Exercise 4.2.7 ($\sigma^{-k}(n) = n^{-k} \sigma_k(n)$)

Question

Prove that $\sigma^{-k}(n) = n^{-k} \sigma_k(n)$.

Richard Ganaye · Accepted Answer

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\begin{proof} 
Let $n$ be any positive integer, and $k \in \R$ as in definition 4.1. By definition of $\sigma_k$ and $\sigma_{-k}$,
\begin{align*}
n^k \sigma_{-k}(n) &= n^k \sum_{d \mid n} d^{-k}\
&=\sum_{d \mid n} \left( \frac{n}{d}ight)^k\
&= \sum_{d\mid n} d^k,
\end{align*}
by the result of Problem 6 applied to $f : x \mapsto x^k$.

This shows that $n^k \sigma_{-k}(n) = \sigma_k(n)$, so
$$\sigma^{-k}(n) = n^{-k} \sigma_k(n).$$
\end{proof}

Exercise 4.2.7 ($\sigma^{-k}(n) = n^{-k} \sigma_k(n)$)

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Exercise 4.2.7 ($\sigma^{-k}(n) = n^{-k} \sigma_k(n)$)

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