Convergence of a recurrent sequence defined by $a_n=(\lambdaR-\lambdaa_{n-1})/(1+a_{n-1})$

Question

I hope someone here can help me. While studying an economics problem I came to a math problem that I can't solve. I'll try to put the problem as succinctly as possible. I came to a sequence where the $n^{th}$ element is defined by:

$$a_n = (\lambda*R-\lambda*a_{n-1})/(1+a_{n-1})$$

where $R$ is a known positive real number. What I want to prove is that the sequence converge for values of $0<\lambda<1$ (negative lambdas do not make economic sense). This seems intuitively true based on the economics behind it. Also, I have run some simulations that tend to support this claim. However, I cannot get the proof right.

From this answer, I got the idea of making $a_n=p_n/q_n$, with $p_0=a_0$ and $q_0=1$ (In all honesty, I don't understand 100% this answer). Then study the system given by:

$$\begin{bmatrix}p_n\\q_n\end{bmatrix}=\begin{bmatrix}-\lambda&\lambda*R\\1&1\end{bmatrix}\begin{bmatrix}p_{n-1}\\q_{n-1}\end{bmatrix}$$

From there I immediately tried to prove the stability based on the trace and determinant of the matrix M: \begin{bmatrix}-\lambda&\lambda*R\\1&1\end{bmatrix}

However, from the stability conditions $tr(M)<0$ and $det(M)>0$ I get the results that $\lambda$ should at the same time be greater than 1 and $<0$, which is fairly easy to see. Those results would suggest that there are no values of $\lambda$ for which the system is stable. This seems to contradict some economic sense and also the simulations I have run.

I am incapable of getting out of this conundrum so far. My best guest is that I'm trying to prove the stability for the system \begin{bmatrix}p_n\\q_n\end{bmatrix} and while those might not be stable, the ratio $p_n/q_n$ might be. And the ratio is actually the sequence I'm interested in.

All the best.

I used my results for $R=3$, $\lambda=\frac 34$, $a_0=1$. The results are correct and converge to the correct limit. — Claude Leibovici, May 14 '19 at 08:26

score 1 · Answer 1 · answered May 14 '19 at 04:25

Considering $$a_n=\lambda\frac{R-a_{n-1}}{1+a_{n-1}}$$ let $$A=\sqrt{\lambda ^2+2 \lambda (1+2 R)+1}\qquad \text{and} \qquad B=\lambda(R+1)$$

If I am not mistaken in my simplifications of the results from a CAS, then $$a_n=\frac 1 {R+1}\frac {K (\lambda +2 R+1-A) \left(\frac{A-\lambda +1}{B}\right)^n+ (-1)^n(\lambda +2 R+1+A) \left(\frac{A+\lambda -1}{B}\right)^n } {K \left(\frac{A-\lambda +1}{B}\right)^{n+1}+ (-1)^{n+1}\left(\frac{A+\lambda -1}{B}\right)^{n+1} }$$ $K$ being fixed by the initial condition.

I get how useful what you've done is, but how exactly did you do it? What exactly is a "CAS"? Sorry for the ignorance and thank you! — Pedro Siqueira Machado, May 14 '19 at 12:36

Convergence of a recurrent sequence defined by $a_n=(\lambda*R-\lambda*a_{n-1})/(1+a_{n-1})$

1 Answers1

Convergence of a recurrent sequence defined by $a_n=(\lambdaR-\lambdaa_{n-1})/(1+a_{n-1})$