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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

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Hint: Show $\sum_{d \mid n} d^q$ is multiplicative and use this proposition:

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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: $\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}$.

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  • $\begingroup$ I proved that $d^q$ is multiplicative but even after extending the formal identity for multiplicative function I am not able to simplify it any further... $\endgroup$ – Cloverr May 30 '15 at 3:42
  • $\begingroup$ @Nilanjan I've edited to include a complete answer. $\endgroup$ – André 3000 Jun 9 '15 at 2:03
  • $\begingroup$ Ty for the solution, I tried it a lot of times earlier but later had to give up $\endgroup$ – Cloverr Jun 9 '15 at 3:56

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