The Cauchy Product of Power Series

# The Cauchy Product of Power Series

Recall from Convergence of Cauchy Products 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 both absolutely convergent series that converge to $A$ and $B$ respectively, then the Cauchy product $\displaystyle{\sum_{n=0}^{\infty} c_n}$ converges absolutely to $AB$.
• If $\displaystyle{\sum_{n=0}^{\infty} a_n}$ is absolutely convergent and converges to $A$, while $\displaystyle{\sum_{n=0}^{\infty} b_n}$ is conditionally convergent and converges to $B$, then the Cauchy product $\displaystyle{\sum_{n=0}^{\infty} c_n}$ converges (not necessarily absolutely) to $AB$.

We will now look at a nice consequence of the first part of this theorem with regards to power series.

 Theorem 1: Let $\displaystyle{\sum_{n=0}^{\infty} a_nx^n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_nx^n}$ be two power series that are absolutely convergent and converge to $A(x)$ and $B(x)$ respectively. Then the Cauchy product of these power series is $\displaystyle{\sum_{n=0}^{\infty} c_nx^n}$ and it converges to $A(x)B(x)$.
• Proof: Suppose that $\displaystyle{A(x) = \sum_{n=0}^{\infty} a_nx^n}$ and $\displaystyle{B(x) = \sum_{n=0}^{\infty} b_nx^n}$. Let $\displaystyle{\sum_{n=0}^{\infty} C_n}$ denote the Cauchy product for these two series. Then for each $n \in \mathbb{N}$ we have that:
(1)
\begin{align} \quad C_n = \sum_{k=0}^{n} (a_kx^k)(b_{n-k}x^{n-k}) = \sum_{k=0}^{n} a_kb_{n-k}x^n = x^n \sum_{k=0}^{n} a_kb_{n-k} \end{align}
• Let $\displaystyle{\sum_{n=0}^{\infty} c_n}$ denote the Cauchy product for the series $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$. Then:
(2)
\begin{align} \quad \sum_{n=0}^{\infty} C_n = \sum_{n=0}^{\infty} x^n \left ( \sum_{k=0}^{n} a_kb_{n-k} \right ) = \sum_{n=0}^{\infty} c_nx^n \end{align}
• From the absolute convergence of $\displaystyle{\sum_{n=0}^{\infty} a_nx^n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_nx^n}$, from the first result mentioned at the top of this page we must have that then:
(3)