TITLE: Alternate proof of non negative infinite series rearrangement? Tao Proposition 7.4.1.
QUESTION [1 upvotes]: Proposition 7.4.1. Let $\sum_{n=0}^{\infty} a_{n} $ be a convergent series of non-negative real numbers, and let $f:\mathbb N \rightarrow\mathbb N$ be a bijection. Then $\sum_{m=0}^{\infty} a_{f(m)} $ is also convergent, and has the same sum: $\sum_{n=0}^{\infty} a_{n} = \sum_{m=0}^{\infty} a_{f(m)} $

Tao's start of the proof:

We introduce the partial sums $S_N := \sum_{n=0}^{N} a_{n}$ and $T_M := \sum_{m=0}^{M} a_{f(m)}$. We know that the sequences $(S_N)_{n=0}^{\infty}$ and $(T_M)_{m=0}^{\infty}$ are increasing. Write $L := sup (S_N)_{n=0}^{\infty}$ and $L' := sup (T_M)_{m=0}^{\infty}$ by Proposition 6.3.8 we know that $L$ is finite and in fact $L = \sum_{n=0}^{\infty} a_{n} $ we will now show that $L = L'$

Tao then proceeds to show $L \leq L'$ and $L' \leq L$ what i am gonna do is suppose that $L$ does not equal $L'$ and suppose without loss of generality $L' > L$ this means there exists some $M' \geq 0$ such that $L' \geq T_{M'} > L$
Now we have already shown on a previous chapter that:

$\sum_{i=m}^{n} a_{i} = \sum_{i=m}^{n} a_{f(i)}$ for any bijection $f$ from the set $\{i \in \mathbb Z : m \leq i \leq n\}$ to itself.

using this we then get $T_{M'} = \sum_{m=0}^{M'} a_{f(m)} = \sum_{n=0}^{M'} a_{n} = S_{M'}$ which is true since $f:\mathbb N \rightarrow\mathbb N$ is a bijection and so $f:\{i \in \mathbb N : 0 \leq i \leq M'\} \rightarrow \{i \in \mathbb N : 0 \leq i \leq M'\}$ is a bijection   and so $S_{M'} > L$ which is absurd since $S_{M'} \leq L$ for all $M \geq 0$ and we are done. This seems much simpler than what Tao does but is something flawed with it?

REPLY [0 votes]: Your claim is false. Consider the function $f(1) = 2, f(2) = 1, f(3) = 4, f(4) = 3,$ i.e. I swap the even an odd numbers. Then if $M' = 3,$ you find that $f(\{1, 2, 3\}) = \{2, 1, 4\}$ and not $\{1,2,3\}.$
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