How to show that the $\lim_{n \to \infty} x_n$ exists?

Question

Consider the sequence $x_{n+1}=\frac{x_n+3}{3x_n+1}$ where $n \geq 1$ with $0<x_1<1$.

I tried it this way, splitting the terms gives $x_{n+1}-x_n=3(1-x_{n+1}x_n)$.

But that doesn’t seems to be useful. Actually I am trying to find a recurrence relation between $x_{n+1}-x_n$ and $x_n-x_{n-1}$. But don’t know how??

Any other approach??

Thanks for any help!!

Answer 1

Hint:

Rewrite as $x_{n+1}=\frac{1}{3}+\frac{8/3}{3x_n+1}$. Now, argue that

• $0 < x_0 < 1 \Longrightarrow 1 < x_1 < 3$
• $1 < x_1 < 3 \Longrightarrow \frac{3}{5} < x_2 < 1$

We see that the possible range shrinks to $(\frac{3}{5},1)$, "closer" to $1$.

Apply this inductively to deduce that $(x_n)\rightarrow 1$.