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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.

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The function $\sigma_k$ is multiplicative, same proof as for $\sigma$. The function $\sigma_k$ is easy to evaluate at prime powers, you just get the sum of a finite geometric series. Now your desired result follows from the Fundamental Theorem of Arithmetic (Unique Factorization Theorem) by multiplicativity. The result for $\sigma(n)$ does not immediately imply the result for $\sigma_k$. But basically the same proof can be used. – André Nicolas Feb 17 '12 at 20:27
up vote 1 down vote accepted

That theorem is not helpful directly, but your stated theorem for $\sigma_k$ follows for the same reason that this theorem does -- because the function is multiplicative.

Hence the value of $\sigma_k(n)$ at any $n$ is just the product of its value at the prime powers dividing $n$.

To check that it's multiplicative, use the fact that if $f$ and $g$ are multiplicative then

$$ h(n)=\sum_{d\mid n}f(d)g(n/d) $$

is multiplicative.

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Yes, it follows almost immediately. Just write out the product of the sums and pick an arbitrary term from the product, and convince yourself these terms are in one-to-one correspondence with the factors of $n$.

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