Dirichlet series generating function of a sequence To find the dirichlet series generating function of the following sequence $\left\{\sum_{d \backslash n}d^q\right\}_{n=1}^\infty$
The series is like this $\frac{1^q}{1^s} + \frac{1^q+2^q}{2^s} + \frac{1^q+3^q}{3^s} + \ldots$  and it looks like the product of two series similar to Riemann zeta function, one maybe to list the divisors and other to sum them all raise to power q , however I don’t see any way to solve it further
 A: Hint:  Show $\sum_{d \mid n} d^q$ is multiplicative and use this proposition:

from p. 59 of Wilf.
Edit: All right, here's a complete solution.  Let $f(n) = \sum_{d \mid n} d^q$.  Since the divisors of $p^k$ are just $1, p, p^2, \ldots, p^k$, then
\begin{align*}
f(p^k) &= \sum_{d \mid p^k} d^q = 1^q + p^q + p^{2q} + \cdots + p^{kq} = \frac{1 - (p^q)^{k+1}}{1 - p^q}
\end{align*}
where the last equality follows by the formula for a finite geometric sum.  Using the cited theorem, then
\begin{align*}
\sum_{n=1}^\infty \frac{f(n)}{n^s} &= \prod_{p \text{ prime}} \sum_{k=0}^\infty \frac{f(p^k)}{p^{ks}} = \prod_{p \text{ prime}} \frac{1}{1 - p^q} \sum_{k=0}^\infty \frac{1 - (p^q)^{k+1}}{p^{ks}}\\
&= \prod_{p \text{ prime}} \frac{1}{1 - p^q} \left(\sum_{k=0}^\infty (p^{-s})^k - p^q \sum_{k=0}^\infty (p^{q-s})^k \right)\\
&= \prod_{p \text{ prime}} \frac{1}{1 - p^q} \left(\frac{1}{1 - p^{-s}} - \frac{p^q}{1 - p^{q-s}} \right) = \prod_{p \text{ prime}} \frac{1}{1 - p^q} \ \frac{1 - p^{q-s} - p^q + p^{q-s}}{(1 - p^{-s})(1 - p^{q-s})}\\
&=\prod_{p \text{ prime}} \frac{1}{1 - p^q} \ \frac{1 - p^q}{(1 - p^{-s})(1 - p^{q-s})} = \prod_{p \text{ prime}} \frac{1}{(1 - p^{-s})(1 - p^{-(s-q)})}\\
&= \left(\prod_{p \text{ prime}} \frac{1}{1 - p^{-s}} \right) \left(\prod_{p \text{ prime}} \frac{1}{1 - p^{-(s-q)}}\right) = \zeta(s) \zeta(s-q)
\end{align*}
where the last equality holds by the Euler product for the zeta function: $\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}$.
