# Finding the limit of this recursive sequence

Assume $a$, $b$, $x_0$ are all positive and define the recursive sequence $(x_n)$ as $x_n = \frac{a}{1+bx_{n-1}}$. Prove the sequence converges if $ab < 1$ and find its limit.

I've done this so far: Assume the limit exists, denote it $l$, then $\lim x_n = \frac{a}{1+b \lim_{x_{n-1}}} \implies l = \frac{a}{1+bl}$ by the algebraic limit theorem. From this, how can I get the condition that $ab<1$?

• According to the first paragraph you are given that $ab \lt 1$, so you are not being asked to derive that condition (as you ask in the last paragraph). – hardmath Aug 9 '14 at 12:01
• oh right, so how do I show the sequence converges then? – user168811 Aug 9 '14 at 12:05
• @Kelenner: Why not give that as an answer? – String Aug 9 '14 at 12:43
• Let $L$ the positive solution of your equation $l=a/(1+bl)$. Show that $|x_n-L|\leq ab|u_{n-1}-L|$. – Kelenner Aug 9 '14 at 12:43

We immediately have that $x_n\geq 0$ for all $n$. The equation $l=a/(1+bl)$ have two solutions, let $\displaystyle L=\frac{\sqrt{1+4ab}-1}{2b}$ be the positive solution. Then we have:
$$x_n-L=\frac{a}{1+bx_{n-1}}-\frac{a}{1+bL}=ab\frac{L-x_{n-1}}{(1+bx_n)(1+bL)}$$ hence $$|x_n-L|\leq (ab)|x_{n-1}-L|$$ By an easy induction, $|x_n-L|\leq (ab)^n|x_0-L|$, and as $0<ab<1$, we have $x_n\to L$.