TITLE: Why does infinite sum of finite summands converge in this proof $\sum 1/p$ converge?
QUESTION [5 upvotes]: In this proof from  the university of Warwick (pdf) there is a step which assumes the infinite sum of finite summands converges.
Why is this?
I can see why each summand $\sum_p1/p^s$ for $s \geq 2$ converges, but not why an infinite sum of these.

REPLY [2 votes]: That is a good question that is not obvious (at least in my mind). Luckily, it is not to difficult to show something even stronger:
$$\sum_{s=2}^\infty \sum_{p\in\mathbb{P}}\frac{1}{sp^s}<\sum_{s=2}^\infty \sum_{p\in\mathbb{P}}\frac{1}{p^s}<\sum_{s=2}^\infty \sum_{n=2}^\infty \frac{1}{n^s}$$
Since these terms are all positive, if the series converges it must converge absolutely. Thus, we are free to switch the order of summations without effecting the overall convergence or divergence. Then
$$=\sum_{n=2}^\infty \sum_{s=2}^\infty \frac{1}{n^s}=\sum_{n=2}^\infty \frac{1}{n(n-1)}=1$$
We conclude the overall series is bounded and hence converges.