Divisor Sum of Integer

Theorem
Let $$n$$ be an integer such that $$n \ge 2$$, with prime decomposition $$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$$.

Let $$\sigma \left({n}\right)$$ be the sigma function of $$n$$.

That is, let $$\sigma \left({n}\right)$$ be the sum of all positive divisors of $$n$$.

Then:
 * $$\sigma \left({n}\right) = \prod_{1 \le i \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$$

Proof
From Basic Properties of Multiplicative Function, we have:
 * $$f \left({n}\right) = f \left({p_1^{k_1}}\right) f \left({p_2^{k_2}}\right) \ldots f \left({p_r^{k_r}}\right)$$

From Sigma of Power of Prime, we have:
 * $$\sigma \left({p_i^{k_i}}\right) = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$$

Hence the result.