Return to Revisions

I think all your questions are answered by the following calculation (assume $m\geq 1$ and $\Re(s)>1$): $$ \sum_{c=1}^\infty\frac{r_m(c)}{c^{2s}} = \sum_{c=1}^\infty\frac{1}{c^{2s}}\sum_{\substack{\text{$d$ mod $c$}\\{(d,c)=1}}}e\left(m\frac{d}{c}\right) = \sum_{c=1}^\infty\frac{1}{c^{2s}}\sum_{\text{$d$ mod $c$}}e\left(m\frac{d}{c}\right)\sum_{n\mid(c,d)}\mu(n) $$ $$ = \sum_{n=1}^\infty\mu(n)\sum_{\substack{c=1\\{n\mid c}}}^\infty\frac{1}{c^{2s}}\sum_{\substack{\text{$d$ mod $c$}\\{n\mid d}}}e\left(m\frac{d}{c}\right) = \sum_{n=1}^\infty\mu(n)\sum_{c=1}^\infty\frac{1}{(nc)^{2s}}\sum_{\text{$nd$ mod $nc$}}e\left(m\frac{nd}{nc}\right)$$ $$ = \sum_{n=1}^\infty\frac{\mu(n)}{n^{2s}}\sum_{c=1}^\infty\frac{1}{c^{2s}}\sum_{\text{$d$ mod $c$}}e\left(m\frac{d}{c}\right) = \sum_{n=1}^\infty\frac{\mu(n)}{n^{2s}}\sum_{c\mid m}\frac{c}{c^{2s}} = \frac{\sigma_{1-2s}(m)}{\zeta(2s)}.$$