[proofplan] We work entirely in the formal [power series](/page/Power%20Series) ring $R[[X]]$, so no analytic convergence is involved. The product in $R[[X]]$ is defined coefficientwise by the Cauchy convolution rule: the coefficient of $X^n$ in a product is the finite sum over all pairs of indices whose total degree is $n$. Applying this definition to $A(X)$ and $B(X)$ gives exactly the sequence $(c_n)_{n\geq 0}$, which proves that the product is its ordinary [generating function](/page/Generating%20Function). [/proofplan] [step:Expand the two ordinary generating functions in $R[[X]]$] By hypothesis, the ordinary generating functions are the formal power series \begin{align*} A(X)=\sum_{m=0}^{\infty}a_mX^m \end{align*} and \begin{align*} B(X)=\sum_{k=0}^{\infty}b_kX^k. \end{align*} Define the product series $C(X)\in R[[X]]$ by \begin{align*} C(X)=A(X)B(X). \end{align*} For each integer $n\geq 0$, let $d_n=[X^n]C(X)$ denote the coefficient of $X^n$ in $C(X)$. [/step] [step:Compute each coefficient of the product by the formal multiplication rule] The multiplication rule in the formal power series ring $R[[X]]$ gives, for every integer $n\geq 0$, \begin{align*} d_n=[X^n]\bigl(A(X)B(X)\bigr)=\sum_{i=0}^{n}a_i b_{n-i}. \end{align*} The sum is finite, so it is a well-defined element of $R$. By the definition of $c_n$, this says $d_n=c_n$ for every $n\geq 0$. [guided] The key point is that multiplication in $R[[X]]$ is formal rather than analytic. To find the coefficient of $X^n$ in $A(X)B(X)$, we multiply terms \begin{align*} a_iX^i \end{align*} from $A(X)$ with terms \begin{align*} b_jX^j \end{align*} from $B(X)$. Their product is \begin{align*} (a_iX^i)(b_jX^j)=a_i b_j X^{i+j}, \end{align*} using the multiplication in the commutative ring $R$ and the rule $X^iX^j=X^{i+j}$. A term contributes to the coefficient of $X^n$ exactly when $i+j=n$. Since $i,j\geq 0$, this condition is equivalent to choosing an index $i$ with $0\leq i\leq n$ and then setting $j=n-i$. Therefore the coefficient of $X^n$ in the product is \begin{align*} [X^n]\bigl(A(X)B(X)\bigr)=\sum_{i=0}^{n}a_i b_{n-i}. \end{align*} This sum has only $n+1$ terms, so it is legitimate in an arbitrary commutative ring; no infinite summation operation in $R$ is being used. By the definition of the sequence $(c_n)_{n\geq 0}$, the right-hand side is $c_n$. Hence \begin{align*} [X^n]\bigl(A(X)B(X)\bigr)=c_n \end{align*} for every integer $n\geq 0$. [/guided] [/step] [step:Identify the product as the ordinary generating function of the convolution sequence] Since $d_n=c_n$ for every integer $n\geq 0$, the product series $C(X)=A(X)B(X)$ has coefficient $c_n$ in degree $n$. Hence \begin{align*} A(X)B(X)=C(X)=\sum_{n=0}^{\infty}c_nX^n. \end{align*} Therefore $A(X)B(X)$ is the ordinary generating function of the sequence $(c_n)_{n\geq 0}$. [/step]