Q:

Interchanging summands among infinitely many infinite series

I am reading the following lecture notes concerning analytic number theory:
http://www.math.uiuc.edu/~hildebr/ant/main4.pdf
On the pages 111/112 the partial product $P_N(s)$ is defined. Then some mathematical conversions are made, that I don't understand. Note that $s\in \mathbb{C}$. Why is it allowed to interchange the arrangement of the summands in the infinite series not only in each series separately, but also among  them?
It is known that each series is absolutely convergent, but I don't know, why it is allowed to interchange the arrangement of the summands even among the different series. Any references are welcome!
Are there additional problems, when letting $N\rightarrow \infty$ ? It is known that the infinite product converges absolutely, but is it allowed to interchange the arrangement of the summands even among the different series in an INFINITE product?
Thanks for the help.

A:

When you work with series which converges absolutely, you can (essentialy) move as if you have a finite series. So in particular if we have a finite series holds $$\sum_{n_{1}\leq N}a_{n_{1}}\sum_{n_{2}\leq N}a_{n_{2}}=\sum_{n_{1}\leq N}\sum_{n_{2}\leq N}a_{n_{1}}a_{n_{2}}.$$
 Using your text's notation, you have $$P_{N}\left(s\right)=\prod_{p\leq N}\left(1+\sum_{m\geq1}\frac{f\left(p^{m}\right)}{p^{ms}}\right)=\prod_{p\leq N}\left(\sum_{m\geq0}\frac{f\left(p^{m}\right)}{p^{ms}}\right)=\sum_{m_{1}\geq0}\frac{f\left(p_{1}^{m_{1}}\right)}{p_{1}^{m_{1}s}}\cdots\sum_{m_{k}\geq0}\frac{f\left(p_{k}^{m_{k}}\right)}{p_{k}^{m_{k}s}}$$
 and, for the hypothesis, each of these series converges absloutely. So you can rearrange (using multiplicativity) as$$\sum_{m_{1}\geq0}\dots\sum_{m_{k}\geq0}\frac{f\left(p_{1}^{m_{1}}\cdots p_{k}^{m_{k}}\right)}{p_{1}^{m_{1}}\cdots p_{k}^{m_{k}}}=\sum_{n\in A\left(N\right)}\frac{f\left(n\right)}{n^{s}}.$$
 For the other question, no, you haven't problems with $N\rightarrow\infty$
 , you now work with the identity$$P_{N}\left(s\right)=\sum_{n\in A\left(N\right)}\frac{f\left(n\right)}{n^{s}}$$
 so now we have no rearrangement, only this identity.

