# Is it true that $\sum\limits_{n=1}^\infty\frac{\chi_{k,1}(n)}{n^s}=\zeta(s)\sum\limits_{d|k}\mu(d)\,d^{-s}$?

Question 1: Is the relationship illustrated in (1) below true where $\chi_{k,1}(n)$ corresponds to the ordering of Dirichlet characters implemented by Mathematica?

(1) $\quad\sum\limits_{n=1}^\infty\frac{\chi_{k,1}(n)}{n^s}=\zeta(s)\sum\limits_{d|k}\mu(d)\,d^{-s},\quad Re[s]>1$

The first 12 Mathematica Dirichlet characters $\chi_{k,1}(n)$ and their associated Dirichlet series given by relationship (1) above are as follows.

$\begin{array}{cc} \{1\} & \zeta (s) \\ \{1,0\} & \left(1-2^{-s}\right) \zeta (s) \\ \{1,1,0\} & \left(1-3^{-s}\right) \zeta (s) \\ \{1,0,1,0\} & \left(1-2^{-s}\right) \zeta (s) \\ \{1,1,1,1,0\} & \left(1-5^{-s}\right) \zeta (s) \\ \{1,0,0,0,1,0\} & \left(1-2^{-s}-3^{-s}+6^{-s}\right) \zeta (s) \\ \{1,1,1,1,1,1,0\} & \left(1-7^{-s}\right) \zeta (s) \\ \{1,0,1,0,1,0,1,0\} & \left(1-2^{-s}\right) \zeta (s) \\ \{1,1,0,1,1,0,1,1,0\} & \left(1-3^{-s}\right) \zeta (s) \\ \{1,0,1,0,0,0,1,0,1,0\} & \left(1-2^{-s}-5^{-s}+10^{-s}\right) \zeta (s) \\ \{1,1,1,1,1,1,1,1,1,1,0\} & \left(1-11^{-s}\right) \zeta (s) \\ \{1,0,0,0,1,0,1,0,0,0,1,0\} & \left(1-2^{-s}-3^{-s}+6^{-s}\right) \zeta (s) \\ \end{array}$

I believe $\chi_{k,1}(n)$ corresponds to the principal character modulo $k$ which can be evaluated as $\chi_{k,1}(n)=\text{If}[\gcd (k,n)=1,1,0]$.

• Try to look at "Titchmarsh - The theory of the Riemann zeta-function" p.10 (1.5. Ramanujan's sums). Maybe that helps. – Diger Sep 17 '18 at 23:33
• I guess for $k$ prime is clear? – Diger Sep 18 '18 at 18:19

1. Multiply your equality by $$k^s$$

2. Calculate the inverse Mellin transform for $$x>0$$ $${\cal M}_f^{-1}(x)=\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} f(s) \, x^{s} \, {\rm d}s$$ for the LHS and RHS respectively.

3. $$\forall m\in{\mathbb N}$$ apply $$\sum_{k|m}$$ to each transform.

LHS:

1. The Mellin transform gives $$\sum_{n=1}^\infty \chi_{k,1}(n) \, \delta\left(\log\frac{kx}{n}\right) = x\sum_{n=1}^\infty \left[(n,k)=1\right] \, \delta\left(x-\frac{n}{k}\right)$$

2. and the summation yields $$x \sum_{k|m} \sum_{n=1}^\infty \left[(n,k)=1\right] \, \delta\left(x-\frac{n}{k}\right) \, .$$ The result is just the decomposition of $$x\sum_{n=1}^\infty \delta\left(x-\frac{n}{m}\right)$$ since the sets $${\cal S}_d = \left\{{n}/{d} \, \big| \, n\in {\mathbb N}, (n,d)=1\right\}$$ are disjoint for different divisors $$d$$ of $$m$$ and $${\cal S} = \bigcup_{d|m} {\cal S}_d =\left\{ n/m \, \big| \, n \in {\mathbb N} \right\} \, .$$

RHS:

1. The Mellin transform gives $$\sum_{n=1}^\infty \sum_{d|k} \mu(d) \, \delta\left(\log\frac{kx}{nd}\right) = x\sum_{n=1}^\infty \sum_{d|k} \mu(d) \, \delta\left(x-\frac{nd}{k}\right)$$

2. and the summation yields $$x\sum_{n=1}^\infty \sum_{k|m}\sum_{d|k} \mu(d) \, \delta\left(x-\frac{nd}{k}\right) = x \sum_{n=1}^\infty \delta\left(x-\frac{n}{m}\right)$$ where the möbius inversion formula was used.

So obviously LHS and RHS are equal as the above expression is valid $$\forall m\in{\mathbb N}$$ and the original expression can be retrieved by another möbius transform.