Prove that $\lim\limits_{n\to\infty}y_{n}=\frac{1}{x}$ if $ y_{n}=y_{n-1}(2-xy_{n-1}), y_{i}>0,(i\in \mathbb{N}),x>0$

Question

I have been asked to prove that $\lim\limits_{n\to\infty}y_{n}=\frac{1}{x}$ if $ y_{n}=y_{n-1}(2-xy_{n-1}), y_{i}>0,(i\in \mathbb{N}),x>0.$

In particular, what I would like to know is if it is possible to prove this by using the inequality between arithmetic and geometric means?

Answer 1

If there is any limit at all, then the limit of $y_n$ is the same as the limit of $y_{n-1}$ so the equation $y=y(2-xy)$ immediately implies that $0$ or $\frac{1}{x}$ are the only possibilities. The first case, $0$, can be excluded because $y_n$ is growing whenever $y_n<\frac{1}{x}$.

But the limit may not exists if $x$ is large, for example. Think of $y_1=1$ and $x=10$, the sequence then diverges very quickly. The usual way of proving convergence for some $x$ is to show that the map $u\mapsto u(2-xu)$ is a contraction on suitable subset of $\Bbb R$, for example. (This is indeed the case for $x$ smaller than $2/y_1$.)

Answer 2

If we let $y_n' = xy_n$, we see that we can write the equation as $y_{n+1}' = y_n'(2-y_n')$, so we can just assume that $x=1$ and work with the equation $y_{n+1} = f(y_n)$, where $f(y) = y(2-y)$.

If we look at the graph of $f$, we see that $f(y) >0$ iff $y \in (0,2)$ and it is easy to check that (i) $f(y) \le 1$ for all $y$, (ii) $y < f(y) \le 1$ for all $y \in (0,1]$, (iii) $f(y) \in (0,1)$ for all $y \in (1,2)$.

We note that if $y_n \in (0,1]$, then $y_n \le y_{n+1} \le 1$.

Since $y_2 >0$, we must have $y_1 \in (0,2)$. If $y_1 \in (1,2)$, then $y_2 \in (0,1]$. Hence, with the possible exception of the initial $y_1$ we see that $y_n$ is a non decreasing sequence bounded above by one, hence it converges to some value $y \in [y_2,1]$. Since $f$ is continuous, we see that $y=f(y)$ and so $y=1$.