I want to prove that $\frac{1}{n}S_{n} \overset{p}{\longrightarrow} p\:$ as $\:n \longrightarrow \infty$ from first principles,
where $S_{n}$ denotes the number of successes in $n$ binomial trials with success probability $p$.
From the definition of convergence in probability.
A random sequence $Y_{n}$ converges in probability to $c$ if:
$$P(|Y_{n} - c| < \epsilon) \longrightarrow 1\text{ as } n \longrightarrow \infty$$
For every $\epsilon > 0$.
We can do the following:
$$\begin{align}P\left(\left|\frac{S_{n}}{n} - p\right| < \epsilon\right) &= P\left(\frac{S_{n}}{n} - p < \epsilon\right) - P\left(\frac{S_{n}}{n} - p < -\epsilon\right) \\&= P(S_{n} < n(p + \epsilon)) - P(S_{n} < n(p - \epsilon))\end{align}$$
I assume that the result follows the binomial cdf. But im not sure how to proceed.
$P(|\frac {S_n} n-p|>\epsilon) \leq \frac {E(\frac {S_n} n-p)^{2}} {\epsilon^{2}}$.
$E(\frac {S_n} n-p)^{2}=\frac 1 {n^{2}} E( \sum_{i=1}^{n} (X_i-p))^{2}$. Expand the square and note that $E(X_i-p)(X_j-p)=0$ for $i \neq j$. Compute $E(X_i-p)^{2}$ and finally note that $\frac {p(1-p)} n\epsilon^{2} \to 0$ as $ n \to \infty$.