Limit of ratio of consecutive terms of a recurrence relation.

Question

My question is:

The sequence $a_n$ is defined by $$\sum_{r=1}^{k}p_ra_{n+r}=0,\forall n$$ where $p_r$ are fixed constants, and the "initial conditions" (i.e. the given values of $a_1,\ldots,a_{k}$). Let $\lambda_r,r\in\{0,1,\ldots,k-1\}$ be the $k$ roots of the characteristic equation $\sum_{r=1}^{k}p_r\lambda^{r}=0$. I want to prove that no matter what the "initial conditions" are, we have $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lambda_r,\exists r.$$ Or this limit does not exist.

How to prove this? I believe it is true.

There are some more famous specific examples such as the Fibonacci sequence defined by $F_{n+2}=F_n+F_{n+1}$. No matter what the values of $F_1, F_2$ are, the limit $\lim_{n\to\infty}\frac{F_{n+1}}{F_n}$ can either be $-0.618$ or $1.618$ or undefined. I think it is related to the fact that $\varphi^n$ is the only geometric progression that satisfy $F_{n+2}=F_n+F_{n+1}$(yes there is another one, $(-\varphi^{-1})^n$).

So let's return to the general case. How can I decide which $\lambda_r$ will be the limit $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ from the initial conditions? Is there a simple rule?

Answer 1

This is not true as stated.

For example, consider $$ p_1=-1, p_2=0, p_3=1, k=3 $$ so $\lambda=\pm1$, and a solution $$ a_n= \begin{cases} 2 & n\text{ even}\\ 1 & n\text{ odd} \end{cases}. $$ Then $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\tag{1}$$ does not exist, so cannot be equal to either $1$ or $-1$.

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Addendum: If the limit (1) exists, then it is one of the $\lambda_r$s. This is because the general solution of $a_n$ is a sum of $$ q_\lambda(n)\lambda^n $$ where $\lambda$ is a root of the characteristic (or auxiliary) polynomial $\sum_{r=0}^{k-1}p_{r+1}X^r$ and $q_\lambda$ is a polynomial (of degree one less than multiplicity of the root $\lambda$, in particular, is a constant if $\lambda$ is a simple root). So dividing and taking limit: $$ \frac{a_{n+1}}{a_n}=\frac{\sum_\lambda q_\lambda(n+1)\lambda^{n+1}}{\sum_\lambda q_\lambda(n)\lambda^n}\to\lambda_r $$ where $r$ is chosen such that:

• all $\lambda_j$ with $\lvert\lambda_j\rvert>\lvert\lambda_r\rvert$ has $q_{\lambda_j}\equiv 0$, and
• all $\lambda_j$ with $\lvert\lambda_j\rvert=\lambda_r\rvert$ has $\deg q_{\lambda_j}<\deg q_{\lambda_r}$ (the degree of the zero polynomial is -1).