Sigma over Phi

Since $p$ is prime, its divisors are $1$ and $p$ , so we have $\sigma(p) = 1 + p$ . Since all integers from $1$ to $p-1$ are coprime to $p$ , so $\phi(p) = p-1$ . Thus, $\frac{\sigma(p)}{\phi(p)} = \frac{p+1}{p-1} = \frac{p - 1 + 2}{p-1} = 1 + \frac{2}{p-1}$ . Clearly, $\frac{2}{p-1}$ decreases as $p$ increases. Since $2$ is the smallest prime, we have $\frac{2}{p-1} \leq \frac{2}{2-1} = 2$ . Hence $\frac{\sigma(p)}{\phi(p)} \leq 1 + 2 = 3$ .