Unity Sum

Here is the generalization. Assume that $s > 1$ throughout.

If $\omega = e^{\frac{2\pi i}{k}}$ , then $\omega^j = 1$ if and only if $k \big| j$ , and hence $\sum_{b=1}^k \omega^{bj} \; = \; \left\{ \begin{array}{lll} k & \qquad & k \big| j \\ 0 & & \mathrm{o.w.} \end{array} \right.$ Thus $\begin{array}{rcl} S_k & = & \displaystyle k^s\sum_{b =1}^k \sum_{a_1 \ge 1} \sum_{a_2 \ge 1} \cdots \sum_{a_n \ge 1} \frac{\omega^{ba_1a_2\cdots a_n}}{(a_1a_2 \cdots a_n)^s} \\ & = & \displaystyle k^{s+1}\sum_{{a_1,a_2,\ldots,a_n \ge 1} \atop {k \big| a_1a_2\cdots a_n}} \frac{1}{(a_1a_2 \cdots a_n)^s} \\ & = & \displaystyle k \sum_{N \ge 1} \frac{\big| \big\{ (a_1,a_2, \ldots, a_n) \in \mathbb{N}^n \,\big|\,a_1a_2 \cdots a_n = kN \big\} \big|}{(a_1a_2 \cdots a_n)^s} \\ & = & \displaystyle k \frac{\big(1^{\star,n}\big)(kN)}{N^s} \end{array}$ where $1^{\star,n}$ is the $n$ -fold Dirichlet convolution of the constant function $1$ with itself (so that, for instance, $1^{\star,2} = \sigma_0$ is the ``number of divisors" function, and $1^{\star,3}$ is what was referred to previously as $1 \star 1 \star 1$ ).

A simple induction shows us that $\big(1^{\star,n}\big)(p^a) \,=\, {a + n-1 \choose a}$ for any prime $p$ and any integers $n \ge 1$ and $a \ge 0$ , and hence $\sum_{a \ge 0} \frac{\big(1^{\star,n}\big)(p^a)}{p^{as}} \; = \; \sum_{a \ge 0} \binom{a + n - 1}{a}\big(p^{-s}\big)^a \; = \; \big(1 - p^{-s}\big)^{-n}$ for all primes $p$ and all integers $n \ge 1$ .

Suppose that the integer $k$ has prime factorization $k \,=\, p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m}$ for primes $p_1,p_2,\ldots,p_m$ and positive integers $\alpha_1,\alpha_2,\ldots,\alpha_m$ . Let $S(k)$ be the set of positive integers which are coprime to $k$ . Then $\begin{array}{rcl} \zeta(s)^n & = & \displaystyle \sum_{N \ge 1} \frac{\big(1^{\star,n}\big)(N)}{N^s} \\ & = & \displaystyle \sum_{\beta_1\ge 0} \sum_{\beta_2 \ge 0} \cdots \sum_{\beta_m \ge 0} \sum_{c \in S(k)} \frac{\big(1^{\star,n}\big)(p_1^{\beta_1} p_2^{\beta_2} \cdots p_m^{\beta_m} c)}{p_1^{\beta_1s} p_2^{\beta_2s} \cdots p_m^{\beta_m s} c^s} \\ & = & \displaystyle \prod_{j=1}^m \left(\sum_{\beta \ge 0} \frac{\big(1^{\star,n}\big)(p_j^{\beta})}{p_j^{\beta s}}\right) \times \left(\sum_{c \in S(k)} \frac{\big(1^{\star,n}\big)(c)}{c^s}\right) \\ & = & \displaystyle \left(\prod_{j=1}^m \big(1 - p_j^{-s}\big)^{-n}\right) \times \left(\sum_{c \in S(k)} \frac{\big(1^{\star,n}\big)(c)}{c^s}\right) \end{array}$ We also note the identity $(1-x)^n\sum_{a \ge 0} {a + b + n - 1 \choose a+b}x^{a+b} \; = \; b {b+n-1 \choose b} B_x(b, n) \qquad |x| < 1$ for integers $b,n \ge 1$ , where $B_x(b,n)$ is the incomplete Beta function. Thus $\begin{array}{rcl} S_k & = & \displaystyle k \sum_{N \ge 1}\frac{\big(1^{\star,n}\big)(kN)}{N^s} \\ & = & \displaystyle k \sum_{\beta_1\ge0}\sum_{\beta_2\ge0} \cdots \sum_{\beta_m \ge 0} \sum_{c \in S(k)} \frac{\big(1^{\star,n}\big)(p_1^{\alpha_1+\beta_1} p_2^{\alpha_2+\beta_2} \cdots p_m^{\alpha_m+\beta_m}c)}{p_1^{\beta_1s} p_2^{\beta_2 s} \cdots p_m^{\beta_m s} c^s} \\ & = & \displaystyle k \prod_{j=1}^m \left(\sum_{\beta \ge 0} \frac{\big(1^{\star,n}\big)(p_j^{\alpha_j+\beta})}{p_j^{\beta s}}\right) \times \left(\sum_{c \in S(k)} \frac{\big(1^{\star,n}\big)(c)}{c^s}\right) \\ & = & \displaystyle k \prod_{j=1}^m \left[\sum_{\beta \ge 0} \binom{\alpha_j + \beta + n - 1}{\alpha_j + \beta}\big(p_j^{-s}\big)^\beta\right] \times \left(\sum_{c \in S(k)} \frac{\big(1^{\star,n}\big)(c)}{c^s}\right) \\ & = & \displaystyle k^{1+s} \prod_{j=1}^m \left[\sum_{\beta \ge 0} \binom{\alpha_j + \beta + n - 1}{\alpha_j + \beta}\big(p_j^{-s}\big)^{\alpha_j+\beta}\right] \times \left(\sum_{c \in S(k)} \frac{\big(1^{\star,n}\big)(c)}{c^s}\right) \\ & = & \displaystyle k^{1+s} \zeta(s)^n \prod_{j=1}^m \left[\big(1 - p_j^{-s}\big)^n \sum_{\beta \ge 0} {\alpha_j + \beta + n - 1 \choose \alpha_j + \beta}\big(p_j^{-s}\big)^{\alpha_j+\beta}\right] \\ & = & \displaystyle k^{1+s} \zeta(s)^n \prod_{j=1}^m \left[ \alpha_j \binom{\alpha_j + n - 1}{\alpha_j} B_{p_j^{-s}}(\alpha_j,n)\right] \;. \end{array}$

Hi @Mark Hennings we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.