\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \)
deepdream of
          a sidewalk
Show Question
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \)
Math and science::Analysis::Tao::07. Series

Rearrangement of infinite series

A feature of finite series which we will recap here is that any rearrangement of the terms of the series does not affect the sum. For example:

\[ a_1 + a_2 + a_3 + a_4 = a_4 + a_1 + a_3 + a_2 \]

This comes from the first property of substitution:

If \( X \) is a finite set, \( f: X \rightarrow R \) is a function, and \( g : Y \rightarrow X \) is a bijection, then:

\[ \sum_{x \in X}f(x) = \sum_{y \in Y}f(g(y)) \]

If we consider any bijection \( g \) from-to the same set \( \{ i \in \mathbb{Z} : n \le i \le m \} \), then we can say:

\[ \sum_{i=n}^{m} a_i = \sum_{i=n}^{m} a_{g(i)} \]

which is the basis for the rearrangement example above.

Can we rearrange the terms of an infinite series and get the same result? Yes and no.

  • An absolutely convergent series: any rearrangement of terms produces the same sum.
  • Conditionally, but not absolutely convergent series: the terms can be arranged to make the series converge to any value, or to diverge.

In summary: rearrangement of absolute convergent series is safe, rearrangement of a non-absolutely convergent series is risky. The latter is due to being able to manipulate the occurances of positives and negatives which, in the original series, are balancing each other out in order to achieve convergence.

The full proposition for absolutely convergent series is as follows.

Rearrangement of infinite series

Let \( \sum_{n=0}^{\infty}a_n \) be an absolutely convergent series of real numbers and let \( f: N \rightarrow N \) be a bijection. Then \( \sum_{m=0}^{\infty} a_{f(m)} \) is also absolutely convergent, and has the same sum:

\[ \sum_{n=0}^{\infty} a_n = \sum_{n=0}^{\infty} a_{f(m)} \]