0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $a'_i$. $\endgroup$ – Steven Stadnicki Jul 26 '14 at 8:07
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

This does not say anything about conditionally convergent series, since a series of positive numbers can never be conditionally convergent (as convergence is the same thing as absolute convergence here, since the absolute value of a positive number is positive).

$\endgroup$
  • $\begingroup$ Your discussion is not rigorous! $\endgroup$ – Mohammad Khosravi Jul 26 '14 at 9:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.