Solving $a_{n+1} = c_n a_n$ using generating functions

Question

I need to solve a recurrence relation of the form for $a_n$:

$$a_{n+1} = c_n a_n,\quad n\ge 0$$

where the constants $c_n$ and the initial condition $a_0$ are given. However I do not know the $c_n$ a priori, so I would like to write down a solution for $a_n$, in some form, that shows the dependence on $c_n$. I tried the generating function approach, but the only way I see here is to define two generating functions:

$$A(x) = \sum_{n\ge0}a_n x^n,\quad B(x) = \sum_{n\ge0}c_n a_n x^n$$

Then we have:

$$\frac{A(x) - a_0}{x} = B(x)$$

This does not get me very far, because I have two generating functions and only one equation. Is there a way to tackle this problem?

Answer 1

You have $$a_{n+1}=c_{n}a_{n}$$ Devide by $\prod_{k=0}^{n}c_{k}$ to give $$\frac{a_{n+1}}{\prod_{k=0}^{n}c_{k}}=\frac{a_{n}}{\prod_{k=0}^{n-1}c_{k}}$$ Let $$\frac{a_{n}}{\prod_{k=0}^{n-1}c_{k}}=A_{n}$$ Then $$A_{n+1}=A_{n}$$ Hence $A_{n}=A$, where $A$ is constant $$A=\frac{a_{n}}{\prod_{k=0}^{n-1}c_{k}}$$ So $$a_{n}=A\prod_{k=0}^{n-1}c_{k}$$

Answer 2

Since \begin{align*} a_{n+1}&=c_na_n\\ a_n&=c_{n-1}a_{n-1}\\ &\ \ \,\vdots\\ a_1&=c_0a_0 \end{align*}

multiplication of LHS and RHS gives \begin{align*} \prod_{k=1}^{n+1}a_k&=\left(\prod_{k=0}^n c_k\right)\left(\prod_{k=0}^na_k\right) \end{align*}

Given that $a_n\ne 0$ for $n\geq 0$ we can divide by $\prod_{k=1}^na_k$ and obtain \begin{align*} a_{n+1}=a_0\prod_{k=0}^nc_k\qquad\qquad n\geq 0 \end{align*}