GCD function and relation between Hurwitz and Riemann zeta function

Question

Does anyone know how to show the following:

$\sum_{k=1}^n\gcd(n,k)\zeta(s,\frac{k}{n})=\left(\sum_{k=1}^n\gcd(n,k)\right)\zeta(s)$

Answer 1

Start by re-writing your LHS sum as follows: $$\sum_{k=1}^n \gcd(n,k) \sum_{d\ge 0} \frac{1}{(d + k/n)^s} = n^s \sum_{k=1}^n \gcd(n,k) \sum_{d\ge 0} \frac{1}{(nd + k)^s}.$$ Now by inspection the sum part is clearly seen to be $$L(s) = \sum_{m\ge 1} \frac{\gcd(n,m)}{m^s}.$$ Let $p$ be the primes that divide $n$ and $q$ those that do not. Let $v$ be the exponent of $p$ in the factorization of $n.$ Then $L(s)$ has the following Euler product: $$\prod_q \frac{1}{1-1/q^s} \prod_p \left(1 + \frac{p}{p^s} + \frac{p^2}{p^{2s}} +\cdots+ \frac{p^v}{p^{vs}} \left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\cdots\right)\right).$$ This is $$\prod_q \frac{1}{1-1/q^s} \prod_p \left(\frac{1-1/p^{(s-1)v}}{1-1/p^{s-1}} +\frac{p^v}{p^{vs}} \frac{1}{1-1/p^s}\right).$$ which is in turn $$\zeta(s) \prod_p \left(\frac{1-1/p^{(s-1)v}}{1-1/p^{s-1}} (1-1/p^s) +\frac{1}{p^{v(s-1)}}\right).$$

It turns out that re-expansion is the right thing to do at this step. We get $$\zeta(s) \prod_p \left(1-\frac{1}{p^s}+\frac{p}{p^s}-\frac{p}{p^{2s}} +\frac{p^2}{p^{2s}}-\frac{p^2}{p^{3s}} + \cdots + \frac{p^{v-1}}{p^{s(v-1)}} -\frac{p^{v-1}}{p^{sv}} +\frac{p^v}{p^{sv}}\right).$$ This yields $$\zeta(s) \prod_p \left(1+\frac{p-1}{p^s}+\frac{p^2-p}{p^{2s}} +\frac{p^3-p^2}{p^{3s}}+\cdots +\frac{p^v-p^{v-1}}{p^{vs}}\right).$$

We recognize a truncated Euler product for the Euler totient function which is $$\zeta(s)\sum_{d|n} \frac{\varphi(d)}{d^s}.$$ Now recall that we actually need $n^s L(s)$ which is $$\zeta(s)\sum_{d|n} \varphi(d)\left(\frac{n}{d}\right)^s.$$ To conclude observe that $$\sum_{k=1}^n \gcd(n,k)^s = \sum_{d|n} d^s \varphi\left(\frac{n}{d}\right)$$ because $\gcd(n,k)=d$ implies $\gcd(n/d,k/d)=1$ and there are $\varphi \left(\frac{n}{d}\right)$ such values.

Nice identity I have to say.

Addendum as per the comment. Replacing Euler's totient $\varphi(d)$ with Jordan's totient $J_l(d)$ we obtain the generalisation (with $l$ being a positive integer) $$\sum_{k=1}^n \gcd(n,k)^l \zeta\left(s,\frac{k}{n}\right) = \zeta(s) \times \sum_{\mathbb{q}\in \{1,\ldots,n\}^l} \gcd(n,\mathbb{q})^s$$ where $\{1,\ldots,n\}^l$ denotes the set of $l$-tuples with constituents between $1$ and $n.$