 Solution

The weak law of large numbers gives $S_n/n\to p$ in probability. If $\alpha <p$, the event $S_n/n\le \alpha$ implies $|S_n/n-p|\ge p-\alpha$, 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. ThusFor 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.