Vrftjkry

I'm studying the convergence of the recurrence relation with a fixed $r > 0, r in Bbb{R}$ with $u_{n+1}$ given by $u_{n+1} = frac{u_{n}^2}r + 1$ and $u_0 = 0$.

It can be shown that $u_n > 1$ for all $n in Bbb{N}$ and then $(u_n)$ is increasing.

By studying that happens when we pass to the limit, we can use the quadratic formula to show that $l$ exists only when $rgeq4$ (this is when the determinant $1-frac{4}r$ is zero). Then, as $(u_n)$ is increasing and unbounded, we have $u_n longrightarrow infty$.

Now, here's where I've started to get a little bit confused. I can see inherently that $l$ would be equal to $frac{r}2-frac{r}2sqrt{1-frac{4}r}$, but my question is on how to show this for any $rgeq4$ and any $n in Bbb{N}$. Would there be a way to show this by induction? Is there some other way to simplify this value or simplify the proof that would make this upper bound easier to show?

First, I'd check the limit. If we assume that there is a limit $u_{l}$, then:

begin{align*}
u_{l} &= frac{u_l^2}{r} + 1 \
0 &= u_l^2 - ru_l + r
end{align*}

So the quadratic formula says this:

$$
u_l = frac{r pm sqrt{r^2-4r}}{2} = frac{r}{2} pm frac{r}{2}sqrt{1 - frac{4}{r}}
$$

As you've noted, starting from $u_0 = 0$, we'd expect convergence to the lower limit, $frac{r}{2} - frac{r}{2}sqrt{1 - frac{4}{r}}$. Let's call that lower limit $b$, and define the series $t_n$ as $t_n = b - u_n$, and see where the recurrence relation given takes us:

begin{align*}
b - t_{n+1} &= frac{(b-t_n)^2}{r} + 1 \
br - rt_{n+1} &= (b-t_n)^2 + r \
br - rt_{n+1} &= b^2 - 2 b t_n + t_n^2 + r \
- rt_{n+1} &= t_n(1 - 2b) + b^2 - br + r
end{align*}

But by construction, we know that $b^2 - br + r = 0$, so:

begin{align*}
- rt_{n+1} &= t_n(1 - 2b) + 0 \
t_{n+1} &= t_nfrac{(2b - 1)}{r}
end{align*}

And from there, it's straightforward that $t_n to 0$, and therefore $u_n to b$.

In general in these types of problems, I find that once you have a putative limit reframing the problem in terms of a new series defined as (limit - old series) usually makes the problem easier.

If the sequence converges, we have

$$u_{infty}=frac{u_{infty}}r+1$$ or

$$u_{infty}=frac r{r-1}.$$

Obviously, convergence is impossible for $r<1$ as the terms remain positive. Then for $r>1$,

$$u_{n+1}-u_infty=frac{u_n}{r}+1-u_infty=frac{u_n-u_infty}r.$$

By induction,

$$u_n-u_infty=(u_0-u_infty)r^{-n}.$$