Zeta question - prime zeta. Basic calculus In one of the steps to get the prime zeta
$$\log \zeta(s) = \sum_{p \in P} p^{-s} + \sum_{p \in P}\sum_{n \geq 2} \frac{p^{-sn}}{n} $$
$$\log \zeta(s) = \sum_{p \in P} p^{-s} + \sum_{n \geq 2} \frac{1}{n}\sum_{p \in P} p^{-sn}$$
Why can we factor the $1/n$ like that? As far as I can tell, the $n$ is varying in the first sum and it is an infinite sum. 
 A: It is a rather well known result that if you have a double sequence $a_{ij}$ of nonnegative terms $$\sum_{j\geqslant 0}\sum_{i \geqslant 0}a_{ij}=\sum_{i \geqslant  0}\sum_{j \geqslant 0}a_{ij}$$
(even in the $+\infty=+\infty$ sense). If any of the two converge, then so does the other, and their sum is equal.
ADD Today I was given a proof of this. I share it here with you. Recall first that if $x_n$ is an increasing sequence, then even in the extended sense $\lim\limits_{n\to\infty}x_n=\sup\limits_{n\geqslant 1}x_n$. Thus, it follows that for each fixed $i$, $$b_i=\sum_{j\geqslant 1}a_{ij}=\sup_{n_i\geqslant 1}\sum_{j=1}^{n_i}a_{ij}$$
Now, set $s_{m,n}=\sum_{i=1}^m\sum_{j=1}^n a_{ij}$ and $s=\sup_{ m,n\geqslant 1}s_{m,n}$. Since $$s_{m,n}\leqslant \sum_{i=1}^m\sum_{j=1}^\infty\leqslant \sum_{i \geqslant  1}\sum_{j \geqslant 1}a_{ij}=t$$ we have $$s\leqslant t$$
Now pick $t'<t$. Then there must exist $m$ such that $$\sum_{i=1}^m\sum_{j=1}^\infty a_{ij}>t'$$
But as before, for each $i=1,2,\ldots,m$, we have that $$\sum_{j=1}^\infty a_{ij}=\sup_{n_i\geqslant 1}\sum_{j=1}^{n_i}a_{ij}$$
Since taking the supremum always commutes with summing, $$\sum_{i=1}^m\sup_{n_i\geqslant 1}\sum_{j=1}^{n_i}a_{ij}=\sup_{n_1,\ldots,n_m\geqslant 1}\sum_{i=1}^m\sum_{j=1}^{n_i}a_{ij}>t'$$
But this means there must exist an $m$-uple $(k_1,\ldots,k_m)$ such that $$\sum_{i=1}^{m}\sum_{j=1}^{k_i} a_{ij}>t'$$
But if $n=\max k_i$ we have $$\sum_{i=1}^m\sum_{j=1}^n a_{ij}>t'$$ so that $$s>t'$$
But $t'$ was arbitrary, so $s\geqslant t$, $s=t$. By symmetry, the claim follows. $\blacktriangle$
