# $X_1, X_2, \dots$ uncorrelated, $\frac{Var[X_i]}{i} \rightarrow 0$, then $\frac{S_n}{n} - \frac{\mathbb{E}[S_n]}{n} \rightarrow 0$ in $L^2$

Let $X_1, X_2, \dots$ be uncorrelated random variables with $\mathbb{E}[X_i]= \mu_i$ and $\displaystyle\frac{Var[X_i]}{i} \rightarrow 0$, when $i \rightarrow +\infty$. Show that $\displaystyle\frac{S_n}{n} - \frac{\mathbb{E}[S_n]}{n} \rightarrow 0$, as $n \rightarrow +\infty$ in $L^2$ and probability.

As convergence in $L^2$ implies convergence in probability, we only need to show convergence in $L^2$. So, we have to show that $\displaystyle\mathbb{E}\left[\left|\frac{S_n}{n} - \frac{\mathbb{E}[S_n]}{n}\right|^2\right] \rightarrow 0$,as $n \rightarrow +\infty$.

As $\displaystyle\frac{Var[X_i]}{i}$, given $\epsilon > 0$, there is $n_0$ such that for $n>n_0$, $\displaystyle\frac{Var[X_i]}{i} < \epsilon$.

Let $C = \displaystyle\max\left\{\frac{Var[X_1]}{1}, \dots, \frac{Var[X_{n_0}]}{n_0}, \epsilon\right\}$. Then, $\displaystyle\frac{Var[X_i]}{i} < C$, for all $i$.

$\displaystyle\mathbb{E}\left[\left|\frac{S_n}{n} - \frac{\mathbb{E}[S_n]}{n}\right|^2\right] = \frac{1}{n^2} \mathbb{E}[|S_n - \mathbb{E}[S_n]|^2] = \frac{1}{n^2}Var[S_n] = \frac{1}{n^2}\sum_{k=1}^n Var[X_k]$

Above, we used that $X_1, X_2, \dots$ are uncorrelated.

Continuing:

$\displaystyle\frac{1}{n^2}\sum_{k=1}^n Var[X_k] = \frac{1}{n}\sum_{k=1}^n \frac{Var[X_k]}{n} \leq \frac{1}{n}\sum_{k=1}^{n} \frac{Var[X_k]}{k} \leq \frac{1}{n} \sum_{k=1}^{n} C$ = C

And it gives me nothing... Perhaps I need to find another upper bound!

Can someone help me?

You simply threw away the fact that $\sigma_j^2:=Var(X_j)/j\to0$ and used only the fact that it's bounded by $C$. Do everything just the way you did til the very end. Then say $$\frac1n\sum_{j=1}^n\sigma_j^2\le\frac1n\left(\sum_{j=1}^{n_0}C+\sum_{j=n_0+1}^n\epsilon\right)\le\frac{C n_0}{n}+\epsilon.$$Now letting $n\to\infty$ shows that $$\limsup_{n\to\infty}\frac1n\sum_{j=1}^nv_j\le\epsilon.$$