Convergence of Cauchy Products of Two Series of Real Numbers

# Convergence of Cauchy Products of Two Series of Real Numbers

Recall from the The Cauchy 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 define the Cauchy product of these series as:

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

Where $\displaystyle{c_n = \sum_{k=0}^{n} a_kb_{n-k} = a_0b_n + a_1b_{n-1} + ... + a_{n-1}b_1 + a_nb_0}$.

We will now look at some nice results regarding the convergence of the Cauchy product of two series of real numbers.

Theorem 1: Let $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ be two series of real numbers and let $\displaystyle{\sum_{n=0}^{\infty} c_n}$ be their Cauchy product.a) If $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ both converge absolutely to $A$ and $B$ respectively, then $\displaystyle{\sum_{n=0}^{\infty} c_n}$ converges absolutely to $AB$.b) If $\displaystyle{\sum_{n=0}^{\infty} a_n}$ converges absolutely to $A$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ converges conditionally to $B$, then $\displaystyle{\sum_{n=0}^{\infty} c_n}$ converges (not necessarily absolutely) to $AB$. |

**Proof of a)**Suppose that $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ converge absolutely to $A$ and $B$ respectively. Let $(C_n)_{n=0}^{\infty}$ denote the sequence of partial sums of the Cauchy product $\displaystyle{\sum_{n=0}^{\infty} c_n}$, and let $(\mid C_n \mid)_{n=0}^{\infty}$ denote the sequence of partial sums of $\displaystyle{\sum_{n=0}^{\infty} \mid c_n \mid}$. Then:

\begin{align} \quad \mid C_n \mid & = \sum_{k=0}^{n} \mid c_k \mid \\ & \leq \left ( \sum_{k=0}^{\infty} \mid a_k \mid \right ) \left ( \sum_{k=0}^{\infty} \mid b_k \mid \right ) \\ \end{align}

- Since $\displaystyle{\sum_{n=0}^{\infty} a_n}$ converges absolutely, the sequence of partial sums $(\mid A_n \mid)_{n=0}^{\infty}$ is bounded above for all $n \in \{ 0, 1, 2, ... \}$ by some $M_1 \in \mathbb{R}$. Similarly, since $\displaystyle{\sum_{n=0}^{\infty} b_n}$ converges absolutely, the sequence of partial sums $(\mid B_n \mid)_{n=0}^{\infty}$ is bounded above for all $n \in \{ 0, 1, 2, ... \}$ by some $M_2 \in \mathbb{R}$. So for all $n \in \{ 0, 1, 2, ... \}$ we have that:

\begin{align} \quad \mid C_n \mid \leq M_1M_2 \end{align}

- Notice that the sequence of partial sums $(\mid C_n \mid)_{n=0}^{\infty}$ is increasing and bounded above. So $(\mid C_n \mid)_{n=0}^{\infty}$ converges, so $\displaystyle{\sum_{n=0}^{\infty} \mid c_n \mid}$ converges, which shows that the Cauchy product $\displaystyle{\sum_{n=0}^{\infty} c_n}$ converges absolutely.

- Furthermore, since $\displaystyle{\sum_{n=0}^{\infty} c_n}$ is absolutely convergent we must have that any rearrangement of the sum of terms is equal. But the regular product of the two given series converges to $AB$. So $\displaystyle{\sum_{n=0}^{\infty} c_n}$ converges to $AB$. $\blacksquare$

The proof of (b) is more technical, so we will omit the proof for the time being.