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adding and removing parentheses in series

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 parent entry.)

A divergent series can become convergent if one adds an infinite amount of parentheses; e.g.
$11+11+11+\ldots$ diverges but $(11)+(11)+(11)+\ldots$ converges. 
A convergent series can become divergent if one removes an infinite amount of parentheses; cf. the preceding example.

If a series contains 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 $kr<n\leqq(k\!+\!1)r$. 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.
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