Let $\sigma_{k}(n)$ denote the sum of the $k$th powers of the divisors of $n$, so that $\sigma_{k}(n)=\sum_{d|n}d^{k}$. Note that $\sigma_{1}(n)=\sigma(n)$.
In a previous exercise, I was asked to show that if $p$ is prime and both $a$ and $k$ are a positive integers, then$\sigma_{k}(p^{a})=1+p^{k}+p^{2k}+\cdots+p^{ak}=\frac{p^{k(a+1)}-1}{p^{k}-1}.$I proved that by induction, and I now want to prove it for $\sigma_{k}(n)$, where $n$ has prime factorization $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{m}^{a_{m}}$. Whilst working on this, I ran into the following theorem:
Theorem. Let the positive integer $n$ have prime factorization $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{s}^{a_{s}}$. Then$\sigma(n)=\frac{p_{1}^{a_{1}+1}-1}{p_{1}-1}\cdot\frac{p_{2}^{a_{2}+1}-1}{p_{2}-1}\cdot\cdots\cdot\frac{p_{s}^{a_{s}+1}-1}{p_{s}-1}=\prod_{j=1}^{s}\frac{p_{j}^{a_{j}+1}-1}{p_{j}-1}.$
Would this then immediately imply that$\sigma_{k}(n)=\prod_{j=1}^{m}\frac{p_{j}^{k(a_{j}+1)}-1}{p_{j}^{k}-1}?$I have the feeling that I may be missing something, however.