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 parent 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 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

  (1)

  
  converges and

  (2)

  
  then also the series

  (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 $$a_1+\ldots+a_n \;=\; \underbrace{(a_1+\ldots+a_{kr})}_{s}+\underbrace{(a_{kr+1}+\ldots+a_n)}_{s'}.$$ When $n \to \infty$ , we have $$s \to S$$ by the convergence of (1) to $S$ , and $$s' \to 0$$ 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.