TITLE: Prove that $n^{-2}\sum_{k=1}^nS_k$ converges in probability
QUESTION [5 upvotes]: Let $\{X_n\}$ be a sequence of uncorrelated random variables with common mean $\mu$, such that $\sup_n$Var$(X_n)<\infty$. If $S_n=\sum_{k=1}^n X_k$, show that
$n^{-2}\sum_{k=1}^nS_k$ converges in probability as $n\to\infty$ and identify the limit.
I know that using Weak Laws of Large Numbers, it is easily proven that $\frac{S_n}{n}\to\mu$ in $L^2$, and thus converges to $\mu$ in probability as well. But I'm having trouble using this result to prove that $n^{-2}\sum_{k=1}^nS_k$ converges in probability. I think it would help to know the limit (at least intuitively) to prove this, but I'm a bit stuck on this problem.
My attempt:
First note that $$\frac{\sum_{k=1}^nS_k}{n^2}\leq\frac{nS_n}{n^2}=\frac{S_n}{n}$$
Then since we know $E(\lvert \frac{S_n}{n}-\mu\rvert^2)\to 0$, $E(\lvert\frac{\sum_{k=1}^nS_k}{n^2}-\mu \rvert^2)\leq E(\lvert \frac{S_n}{n}-\mu\rvert^2)$ by monotonicity (since expectations are integrals, and monotonicity holds for integals), and thus $E(\lvert\frac{\sum_{k=1}^nS_k}{n^2}-\mu \rvert^2)\to 0$ as $n\to\infty$. Now using Markov's inequality, I can conclude that $\frac{\sum_{k=1}^nS_k}{n^2}\to\mu$ in probability.
I kind of get the feeling that my answer may be incorrect. I would love for someone to provide some hints/point me in the right direction. Thanks

REPLY [2 votes]: Let $T_n=\sum_{k=1}^{n} S_k$. Then $\frac {T_n} {n^{2}}=\frac {1(S_1-1\mu)/1+\cdots+(n(S_n-n\mu)/n} {n^{2}}+\mu \frac {n(n+1)} {2n^{2}}$. By triangle inequality for $L^{2}$ norm it is easy to see that the first term tends to $0$ in  $L^{2}$ norm. Hence the limit (in  $L^{2}$ norm, hence also in probability) is $\mu /2$.