Exercise 2.26

Proof. Without any consideration of convergence :

(a)

$$\begin{array}{llll}\hfill \zeta (s)\underset{m=1}{\overset{\infty }{\sum }}\frac{\mu (m)}{m^s}& =\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n^s}\underset{m=1}{\overset{\infty }{\sum }}\frac{\mu (m)}{m^s}& \hfill & \\ \hfill & =\underset{n,m\ge 1}{\sum }\frac{\mu (m)}{n^s{m}^s}& \hfill & \\ \hfill & =\underset{u=1}{\overset{\infty }{\sum }}\underset{m\mid u}{\sum }\mu (m)\frac{1}{u^s}(u=\mathit{nm})& \hfill & \\ \hfill & =\underset{u=1}{\overset{\infty }{\sum }}\frac{1}{u^s}\underset{m\mid u}{\sum }\mu (m)& \hfill & \\ \hfill & =1& \hfill & \end{array}$$

Indeed, ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{m\mid u}\mu (m)=1$ if $u=1$, $0$ otherwise. So

$$\zeta {(s)}^{-1}=\underset{n\in {\mathbb{N}}^{\text{∗}}}{\sum }\mu (n)∕n^s.$$

(b)

$$\begin{array}{llll}\hfill \zeta {(s)}^2& =\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n^s}\underset{m=1}{\overset{\infty }{\sum }}\frac{1}{m^s}& \hfill & \\ \hfill & =\underset{n,m\ge 1}{\sum }\frac{1}{{(\mathit{nm})}^s}& \hfill & \\ \hfill & =\underset{u\ge 1}{\sum }\underset{n\mid u}{\sum }\frac{1}{u^s}& \hfill & \\ \hfill & =\underset{u\ge 1}{\sum }\frac{1}{u^s}\underset{n\mid u}{\sum }1& \hfill & \\ \hfill & =\underset{u\ge 1}{\sum }\frac{1}{u^s}\nu (u)& \hfill & \end{array}$$

So

$$\zeta {(s)}^2=\underset{n=1}{\overset{\infty }{\sum }}\frac{\nu (n)}{n^s}.$$

(c)

For $\mathrm{Re}(s)>2$, $$\begin{array}{llll}\hfill \zeta (s)\zeta (s-1)& =\underset{n\ge 1}{\sum }\frac{1}{n^s}\underset{m\ge 1}{\sum }\frac{1}{m^{s-1}}& \hfill & \\ \hfill & =\underset{m,n\ge 1}{\sum }\frac{m}{{(\mathit{nm})}^s}& \hfill & \\ \hfill & =\underset{u\ge 1}{\sum }\left(\underset{m\mid u}{\sum }m\right)\frac{1}{u^s}& \hfill & \\ \hfill & =\underset{u\ge 1}{\sum }\frac{\sigma (u)}{u^s}& \hfill & \end{array}$$

So

$$\zeta (s)\zeta (s-1)=\underset{n\ge 1}{\sum }\frac{\sigma (n)}{n^s}.$$