Divisor Sum of Integer

Theorem
Let $n$ be an integer such that $n \ge 2$.

Let the prime decomposition of $n$ be:
 * $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$

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

Proof
We have that the Sigma Function is Multiplicative.

From Value of Multiplicative Function is Product of Values of Prime Power Factors, 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:
 * $\displaystyle \sigma \left({p_i^{k_i}}\right) = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

Hence the result.