Find the solution to and limit of $a_{n+1} =\frac{v}{a_n+w} $ with $a_1>0, v > 0, w>0$

Question

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.

Answer 1

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.

Answer 2

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 $v >0$...

Answer 3

Find the solution to $a_{n+1} =\frac{v}{a_n+w} $ with $a_1>0, v > 0, w>0$.

My solution:

The limit $L =\frac{-w+\sqrt{w^2+4v}}{2} $.

If $b_m = a_m-L$, then

$b_m =\frac{1}{c_0(-d)^m+\frac{c}{(d+1)} }$

where

$\begin{array}\\ c &=-1/L\\ d &=1+w/L=v/L^2\\ c_0 &=\frac1{-db_1}+\frac{c}{d(d+1)}\\ &=\frac1{-d(a_1-L)}+\frac{c}{d(d+1)}\\ \end{array} $

Since $d > 1$, as long as $c_0 \ne 0$, $b_m \to 0$ so that $a_m \to L$.

Also, $b_{2m} =\frac{1}{c_0d^{2m}+\frac{c}{(d+1)} }$ and $b_{2m+1} =\frac{1}{-c_0d^{2m+1}+\frac{c}{(d+1)} }$.

Here is what I did.

If it has a limit $L$, then $L =\frac{v}{L+w} $ or $L^2+wL-v=0$ or $L =\frac{-w\pm\sqrt{w^2+4v}}{2} $. To make $L > 0$, choose $L =\frac{-w+\sqrt{w^2+4v}}{2} $.

Let $b_n = a_n-L$. then $b_{n+1}+L =\frac{v}{b_n+L+w} $ or

$\begin{array}\\ b_{n+1} &=\frac{v}{b_n+L+w}-L\\ &=\frac{v-L(b_n+L+w)}{b_n+L+w}\\ &=\frac{v-Lb_n-L^2-wL}{b_n+L+w}\\ &=\frac{-Lb_n}{b_n+L+w} \qquad\text{since }L^2+wL-v=0\\ \end{array} $

Therefore

$\begin{array}\\ \frac1{b_{n+1}} &=\frac{b_n+L+w}{-Lb_n}\\ &=\frac1{-L}-\frac{L+w}{Lb_n}\\ &=c-\frac{d}{b_n} \qquad\text{where } c=-1/L, d=1+w/L=v/L^2\\ \text{so}\\ \frac1{d^{n+1}b_{n+1}} &=\frac{c}{d^{n+1}}-\frac{1}{d^nb_n}\\ \text{or}\\ \frac1{(-d)^{n+1}b_{n+1}} &=\frac{c}{(-d)^{n+1}}+\frac{1}{(-d)^nb_n}\\ \end{array} $

Letting $c_n =\frac{1}{(-d)^nb_n} $, $c_{n+1} =\frac{c}{(-d)^{n+1}}+c_n $ where $c_1 =\frac1{-db_1} =\frac1{-d(a_1-L)} $.

Summing from $1$ to $m-1$,

$\begin{array}\\ c_m-c_1 &=\sum_{n=1}^{m-1}(c_{n+1}-c_n)\\ &=\sum_{n=1}^{m-1}\frac{c}{(-d)^{n+1}}\\ &=\frac{c}{(-d)^2}\sum_{n=1}^{m-1}\frac{1}{(-d)^{n-1}}\\ &=\frac{c}{d^2}\sum_{n=0}^{m-2}\frac{1}{(-d)^{n}}\\ &=\frac{c}{d^2}\frac{1-\frac1{(-d)^{m-1}}}{1-\frac1{-d}}\\ &=\frac{c}{d^2}\frac{1-\frac1{(-d)^{m-1}}}{1+\frac1{d}}\\ &=c\frac{1-\frac1{(-d)^{m-1}}}{d(d+1)}\\ &=c(\frac1{d(d+1)}+\frac1{(-d)^{m}(d+1)})\\ &=\frac{c}{d(d+1)}+\frac{c}{(-d)^{m}(d+1)}\\ \text{or}\\ c_m &=c_1+\frac{c}{d(d+1)}+\frac{c}{(-d)^{m}(d+1)}\\ &=c_0+\frac{c}{(-d)^{m}(d+1)} \qquad\text{where }c_0=c_1+\frac{c}{d(d+1)}\\ \end{array} $

Therefore

$\begin{array}\\ b_m &=\frac1{(-d)^m}\frac{1}{c_m}\\ &=\frac1{(-d)^m}\frac{1}{c_0+\frac{c}{(-d)^{m}(d+1)} }\\ &=\frac{1}{c_0(-d)^m+\frac{c}{(d+1)} }\\ \text{so that}\\ b_{2m} &=\frac{1}{c_0d^{2m}+\frac{c}{(d+1)} }\\ \text{and}\\ b_{2m+1} &=\frac{1}{-c_0d^{2m+1}+\frac{c}{(d+1)} }\\ \end{array} $

We have

$\begin{array}\\ c_0 &=c_1+\frac{c}{d(d+1)}\\ &=\frac1{-d(a_1-L)}-\frac{1}{d(d+1)}\\ &=\frac1{-d}(\frac1{a_1-L}+\frac{1}{d+1})\\ &=\frac1{-d(a_1-L)(d+1)}(d+1+a_1-L)\\ &=\frac1{-d(a_1-L)(d+1)}(2+w/L+a_1-L)\\ \end{array} $

Since $L^2+wL=v$, if $c_0 = 0$ then $a_1 =L-w/L-1 $.

If $|d| > 1$, which will be true if $w > 0$, and $c_0 \ne 0$, then $c_m \to c_0 $ and $b_m \approx \frac1{(-d)^mc_0} \to 0 $ so that $a_m \to L$.

For the question that inspired this, $v=6, w=1$, so $L=2$, so $c=-\frac12$ and $d=1+\frac12 =\frac32$ and $c_1 =\frac1{-d(a_1-L)} =\frac1{-\frac32(1-2)} =\frac23 $ so $c_0 =\frac23+\frac{-\frac12}{\frac32\frac52} =\frac23(1-\frac{1}{5}) =\frac{8}{15} $ and

$\begin{array}\\ b_m &=\frac{1}{c_0(-d)^m+\frac{c}{(d+1)} }\\ &=\frac{1}{\frac{8}{15}(-\frac32)^m+\frac{-\frac12}{\frac52} }\\ &=\frac{1}{\frac{8}{15}(-\frac32)^m-\frac15}\\ &=\frac{15}{8(-\frac32)^m-3}\\ \text{so that}\\ a_m &=2+\frac{15}{8(-\frac32)^m-3}\\ &=\frac{15+2(8(-\frac32)^m-3)}{8(-\frac32)^m-3}\\ &=\frac{9+16(-\frac32)^m}{8(-\frac32)^m-3}\\ \end{array} $

Note that $b_{2m} =\frac{15}{8(-\frac32)^{2m}-3} =\frac{15}{8(\frac94)^{m}-3} $ and $b_{2m+1} =\frac{15}{8(-\frac32)^{2m+1}-3} =\frac{-15}{8\frac32(\frac94)^{m}+3} =\frac{-15}{12(\frac94)^{m}+3} =\frac{-5}{4(\frac94)^{m}+1} $ so that $b_{2m}$ is decreasing and $b_{2m+1}$ is increasing.

Answer 4

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...