adding and removing parentheses in series

We consider series with real or complex terms.

• If one groups the terms of a convergent series by adding parentheses but not changing the order of the terms, the series remains convergent and its sum the same. (See theorem 3 of the http://planetmath.org/node/6517parent entry.)

• A divergent series can become convergent if one adds an infinite amount of parentheses; e.g.
$1-1+1-1+1-1+-\ldots$ diverges but $(1-1)+(1-1)+(1-1)+\ldots$ converges.

• A convergent series can become divergent if one removes an infinite amount of parentheses; cf. the preceding example.

• If a series parentheses, they can be removed if the obtained series converges; in this case also the original series converges and both series have the same sum.

• If the series

 $\displaystyle(a_{1}+\ldots+a_{r})+(a_{r+1}+\ldots+a_{2r})+(a_{2r+1}+\ldots+a_{% 3r})+\ldots$ (1)

converges and

 $\displaystyle\lim_{n\to\infty}a_{n}\;=\;0,$ (2)

then also the series

 $\displaystyle a_{1}+a_{2}+a_{3}\ldots$ (3)

converges and has the same sum as (1).

Proof.  Let $S$ be the sum of the (1).  Then for each positive integer $n$, there exists an integer $k$ such that  $kr.  The partial sum of (3) may be written

 $a_{1}+\ldots+a_{n}\;=\;\underbrace{(a_{1}+\ldots+a_{kr})}_{s}+\underbrace{(a_{% kr+1}+\ldots+a_{n})}_{s^{\prime}}.$

When  $n\to\infty$, we have

 $s\to S$

by the convergence of (1) to $S$, and

 $s^{\prime}\to 0$

by the condition (2).  Therefore the whole partial sum will tend to $S$, Q.E.D.

Note.  The parenthesis expressions in (1) need not be “equally long” — it suffices that their lengths are under an finite bound.

Title adding and removing parentheses in series AddingAndRemovingParenthesesInSeries 2013-03-22 18:54:09 2013-03-22 18:54:09 pahio (2872) pahio (2872) 13 pahio (2872) Topic msc 40A05 EmptySum