The Product of Two Series of Real Numbers

# The Product of Two Series of Real Numbers

Consider two series of real numbers, $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$. Then we can consider the product of these two series:

(1)
\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 + ...)(b_0 + b_1 + b_2 + ... ) \end{align}

We can organize the terms of the expanding the product on the righthand side above in the following array:

(2)
\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}

It is not clear how exactly we should sum up the terms in this array since the original multiplication involved multiplication of infinite sums. One such way is defined (generically) below.

 Definition: Let $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ be two series of real numbers. The Product of these two series denoted $\displaystyle{\left ( \sum_{n=0}^{\infty} a_n \right ) \left ( \sum_{n=0}^{\infty} b_n \right )}$ is given by the partial sum sequence $\displaystyle{S_n = \sum_{i=0}^{n} \left ( \sum_{j=0}^{n} a_ib_j \right ) = \left ( \sum_{i=0}^{n} a_i \right ) \left ( \sum_{j=0}^{n} b_j \right )}$.

This type of series multiplication tells us exactly how the add the terms in the array above. $S_0$ is given by summing up the red terms below:

(3)
\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}

$S_1$ is given by summing up the orange terms below:

(4)
\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} & \color{orange}{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}

$S_2$ is given by summing up the yellow terms below:

(5)
\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} & \color{gold}{a_1b_2} & \cdots & a_1b_n & \cdots \\ \color{gold}{a_2b_0} & \color{gold}{a_2b_1} & \color{gold}{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, $S_n$ is given by summing up all terms in the $n + 1$ by $n+1$ top-left subarray above.

We will now look at a nice theorem which tells us if the two series in the product converge to $A$ and $B$, then the product of the two series will also converge to the product, $AB$.

 Theorem 1: If $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ converge to $A$ and $B$ respectively, then the product $\displaystyle{\left ( \sum_{n=0}^{\infty} a_n \right ) \left ( \sum_{n=0}^{\infty} b_n \right )}$ converges to $AB$.
• Proof: Let $(A_n)_{n=0}^{\infty}$ and $(B_n)_{n=0}^{\infty}$ be the sequences of partial sums for $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ respectively. If $S_n$ is the sequence of partial sums for the product of these two series, then:
(6)
\begin{align} \quad S_n = \left ( \sum_{i=0}^{n} a_i \right ) \left ( \sum_{j=0}^{n} b_j \right ) = A_nB_n \end{align}
• Since the sequences $A_n$ and $B_n$ converge to $A$ and $B$ as $n \to \infty$, we have that $S_n$ converges to $AB$ as $n \to \infty$. Therefore:
(7)
\begin{align} \quad \left ( \sum_{n=0}^{\infty} a_n \right ) \left ( \sum_{n=0}^{\infty} b_n \right ) = AB \quad \blacksquare \end{align}