Let $X_1,X_2,...$ be iid with finite mean $\mu$ and finite variance $\sigma^2$
Prove that
$\frac{1}{n(n-1)}\sum_{i=1}^nX_i^2 \to 0$ in probability.
I tried to use variance to prove this and also tried to prove using the law of large numbers but I think I am missing something.
Since the random variables $X_i^2$, $i \geq 1$, are independent, identically distributed and have finite mean it follows from the strong law of large numbers that
$$\frac{1}{n} \sum_{i=1}^n X_i^2 \to \mathbb{E}(X_1^2) < \infty \quad \text{a.s.}$$
This implies, in particular,
$$\frac{1}{n(n-1)} \sum_{i=1}^n X_i^2 \to 0 \quad \text{a.s.}$$
Alternative reasoning to prove convergence in probability: By Markov's inequality, we have
$$\mathbb{P} \left(\left| \frac{1}{n(n-1)} \sum_{i=1}^n X_i^2 \right| \geq \epsilon \right) \leq \frac{1}{\epsilon \, n \, (n-1)} \sum_{i=1}^n \underbrace{\mathbb{E}(X_i^2)}_{=\mathbb{E}(X_1^2)=\sigma^2+\mu^2} = \frac{\sigma^2+\mu^2}{\epsilon} \frac{1}{n-1} \xrightarrow[]{n \to \infty} 0$$
for any $\epsilon>0$.