Prime Number is Deficient/Proof 2

Proof
Let $p$ be a prime number.

From Sigma Function of Prime Number:
 * $\sigma \left({p}\right) = p + 1$

and so:
 * $\dfrac {\sigma \left({p}\right)} p = \dfrac {p + 1} p = 1 + \dfrac 1 p$

As $p > 1$ it follows that $\dfrac 1 p < 1$.

Hence:
 * $\dfrac {\sigma \left({p}\right)} p < 2$

The result follows by definition of deficient.