Problem with proof that positive infinite series are commuative

Proof from real analysis book:

Let $\sum_{n=0}^{\infty}a_n$ converge, where $a_n \geq 0, n \in \mathbb{N}$.

Then the series

$$\sum_{k=1}^{\infty}a'_k = a'_1 + a'_2 + \cdots + a'_k + \cdots$$

Is constructed by rearranging the elements of the original series $$a'_1 = a_{n_1}, \ a'_2 = a_{n_2}, \ \ldots, \ a'_k = a_{n_k}, \ \ldots$$ where $(n_1,n_2,\ldots,n_k,\ldots)$ is a permutation of the set $\mathbb{N}$.

We must now prove that $\sum_{k=1}^{\infty}a'_k$ converges and has the same sum as the original series.

Let $s_n = \sum_{k=0}^{n}a_k$, $s'_k = \sum_{i=1}^{k}a'_i$ and $\lim_{n \to \infty}s_n = s$.

If for some $k\in\mathbb{N}$ $$max\{n_1,n_2,\ldots,n_k\} = n_0$$

then $s'_k \leq s_{n_0}$, it follows that $s'_k \leq s$ for all $k\in \mathbb{N}$. So the sequence $(s'_k)_{k\in\mathbb{N}}$ is bounded and increasing so it converges, $\lim_{k \to \infty}s'_k = s'$. It follows that $s' \leq s$.

The following part of the proof where the author proves that $s' \geq s$ confuses me:

Because the series $\sum_{n=0}^{\infty}a_n$ can be constructed from the series $\sum_{k=1}^{\infty}a'_k$ by rearranging it's elements, it follows that $s' \geq s$.

If we can use this kind of "rearrangement argument" couldn't we use it both ways? i.e. Because $\sum_{k=1}^{\infty}a'_k$ is constructed by rearranging elements of $\sum_{n=0}^{\infty}a_n$ it follows that $s'_k \leq s$.

If this argument could be used it would lead to a false conclusion that conditionally convergent series are also commutative.

• The 'following part' that confuses you is just a symmetric version of the initial part - consider swapping the roles of the $a_i$ and the $a'_i$. – Steven Stadnicki Jul 26 '14 at 8:07

Theorem: Let $a_i\ge0$ for any $i\in {\mathbb N}$ and $s_n = \sum_{i=1}^na_i$ for any $n\in {\mathbb N}$. If $s_i$s are convergent to $s\in{\mathbb R}$ then for any permutation $\pi :{\mathbb N}\rightarrow {\mathbb N}$, the series related to $a_{\pi(i)}$ is also convergent to $s$.
Proof: Let $t_m = \sum_{i=1}^ma_{\pi(i)}$, $p_m=\max\{ \pi(1),\ldots,\pi(m)\}$ and $q_m=\max\{ \pi^{-1}(1), \ldots,\pi^{-1}(m)\}$. Then one can easily see $t_m \le s_{p_m}\le s$ and $s_n \le t_{q_n}$. Thus as $\{t_m\}_{m\in {\mathbb N}}$ is increasing and bounded it has some limit $t$ such that $t_m\le t$. Therefore $s_n\le t$.
Having $s_n\le t$ and $t_m\le s$ for any $n,m\in{\mathbb N}$, the result is straightforward.
The argument is symmetric in the $a_k$ series and the $a_k'$ series. The point is that both of these series are series of positive elements, and either can be seen as the other with their elements rearranged.