Inserting and Removing Parentheses in Series of Real Numbers

# Inserting and Removing Parentheses in Series of Real Numbers

 Definition: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be a series and let $p : \mathbb{N} \to \mathbb{N}$ be a strictly increasing function. Let $\displaystyle{b_n = \sum_{k=p(n)+1}^{p(n + 1)} a_k}$ where we denote $p(0) = 1$. Then the series $\displaystyle{\sum_{n=1}^{\infty} b_n}$ is said to result from Inserting Parentheses into $\displaystyle{\sum_{n=1}^{\infty} a_n}$, and $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is said to result from Removing Parentheses from $\displaystyle{\sum_{n=1}^{\infty} b_n}$.

While the definition above seems rather wordy, the intuition behind it is clear. Let $p : \mathbb{N} \to \mathbb{N}$ be a strictly increasing function. Denote the term $b_1$ to be the sum of the first $p(1)$ terms of $(a_n)_{n=1}^{\infty}$. Then denote the term $b_2$ to be the sum of the terms of the sequence $(a_n)_{n=p(1) + 1)}^{p(n + 1)}$. More generally we have that:

(1)
\begin{align} \quad \sum_{n=1}^{\infty} a_n = \underbrace{(a_1 + a_2 + … + a_{p(1)})}_{b_1} + \underbrace{(a_{p(1) + 1} + a_{p(1) + 2} + … + a_{p(2)})}_{b_2} + … + \underbrace{(a_{p(n - 1) + 1} + a_{p(n - 1)+2} + … + a_{p(n)})}_{b_n} + ... \end{align}

In the following theorem we will see that if $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is a convergent series then any series $\displaystyle{\sum_{n=1}^{\infty} b_n}$ obtained from it by inserting parentheses is also convergent to the same sum!

 Theorem 1: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be a series of real numbers that converges to $s$. If $\displaystyle{\sum_{n=1}^{\infty} b_n}$ is a series obtained by inserting parentheses into $\displaystyle{\sum_{n=1}^{\infty} a_n}$ then $\displaystyle{\sum_{n=1}^{\infty} b_n}$ also converges to $s$.
• Proof: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converge to $s$. Then by Cauchy’s condition, for all $\epsilon > 0$ there exists an $N_0 \in \mathbb{N}$ such that if $n \geq N_0$ then for all $p \in \mathbb{N}$:
(2)
\begin{align} \quad \mid a_{n+1} + a_{n+2} + … + a_{n+p} \mid < \epsilon \end{align}
• Take $N \in \mathbb{N}$ such that $N \geq N_0$. Then for all $n \geq N$ we have that $a_{p(n) + 1} \geq a_{n+1}$ since $p$ is a strictly increasing function. So if $n \geq N$ and for all $k \in \mathbb{N}$ we see that:
(3)
\begin{align} \quad \mid b_{n+1} + b_{n+2} + … + b_{n+k} \mid = \mid (a_{p(n) + 1} + … + a_{p(n+1)}) + (a_{p(n+1) + 1} + … + a_{p(n+2)}) + … + (a_{p(n + k - 1) + 1} + … + a_{p(n + k)})] < \epsilon \end{align}
• So, Cauchy’s condition is satisfied for the series $\displaystyle{\sum_{n=1}^{\infty} b_n}$, so this series $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges.
• All that remains to show is that $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges to $s$. Let $(s_n)_{n=1}^{\infty}$ be the sequence of partial sums of $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and let $(s_n’)_{n=1}^{\infty}$ be the sequence of partial sums of $\displaystyle{\sum_{n=1}^{\infty} b_n}$. Then $(s_n)_{n=1}^{\infty}$ converges to $s$, and since $(s_n’)_{n=1}^{\infty}$ is a subsequence of $(s_n)_{n=1}^{\infty}$ we must have that $(s_n’)_{n=1}^{\infty}$ also converges to $s$. Hence:
(4)
\begin{align} \quad \sum_{n=1}^{\infty} b_n = s \end{align}

It is very important to note that the converse of Theorem 1 is not true in general, i.e., if $\displaystyle{\sum_{n=1}^{\infty} b_n}$ is a series obtained by inserting parentheses into $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges, then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ may converge or diverge.

An easy example of this is the following series:

(5)
\begin{align} \quad \sum_{n=1}^{\infty} a_n = \displaystyle{\sum_{n=1}^{\infty} (-1)^n} \end{align}

This series clearly converges. If we insert parentheses by letting $b_n = a_{2n-1} + a_{2n}$, then for all $n \in \mathbb{N}$ we have that $b_n = (-1)^{2n - 1} + (-1)^{2n} = -1 + 1 + 0$, and so the following series obtained by inserting parentheses is convergent:

(6)
\begin{align} \quad \sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} = 0 \end{align}