If we know that $W_m(k)$ is multiplicative over $k$ (if $\gcd(k_1,k_2)=1$ then $W_m(k_1k_2)=W_m(k_1)W_m(k_2)$), and $W_m(k)=0$ if $p^2|k$, and that $|W_m(k)|\le c\cdot \frac{m}{k^{3/2}}$. So that the following series $$S(m):=\sum_{k\ge 1} W_m(k)$$ is absolute convergence. I have read that for the absolute convergence series, we may rearrange terms in the sum and obtain the same convergence. This means we can rearrange it and group as a product like this immediately? $$S(m)=\prod_{\text{ prime }p}(1+W_m(p)).$$
Furthermore, if the above equals $$\prod_{p|m}\left(1+\dfrac{1}{p-1}\right)\prod_{p\not| \text{ }m}\left(1-\dfrac{1}{(p-1)^2}\right)$$ how can we certain to multiply and rearrange terms in the product, like below? Or is it the consequence of being absolute convergence as well? I mean that the above becomes, if $2|m$, (by the book) $$2\prod_{\substack{p|m\\p>2}}\dfrac{1+\dfrac{1}{p-1}}{1-\dfrac{1}{(p-1)^2}}\prod_{p>2}\left(1-\dfrac{1}{(p-1)^2}\right)$$