Discrete integrating factor ($U_n^{-1}$)

For $x_{n+1}=a_n{x}_n+b_n$, put $U_0=1$ and $U_n={\prod }_{j=1}^n{a}_j$, with $a_j\ne 0$. Dividing by this accumulated homogeneous growth gives $z_n=x_n/U_{n-1}$ and $z_{n+1}-z_n=b_n/U_n$. A telescoping series then provesThis is the discrete counterpart of the integrating factor for a first-order linear differential equation.