Exercise 4.2.8 (Formula for $\sigma_k(n)$)

Question

Find a formula for $\sigma_k(n)$.

Richard Ganaye · Accepted Answer

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
We compute first $\sigma_k(p^)$, where $p$ is a prime number and $ \in \N$.

Since the divisors of $p^\alpha$ are $1,p,p^2,\ldots,p^\alpha$,
\begin{align*}
\sigma_k(p^\alpha) &= \sum_{d \mid p^\alpha} d^k\
&= \sum_{i=0}^\alpha p^{ki}\
&= \frac{p^{k(\alpha+1)} - 1}{p^k-1}.
\end{align*}
Since $\sigma_k$ is a multiplicative function,
$$\sigma_k(n) = \prod_{p^\alpha \mid \mid n }  \frac{p^{k(\alpha+1)} - 1}{p^k-1}.$$

\end{proof}

Exercise 4.2.8 (Formula for $\sigma_k(n)$)

Answers

Comments

Exercise 4.2.8 (Formula for $\sigma_k(n)$)

Answers

Comments

Add answer