Prove the sequence is monotone decreasing and find the limit?

Question

Let $a,b$ be positive real number. Set $x_0 = a$ and $x_{n+1} = 1/[(1/x_n) + b]$ for $n ≥ 0$

(a) Prove that $x_n$ is monotone decreasing.

(b) Prove that the limit exists and find it.

My work:

(a) By given premises, $x_n ≥ 0$ for each n,

$x_1 = x_{0+1} = 1/[(1/x_0) + b] = 1/[(1/a) + b]$

$x_2 = x_{1+1} = 1/[(1/x_1) + b] = 1/[(1/a) + b + b] = 1/[(1/a) + 2b]$

and then continues.

we can see as $n$ increases, $x$ dereases as denominator increases.

$n→∞, nb→∞$, the sequences decreases

(then i dont know how to continue with induction)

(b) By Monotone Convergence Theorem, the sequence ($x_n$) is convergent, bounded below, limit exists

....But I dont know how to find that limit ....

Thank you guys!!Please

Answer 1

Let us compute: $$ x-f(x)=x-\frac{x}{1+bx}=\frac{bx^2}{1+bx}. $$ So we see that $x-f(x)>0$ for all $x>0$.

An easy induction proof shows that $x_n>0$ for all $n$.

So for all $n$, $$ x_n-f(x_n)=x_n-x_{n+1}>0. $$

Hencee, indeed, the sequence $(x_n)$ is decreasing and bounded below by $0$. So it converges to some limit $L\geq 0$.

Now since $f$ is continuous, we must have at the limit $$ L=f(L)\quad\Leftrightarrow\quad L=\frac{L}{1+bL}\quad\Leftrightarrow\quad L=0. $$

So $(x_n)$ converges to $0$.