Divisor Sum of Integer

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

Let $\map {\sigma_1} n$ be the divisor sum of $n$.

That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.

Let the prime decomposition of $n$ be:
 * $\displaystyle n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$

Then:
 * $\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

Proof
We have that the Divisor Sum Function is Multiplicative.

From Value of Multiplicative Function is Product of Values of Prime Power Factors, we have:
 * $\map f n = \map f {p_1^{k_1} } \map f {p_2^{k_2} } \ldots \map f {p_r^{k_r} }$

From Divisor Sum of Power of Prime, we have:
 * $\ds \map {\sigma_1} {p_i^{k_i} } = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

Hence the result.