Rearrangements of Terms in Series of Real Numbers

Definition: Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers and let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be the corresponding series. Furthermore, let $f : \mathbb{N} \to \mathbb{N}$ be a bijection. Then a Rearrangement of $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is the series $\displaystyle{\sum_{n=1}^{\infty} b_n}$ where $b_n = a_{f(n)}$ for all $n \in \mathbb{N}$.

It is very important that for $\displaystyle{\sum_{n=1}^{\infty} b_n}$ to be considered a rearrangement of $\displaystyle{\sum_{n=1}^{\infty} a_n}$, we must have that $f$ is indeed a bijection. Injectivity of $f$ ensures that no term in the sum is counted more than once, while surjectivity of $f$ ensures that we do not miss any term in the rearrangement sum!

For example, consider the function $f : \mathbb{N} \to \mathbb{N}$ defined for all $n \in \mathbb{N}$ by:

(1)

\begin{align} \quad f(n) = \left\{\begin{matrix} n + 1 & \mathrm{if} \: n \: \mathrm{is \: odd.}\\ n - 1 & \mathrm{if} \: n \: \mathrm{is \: even.} \end{matrix}\right. \end{align}

Then $f$ is a bijection. To show this, suppose that $f(x) = f(y)$. If $x$ is odd, then $f(x) = x + 1$. But then $f(y) = x + 1$ which implies that $y$ is odd, and $f(y) = y + 1$. Therefore:

(2)

\begin{align} \quad f(x) & = f(y) \\ \quad x + 1 & = y + 1 \\ \quad x & = y \end{align}

Suppose instead that $x$ is even. Then $f(x) = x - 1$. But then $f(y) = x - 1$ which implies that $y$ is even, and $f(y) = y + 1$. Therefore:

(3)

\begin{align} \quad f(x) & = f(y) \\ \quad x - 1 & = y - 1 \\ \quad x & = y \end{align}

In either case we see that if $f(x) = f(y)$ then $x = y$, so $f$ is injective.

We now show that $f$ is surjective. Let $n \in \mathbb{N}$. Then either $n$ is odd or $n$ is even. If $n$ is odd, then let $m = n + 1$. We see that:

(4)

\begin{align} \quad f(m) = f(n+1) = (n+1) - 1 = n \end{align}

Now suppose that $n$ is even and let $m = n - 1$. Then:

(5)

\begin{align} \quad f(m) = f(n - 1) = (n - 1) + 1 = n \end{align}

So $f$ is surjective. Since $f$ is both injective and surjective we conclude that $f$ is bijective.

Now consider the series $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{2^n}}$. The original series can be expanded as:

(6)

\begin{align} \quad \sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + … \end{align}

The rearrangement of this series defined by the bijection $f$ above is:

(7)

\begin{align} \quad \sum_{n=1}^{\infty} \frac{1}{2^{f(n)}} = \frac{1}{4} + \frac{1}{2} + \frac{1}{16} + \frac{1}{8} + \frac{1}{64} + \frac{1}{32} + … \end{align}

In this particular example, the rearrangement above converges to $1$ as does the original series. However, in general, this is surprisingly not the case! We will look more into this subsequently.