adding and removing parentheses in series

We consider series with real or complex terms.

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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.)
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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.
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A convergent series can become divergent if one removes an infinite amount of parentheses; cf. the preceding example.
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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.
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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 $kr<n\leqq(k\!+\!1)r$ . 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
Canonical name	AddingAndRemovingParenthesesInSeries
Date of creation	2013-03-22 18:54:09
Last modified on	2013-03-22 18:54:09
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	13
Author	pahio (2872)
Entry type	Topic
Classification	msc 40A05
Related topic	EmptySum