= Solution
The <weak law of large numbers> gives $S_n/n\to p$ in probability. If $\alpha<p$, the event $S_n/n\leq\alpha$ implies $|S_n/n-p|\geq p-\alpha$, whose probability tends to zero. If $\alpha>p$, its complement $S_n/n>\alpha$ implies $|S_n/n-p|>\alpha-p$, whose probability also tends to zero. Thus
$$
\boxed{\lim_{n\to\infty}\mathbb P(S_n/n\leq\alpha)=\begin{cases}0&\alpha<p,\\1&\alpha>p.\end{cases}}
$$
For an explicit independent bound, <Chebyshev's inequality> bounds either unwanted probability by $p(1-p)/[n(\alpha-p)^2]$. The boundary $\alpha=p$ is outside the two requested cases and is not decided by the <weak law of large numbers> alone.
Back to article page
