely. Then all variances and covariances are $0$, while $\sum_{i=1}^n X_i/n^\alpha = (2^{n+1}-2)/n^\alpha \to +\infty$, not $0$.

Answer 2

I assume that $E(X_i)=0$, for $i=1, \ldots, \infty$. Then, you need only to note that \begin{align*} E\bigg(\Big(\sum_{i=1}^n X_i \Big)^2 \bigg) &= \sum_{i=1}^n E\big( X_i^2\big) +\sum_{i, j=1, i\ne j}^n E(X_iX_j)\\ &= \sum_{i=1}^n E\big( X_i^2\big)\\ &\le nc. \end{align*}