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.
  $1-1+1-1+1-1+-\mathrm{\dots }$ diverges but $(1-1)+(1-1)+(1-1)+\mathrm{\dots }$ 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

  $(a_1+\mathrm{\dots }+a_r)+(a_{r+1}+\mathrm{\dots }+a_{2r})+(a_{2r+1}+\mathrm{\dots }+a_{3r})+\mathrm{\dots }$(1)

  converges and

  $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\mathrm{\hspace{0.33em}0},$(2)

  then also the series

  $a_1+a_2+a_3\mathrm{\dots }$(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  .  The partial sum of (3) may be written

  $$a_1+\mathrm{\dots }+a_n=\underset{s}{\underbrace{(a_1+\mathrm{\dots }+a_{kr})}}+\underset{s^{\prime }}{\underbrace{(a_{kr+1}+\mathrm{\dots }+a_n)}}.$$

  When  $n\to \mathrm{\infty }$, 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.