Convergence to 0 in probability for non-iid random variables Assume $U_k$ are correlated standard normal random variables.
Let $R_k := a_k U_k^2$, with $a_k > 0$ and $\sum_{k=1}^{\infty} a_k < \infty$.
How can we prove that $S_p:= \frac{1}{p}\sum_{k=1}^{p}R_k$ converge to 0 in probability?
 A: The first thing to try is always Chebyshev's inequality -- especially with well-behaved distributions where moments are likely to be enough to decide the question.
$$\mathrm{var}[S_p]=\frac{1}{p^2}\sum_{i,j=1}^p \mathrm{cov}[a_iU_i^2,a_jU_j^2]$$
Now, correlations are bounded by 1,so:
$$\left|\mathrm{cov}[a_iU_i^2,a_jU_j^2]\right|\leq \sqrt{\mathrm{var}[a_iU_i^2]\mathrm{var}[a_jU_j^2]}=|a_ia_j|C$$
for some finite $C$ that we could work out if we cared.
So,
$$\mathrm{var}[S_p]\leq C\sum_{i,j=1}^p|a_ia_j|$$
and this bound is sharp; it is achieved when the $U_i$ are perfectly correlated.  We need only ask if
$$\lim_{p\to\infty}\frac{1}{p^2}\sum_{i,j=1}^p|a_ia_j|= 0$$
There can be only finitely many $a_i>1$; we can delete an initial segment of the series if necessary and assume there are none, so $a_ia_j<a_i$.  Then
$$\frac{1}{p^2}\sum_{i,j=1}^p|a_ia_j|< \frac{1}{p^2}\sum_{i=1}^pp|a_i|=\frac{1}{p}\sum_{i=1}^p|a_i|$$
Since $\sum_{i=1}^p|a_i|$ is bounded, $$\frac{1}{p}\sum_{i=1}^p|a_i|$$
converges to zero.
So, $S_p$ converges to zero in mean square and thus in probability.
