Let $\displaystyle \sigma_k(n) = \sum_{d|n}d^k$ (Note: $\displaystyle \sigma_1(n) = \sigma(n)$) the sum of the $\displaystyle k$th powers of divisors of n.

Is $\displaystyle \sigma_3(2) = 1^3 + 2^3 = 9 $?

Also I need so help on the following:- Give a formula for $\displaystyle \sigma_k(p)$ and $\displaystyle \sigma_k(p^a)$ where p is prime, a is positive integer.

My answer is $\displaystyle \sigma_k(p) = 1 + p^k $ and $\displaystyle \sigma_k(p^a) = 1+ \sum_{i=1}^a p^i$, right?

- Show the function $\displaystyle \sigma_k$ is multiplicative, but not completely multiplicative.
- Find a formula for $\displaystyle \sigma_k(n)$, where n has prime power factorization $\displaystyle n = p_1^{a_1}p_2^{a_2}...p_m^{a_m}$.

Thank you for your help.