The Cauchy Product of Two Series of Real Numbers

# The Cauchy Product of Two Series of Real Numbers

Recall from The Product of Two Series of Real Numbers page that if $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ are two series of real numbers then we can define the product of these series $\displaystyle{\left ( \sum_{n=0}^{\infty} a_n \right ) \left ( \sum_{n=0}^{\infty} b_n \right )}$ to have the following sequence of partial sums:

(1)
\begin{align} \quad S_n = \left ( \sum_{i=0}^{n} a_i \right ) \left ( \sum_{j=0}^{n} b_j \right ) \end{align}

Of course, there are other ways in which we can add the terms in the expanded product of multiplying the series $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ together.

Recall that from expanding the infinite products directly we have that:

(2)
\begin{align} \quad \left ( \sum_{n=0}^{\infty} a_n \right ) \left ( \sum_{n=0}^{\infty} b_n \right ) = (a_0 + a_1 + a_2 + ... + a_n + ...)(b_0 + b_1 + b_2 + ... + b_n + ...) \end{align}

And we can organize the terms in this expansion in the following array:

(3)
\begin{align} \quad \begin{matrix} a_0b_0 & a_0b_1 & a_0b_2 & \cdots & a_0b_n & \cdots \\ a_1b_0 & a_1b_1 & a_1b_2 & \cdots & a_1b_n & \cdots \\ a_2b_0 & a_2b_1 & a_2b_2 & \cdots & a_2b_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \\ a_nb_0 & a_nb_1 & a_nb_2 & \cdots & a_nb_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \end{matrix} \end{align}

Another way that we can multiply two series together (to tell us how to add the terms above) is to consider what is called the Cauchy product of two series which we define below.

 Definition: Let $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ be two series. The Cauchy Product of these two series is defined as the sum $\displaystyle{\sum_{n=1}^{\infty} c_n}$ where $\displaystyle{c_n = \sum_{k=0}^{n} a_kb_{n-k}}$ for all $n \in \{ 0, 1, 2, ... \}$.

While the Cauchy product may seem rather abstract - it is actually a very nice way to sum of the terms of multiplying the series $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ together. If $(C_n)_{n=1}^{\infty}$ denote the partial sums of $(c_n)_{n=1}^{\infty}$ we we have that $C_0$ is given by summing all of the red terms together:

(4)
\begin{align} \quad \begin{matrix} \color{red}{a_0b_0} & a_0b_1 & a_0b_2 & \cdots & a_0b_n & \cdots \\ a_1b_0 & a_1b_1 & a_1b_2 & \cdots & a_1b_n & \cdots \\ a_2b_0 & a_2b_1 & a_2b_2 & \cdots & a_2b_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \\ a_nb_0 & a_nb_1 & a_nb_2 & \cdots & a_nb_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \end{matrix} \end{align}

$C_1$ is given by summing all of the orange terms together:

(5)
\begin{align} \quad \begin{matrix} \color{orange}{a_0b_0} & \color{orange}{a_0b_1} & a_0b_2 & \cdots & a_0b_n & \cdots \\ \color{orange}{a_1b_0} & a_1b_1 & a_1b_2 & \cdots & a_1b_n & \cdots \\ a_2b_0 & a_2b_1 & a_2b_2 & \cdots & a_2b_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \\ a_nb_0 & a_nb_1 & a_nb_2 & \cdots & a_nb_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \end{matrix} \end{align}

$C_2$ is given by summing all of the yellow terms together:

(6)
\begin{align} \quad \begin{matrix} \color{gold}{a_0b_0} & \color{gold}{a_0b_1} & \color{gold}{a_0b_2} & \cdots & a_0b_n & \cdots \\ \color{gold}{a_1b_0} & \color{gold}{a_1b_1} & a_1b_2 & \cdots & a_1b_n & \cdots \\ \color{gold}{a_2b_0} & a_2b_1 & a_2b_2 & \cdots & a_2b_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \\ a_nb_0 & a_nb_1 & a_nb_2 & \cdots & a_nb_n & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \ddots \end{matrix} \end{align}

In general, $C_n$ is obtained by summing the $0^{\mathrm{th}}$, $1^{\mathrm{st}}$, …, and $n^{\mathrm{th}}$ diagonals in the array given above.

Let's now look at an example of expressing the Cauchy product of two series. For example, consider the following series of real numbers:

(7)
\begin{align} \quad \displaystyle{\sum_{n=0}^{\infty} \left ( \frac{1}{2} \right )^n} \end{align}

Suppose that we want to take the Cauchy product of the series above with itself, i.e., take $\displaystyle{\sum_{n=0}^{\infty} \left ( \frac{1}{2} \right )^n} \cdot \displaystyle{\sum_{n=0}^{\infty} \left ( \frac{1}{2} \right )^n}$. Then the sequence $c_n$ is defined as:

(8)
\begin{align} \quad c_n &= \sum_{k=0}^{n} \left ( \frac{1}{2} \right )^k \cdot \left ( \frac{1}{2} \right )^{n - k} \\ \quad c_n &= \sum_{k=0}^{n} \left ( \frac{1}{2} \right )^n \\ \quad c_n &= (n + 1) \left ( \frac{1}{2} \right )^n \end{align}

Then the corresponding Cauchy product is:

(9)
\begin{align} \quad \sum_{n=0}^{\infty} c_n = \sum_{n=0}^{\infty} (n + 1) \left ( \frac{1}{2} \right )^n \end{align}

It turns out the this Cauchy product indeed converges, however, the Cauchy product of two convergent series is not always convergent. We will investigate this more thoroughly soon.