# Why $\lim_{n\rightarrow \infty}\frac{F(n,n)}{F(n-1,n)} =\lim_{n\rightarrow \infty}\frac{F(n+1,n)}{F(n,n)}$?

Let $$\alpha,\beta, \gamma \in \mathbb{R}^+$$ be and the function

$$F(m,n)= \begin{cases} 1, & \text{if m n=0 }; \\ \alpha F(m ,n-1)+ \beta F(m-1,n )+ \gamma F(m-1,n-1), & \text{ if m n>0. }% \end{cases}$$ Please, a proof for $$\lim_{n\rightarrow \infty}\frac{F(n,n)}{F(n-1,n)} =\lim_{n\rightarrow \infty}\frac{F(n+1,n)}{F(n,n)}$$

• Probably won't help. Generating function $$\sum _{m=0}^{\infty } \sum_{n=0}^{\infty }{\it F} \left( n,m \right) {x}^{n}{y}^{m} =-{\frac {\alpha\,xy+\beta\,xy-\alpha\,x-\beta\,y-xy+1}{ \left( y-1 \right) \left( x-1 \right) \left( \gamma\,xy+\alpha\,x+\beta\,y-1 \right) }}$$ Commented Apr 10, 2021 at 11:58
• Actually that's how one would start. Then use the contour-integral formula to get algebraic generating functions for the diagonal series with $m=n$ and $m = n \pm 1$, and use the singularities nearest the origin to recover asymptotic formulas for those series. Commented Apr 17, 2021 at 16:02