Prove that a recursive sequence converges and determine its limit

Question

I've been working on this task for hours now but I just can't prove that this recursive sequence is convergent.

$$ a_0=5 \\ a_{n+1}=\frac15(2+6a_n-{a_n}^2) $$

Feeding some values to the calculator, one can clearly see that this sequence converges to 2. I firstly wanted to show that the sequence is monotone, however when I try doing something like that:

$$ a_{n+1} \geq a_n $$ I end up at: $$ -1 \leq a_n \leq 2$$ which just confuses me even more.

How can I show that this recursive sequence converges to 2?

Thanks very much in advance to all who try to help me.

Edit:

As stated in some comments below, I just need to show additionally that $ a_{n+1} \leq 2 $. But when I plug in the $ a_{n+1} $ from above I get $ \frac15(2+6a_n-{a_n}^2) \leq 2 $ what leaves me stuck again.

Answer 1

So taking

$$f(x)=\frac{1}{5}(2+6x-x^2)$$

$f$ is two order polynomial with a maxima at $x=3$ equal to $\dfrac{11}{5}$

So $a_n$ is minored by $\dfrac{11}{5}$.

Secondly:

$$f(x)-x=\frac{1}{5}(2+x-x^2)$$

which admit minima at $x=\dfrac{1}{2}$ equal to $\dfrac{9}{20}$

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So because $a_{n+1}-a_n=f(x_n)-x_n>0$

$a_n$ growths. It is further majorated by $\dfrac{11}{5}$ (show it from $a_1$ by iteration)

Thus the limit if the positive fix point of $f$ (because $a_n$ is positive) which is at $f(x)-x=0$ i.e $x=2$

Hence we get: $$a_n \rightarrow 2$$

Answer 2

Define $e_n=a_n-2$, therefore$$e_{n+1}={1\over5}(2+6a_n-a_n^2)-2={1\over5}(6a_n-a_n^2-8)=-{a_n-4\over 5}({a_n-2})=-{a_n-4\over 5}e_n$$hence $$|e_{n+1}|=|{a_n-4\over 5}||e_n|\le {2\over 5}|e_n|$$which leads to $e_n\to 0$ or $a_n\to 2$ since for all $n>1$, $2<a_{n+1}<a_{n}$.