uniform convergence of series, equivalent conditions, proof or source Assume $f_n\colon\mathbb{R}\to\mathbb{R}$. TFAE:
(1) the series $\sum |f_n|$ is uniformly convergent
(2) for any permutation $\pi$ of $\mathbb{N}$, the series $\sum f_{\pi(n)}$ is uniformly convergent
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How to prove this or where can I find a complete proof to read?
Thank you in advance.
 A: (1)$\implies$ (2): This is the easy direction. It's very much like the proof that if $a_1,a_2, \dots \ge 0,$ then
$$\sum_{n=1}^{\infty}a_n = \sum_{n=1}^{\infty}a_{\pi(n)}.$$
So I'll leave it to you to prove that (1) implies the uniform convergence of $\sum_{n=1}^{\infty}|f_{\pi(n)}|,$ and hence the uniform convergence of $\sum_{n=1}^{\infty}f_{\pi(n)}.$
(2)$\implies$ (1): Suppose $\sum |f_{n}|$ does not converge uniformly on $\mathbb R.$ Then there exists $\epsilon>0,$ sequences $(m_k),(n_k)$ of integers, and a real sequence $(x_k)$  such that
$$0<m_1<n_1<m_2<n_2 < \cdots \to \infty$$
and
$$\tag 1 \sum_{n=m_k}^{n_k} |f_n(x_k)|\ge \epsilon \text { for all }k.$$
For each $k$ we split $\{m_k,\dots,n_k\}$ into sets of integers $n$ such that $f_n(x_k) >0, f_n(x_k) <0, f_n(x_k)=0.$ Call these sets $A_k,B_k,C_k$ respectively. Then  we can write the sum in $(1)$ as
$$\tag 2 \sum_{n\in A_k} f_n(x_k) - \sum_{n\in B_k} f_n(x_k).$$
It follows that either the first sum in $(2)$ is $\ge \epsilon/2,$ or the second sum in $(2)$ is $\le -\epsilon/2.$
For each $k$ we now reorder the integers in $\{m_k,\dots,n_k\}$ by putting the integers in $A_k$ first, the integers in $B_k$ second, and the integers in $C_k$ last. That defines a a permutation $\pi$ on the union of the sets $\{m_k,\dots,n_k\}.$ On the rest of $\mathbb N,$ simply define $\pi(n)=n.$ Then $\pi$ is a permutation on $\mathbb N.$
The integers in $A_k,B_k$ now occur consecutively in $\sum f_{\pi(n)}.$ Thus by $(2)$ and our "$\epsilon/2$ observation", it's simple to verify that the partial sums of this series are not uniformly Cauchy. Therefore $\sum f_{\pi(n)}$ is not uniformly convergent. That contradicts our assumption, and therefore $\sum |f_{n}|$ converges uniformly as desired.
