[proofplan] We compute $\mathbb{E}[S_n/n]$ and $\operatorname{Var}(S_n/n)$ using linearity and the variance formula for sums of independent variables, then apply Chebyshev's inequality to bound $\mathbb{P}(|S_n/n - \mu| > \varepsilon)$. The bound tends to zero as $n \to \infty$. [/proofplan] [step:Compute $\mathbb{E}[S_n/n] = \mu$ and $\operatorname{Var}(S_n/n) = \sigma^2/n$] By [linearity of expectation](/theorems/1117), \begin{align*} \mathbb{E}\!\left[\frac{S_n}{n}\right] = \frac{1}{n}\,\mathbb{E}[S_n] = \frac{1}{n}\sum_{i=1}^n \mathbb{E}[X_i] = \frac{1}{n} \cdot n\mu = \mu. \end{align*} For the variance, since $X_1, \ldots, X_n$ are independent (and identically distributed), the [variance of a sum](/theorems/1119) of independent random variables gives $\operatorname{Var}(S_n) = \sum_{i=1}^n \operatorname{Var}(X_i) = n\sigma^2$. Using $\operatorname{Var}(cZ) = c^2\operatorname{Var}(Z)$ with $c = 1/n$: \begin{align*} \operatorname{Var}\!\left(\frac{S_n}{n}\right) = \frac{1}{n^2}\operatorname{Var}(S_n) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}. \end{align*} [/step] [step:Apply Chebyshev's inequality and take $n \to \infty$] By [Chebyshev's inequality](/theorems/1126) applied to the random variable $\bar{X}_n = S_n/n$ with $\mathbb{E}[\bar{X}_n] = \mu$ and $\operatorname{Var}(\bar{X}_n) = \sigma^2/n$: \begin{align*} \mathbb{P}\!\left(\left|\frac{S_n}{n} - \mu\right| \ge \varepsilon\right) \le \frac{\operatorname{Var}(S_n/n)}{\varepsilon^2} = \frac{\sigma^2}{n\varepsilon^2}. \end{align*} For fixed $\varepsilon > 0$ and fixed $\sigma^2 < \infty$, \begin{align*} 0 \le \mathbb{P}\!\left(\left|\frac{S_n}{n} - \mu\right| \ge \varepsilon\right) \le \frac{\sigma^2}{n\varepsilon^2} \to 0 \quad \text{as } n \to \infty. \end{align*} By the squeeze theorem (the probability is bounded between $0$ and a quantity tending to $0$), $\mathbb{P}(|S_n/n - \mu| \ge \varepsilon) \to 0$. Since $\mathbb{P}(|S_n/n - \mu| > \varepsilon) \le \mathbb{P}(|S_n/n - \mu| \ge \varepsilon) \to 0$, the result follows. [guided] This is the classic "three-line proof" of the Weak Law, and the strategy is worth understanding in full: 1. **Compute the mean of $\bar{X}_n$:** Linearity of expectation (which requires no independence) gives $\mathbb{E}[\bar{X}_n] = \mu$. This tells us the sample mean is "centred" at the right place. 2. **Compute the variance of $\bar{X}_n$:** Here independence is consumed. The variance of a sum of independent random variables equals the sum of variances: $\operatorname{Var}(S_n) = n\sigma^2$. Dividing by $n^2$ (from the $1/n$ scaling), we get $\operatorname{Var}(\bar{X}_n) = \sigma^2/n$. This is the key calculation — it says the spread of $\bar{X}_n$ shrinks like $1/n$. 3. **Apply Chebyshev:** Chebyshev converts the shrinking variance into a probability bound. The inequality $\mathbb{P}(|\bar{X}_n - \mu| \ge \varepsilon) \le \sigma^2/(n\varepsilon^2)$ gives an explicit rate of convergence. Note that the hypothesis $\sigma^2 < \infty$ is essential: without it, Chebyshev's inequality does not apply, and the Weak Law can fail (there exist distributions with infinite variance for which $S_n/n$ does not converge in probability). However, the Weak Law actually holds under the weaker condition $\mathbb{E}[|X_1|] < \infty$ (without assuming finite variance) — but that proof requires truncation arguments rather than Chebyshev. [/guided] [/step]