Show that the sequence is contractive Let $w_n=(\lambda+w_{n-1})^{-1}\space \forall \space n\in\Bbb{N}$, where $w_0$ and $\lambda$ are given positive numbers. Prove that:
a)$0<w_n<\lambda^{-1} \space \forall \space n\in\Bbb{N};$
b)the sequence {$w_n$} with $n\in\Bbb{N}$ has the contraction property if $\lambda>1$. Hence prove that lim$_{n\to\infty}w_n$ exists for $\lambda>1$ and find its value.
Challenge: Show that the sequence is contractive for some positive values for $\lambda\leq1.$
I am able to do part a) and b), but not the challenge part. I was given hint of using part a) to show that $w_n>\frac{\lambda}{\lambda^2+1}\space \forall\space n\geq2$. Can someone enlighten me on how to get this result from part a)? I used induction to prove part a). Much thanks!
 A: We wish to find some $r \in [0, 1)$ s.t. for all $n$, we have $d(w_{n + 1}, w_{n + 2}) \leq r d(w_n, w_{n + 1})$.
In particular, we may look for the case where $w$ is a constant sequence; that is, where $w_0 = (\lambda + w_0)^{-1}$; that is, where $w_0^2 + \lambda w_0 - 1 = 0$. The solutions to this equation are $w_0 = \frac{-\lambda \pm \sqrt{\lambda^2 + 4}}{2}$.
Simply take $w_0 = \frac{-\lambda + \sqrt{\lambda^2 + 4}}{2}$. Then $w$ is constant and therefore contractive.
Edit: apparently, original poster wants to that there exists $\lambda \leq 1$ such that for all $w_0 > 0$ this holds.
Suppose $n > 2$. Then we have $w_{n - 1} < \lambda^{-1}$. Then $\lambda + w_{n - 1} < \lambda + \lambda^{-1}$. Then $\frac{\lambda}{\lambda^2 + 1} = \frac{1}{\lambda + \lambda^{-1}} < \frac{1}{\lambda + w_{n - 1}} = w_n$.
Now we have $|\frac{d}{dx} \frac{1}{\lambda + x}| = \frac{1}{(\lambda + x)^2}$. Whenever we have $x > \frac{\lambda}{1 + \lambda^2}$, we have $\frac{1}{\lambda + x} < \frac{1}{\lambda + \frac{\lambda}{\lambda^2 + 1}} = \frac{1}{\lambda}\frac{\lambda^2 + 1}{\lambda^2 + 2}$. When $\lambda$ is slightly less than 1, this value too will be less than one. This is sufficient to show the sequence is contractive.
