Find the solution to and limit of $a_{n+1} =\frac{v}{a_n+w} $ with $a_1>0, v > 0, w>0$ Find the solution to
and limit of
$a_{n+1}
=\frac{v}{a_n+w}
$
with
$a_1>0, v > 0, w>0$.
This was inspired by
my answer to
Converging sequence $a_{{n+1}}=6\, \left( a_{{n}}+1 \right) ^{-1}$.
I will give my answer
in two days.
 A: I was mystified by @Tnilk Imaniq's answer in the linked thread, but then I realized that what he was doing was to set $b_n=1/(a_n+c)$ and then get a linear difference equation for $b_n$. So we do the same:
$$b_n=\frac1{a_n+c};\,\,a_n=\frac1{b_n}-c$$
$$\frac1{b_{n+1}}-c=\frac v{\frac1{b_n}-c+w}=\frac{1-cb_{n+1}}{b_{n+1}}=\frac{vb_n}{1+(w-c)b_n}$$
Clearing denominators,
$$vb_nb_{n+1}=1-cb_{n+1}+(w-c)b_n-c(w-c)b_nb_{n+1}$$
Since all we know is linear difference equations, we demand $c^2-wc-v=0$.
$$c=\frac{w\pm\sqrt{w^2+4v}}2=\frac{w-\sqrt{w^2+4v}}2$$
because we anticipate that $$\lim_{n\rightarrow\infty}b_n=\infty$$
and we know that $a_n>0$, so we must have $c<0$. Then our linear difference equation is
$$cb_{n+1}+(c-w)b_n=\frac{w-\sqrt{w^2+4v}}2b_{n+1}+\frac{-w-\sqrt{w^2+4v}}2b_n=1$$
We seek a particular solution $b_{np}=A$, so
$$-A\sqrt{w^2+4v}=1,\,\,b_{np}=\frac{-1}{\sqrt{w^2+4v}}$$
Now we solve the homogeneous equation
$$\frac{w-\sqrt{w^2+4v}}2b_{n+1,h}+\frac{-w-\sqrt{w^2+4v}}2b_{nh}=0$$
This has a solution of the form $b_{nh}=c_1r^k$ if
$$r=\frac{\frac{w+\sqrt{w^2+4v}}2}{\frac{w-\sqrt{w^2+4v}}2}=\frac{\left(w+\sqrt{w^2+4v}\right)^2}{-4v}$$
so
$$b_n=b_{np}+b_{nh}=\frac{-1}{\sqrt{w^2+4v}}+c_1r^n$$
$$b_1=\frac1{a_1+\frac{w-\sqrt{w^2+4v}}2}=\frac{-1}{\sqrt{w^2+4v}}+c_1r$$
From which we obtain
$$c_1r=\frac{a_1+\frac{w+\sqrt{w^2+4v}}2}{\sqrt{w^2+4v}\left(a_1+\frac{w-\sqrt{w^2+4v}}2\right)}$$
So our solution is
$$a_n=\frac{\sqrt{w^2+4v}}{-1+\frac{a_1+\frac{w+\sqrt{w^2+4v}}2}{a_1+\frac{w-\sqrt{w^2+4v}}2}\left(\frac{\left(w+\sqrt{w^2+4v}\right)^2}{-4v}\right)^{n-1}}-\frac{w-\sqrt{w^2+4}}2$$
Since $|r|>1$, it follows that
$$\lim_{n\rightarrow\infty}a_n=\frac{\sqrt{w^2+4v}-w}2$$
I see that the OP has posted an answer while I was typing this up even though he said he would give us two days to reply. Oh well...
A: If the limit exists, it must equal $$L = \frac{\sqrt{w^2 + 4v} - w}{2}$$ as we know it satisfies $L^2 + Lw - v = 0$ and $a_1 > 0 , v > 0, w > 0$ .
We now only need to show the sequence convergence. Following Marty Cohen's approach in the special case, we define $a_n = b_n + L$. After some arranging, we have $b_{n+1} = \frac{b_n (w - \sqrt{w^2 + 4v})}{2 b_n + w + \sqrt{w^2 + 4v}}$  
Differentiating, we see $$(\frac{x (w - \sqrt{w^2 + 4v})}{2 x + w + \sqrt{w^2 + 4v}})' = - \frac{4v}{(2x + \sqrt{w^2 + 4v} + w)^2}$$
$$= \frac{-v}{(x + L + w)^2}$$
We notice our derivative is less than zero for all x as $v>0$.
Some messy computation follows but all that's left is to bound $b_n$ and show that the bounds become tighter on $b_{n+1}$, leaving us with $|\frac{b_{n+1}}{b_n}| < 1$ for $n \ge c$ where c is some constant.
A: If the sequence converges, then the limit $L$ satisfies
$$
L = \frac{v}{L+w} \iff L^2 + Lw -v = 0 \iff L = \frac{-w \pm \sqrt{w^2+4v}}{2}
$$
You are guaranteed the root is real since you are assuming $w >0$ and $v >0$...
