Growth/decay rate of $a_1 = \frac{5}{2}, a_{n + 1} = \frac{1}{5}(a_n^2 + 6)$ this might be a vague question but I am interested in the recurrence relation defined by $a_1 = \frac{5}{2}$ and $a_{n + 1} = \frac{1}{5}(a_n^2 + 6)$. Some properties I have proven include that

*

*$2 < a_n \leq \frac{5}{2}$

*is strictly decreasing

*converges to $2$.

The next thing I did was looking at the difference $r_i = a_i - 2$, for which I found to 0.5, 0.45, 0.4005, 0.35248, 0.3068325, 0.2642952, 0.2254066, 0.1904869, 0.1596466, 0.1328146. Clearly there is some "near-"exponential decay going on here. One more result I have proven is that $\frac{8}{10}r_i < r_{i + 1}\leq \frac{9}{10}r_i$. However I am unable to show any further results about the error term $r_i$ - obviously it tends to $0$ exponentially with $0.8 < r \leq 0.9$ but can we do better?
So my question is, is there more precise bounds for $a_{i + 1} - 2$, whether by relating to $a_i - 2$ or in general?
(Also note that the original quadratic recurrence doesn't seem to be solvable by existing techniques.)
 A: Let $r_n = a_n-2$. One has
\begin{align*}
r_{n+1} &= a_{n+1}-2 \\
&= \frac{1}{5}(a_n^2+6)-2 \\
&= \frac{1}{5}((r_n+2)^2+6)-2\\
&=\frac{1}{5}(r_n^2+4r_n)
\end{align*}
so $(r_n)$ satisfies the relation
$$\boxed{r_{n+1} =\frac{1}{5}(r_n^2+4r_n)}$$
You deduce that
\begin{align*} \log(r_{n+1})-\log(r_n) &= \log\left(\frac{4}{5}\right)+ \log\left(1+\frac{r_n}{4}\right)
\end{align*}
and because $(r_n)$ tends to $0$, one has the following limit
$$\lim_{n \rightarrow +\infty} \log(r_{n+1})-\log(r_n) = \log\left(\frac{4}{5}\right)$$
By Cesaro's theorem, you get that
$$\log(r_n) \sim n\log\left(\frac{4}{5}\right)$$
which gives you
$$\boxed{\lim_{n \rightarrow +\infty} r_n^{1/n} = \frac{4}{5}}$$
A: Thank you @TheSilverDoe for his solution. However just seconds apart I have also come up with my solution, so I wanted to post here to see whether it is correct or not.
Since we have $r_{n + 1} = \frac{1}{5}(r_n^2 + 4r_n)$, we can think of this as a map $r\mapsto \frac{1}{5}(r^2 + 4r) = r + \frac{1}{5}(r^2 - r)$. Now the ridiculous part is we can think of this as a newton-raphson map $x\mapsto x - \frac{f(x)}{f'(x)}$, which by solving a differential equation gives us $f(x) = \frac{cx^5}{{(x - 1)}^5}$. So the $r$-recurrence essentially finds the root of $f$. Since the function $f$ has a root of multiplicative $5$ at $x = 0$, the map converges linearly i.e. asymptotically by a constant factor. Here it is $0.8$ as we have shown. So $r_{n + 1} \approx \big(\frac{4}{5}\big)^n$ as desired.
