<s>[INST] Is the proof to a recursive sequence $a_{n+1}=1+\frac{1}{5}a_n^2$ correct? We have a recursive sequence:
$$a_1 = 1$$
$$a_{n+1}=1+\frac{1}{5}a_n^2$$
We need to show that the sequence is convergent. We also need to analyse what would happen if we would pick a different first term $a_n$
What I did:
$$
a_{n+1} \geq a_n\\
1+\frac{1}{5}a_n^2 \geq a_n\\
a_n^2-5a_n+5\geq 0\\
(a_n-\frac{5-\sqrt5}{2})(a_n -\frac{5+\sqrt5}{2})\geq 0\\
\implies$$
$\text{so this implies that if all the terms are bellow}$ $\frac{5-\sqrt5}{2}$ or above $\frac{5+\sqrt5}{2}$ then the sequence is increasing
We see that $a_1 = 1$ is bellow $\frac{5-\sqrt5}{2}$, so let's prove that $a_n \leq \frac{5-\sqrt5}{2}$ by induction:
$n = 1:$
$$a_1 = 1 \leq \frac{5-\sqrt5}{2}$$
$$\text{holds}$$
$n \rightarrow n+1:$
$$
\frac{5-\sqrt5}{2} \geq a_{n+1} \\
\frac{5-\sqrt5}{2} \geq 1+\frac{1}{5}a_n^2\\
\frac{15-5\sqrt5}{2} \geq a_n^2\\
\text{for this to hold $a_n$ should not be between the positive and the negative root.}\\
\text{But we see that $\frac{5-\sqrt5}{2}$ is more than the $-\sqrt{\frac{15-5\sqrt5}{2}}$ sp this means that the sequence isn't increasing.}
$$
I do not know how to prove this, or if my process is correct and I made a mistake somewhere.
 [/INST] $a_1=1<\frac{5-\sqrt{5}}{2}$
suppose now that $a_k\le \frac{5-\sqrt{5}}{2}$
$$a_{k+1}=1+\frac15 a_k^2\le 1+\frac15 \cdot\left( \frac{5-\sqrt{5}}{2}\right)^2=\frac{5-\sqrt{5}}{2}$$
therefore $a_n\le \frac{5-\sqrt{5}}{2}$ for any $n$.
Edit
As it is monotone increasing and bounded the sequence converges to $a$.
To find $a$ we use the definition of the sequence
$$a=\frac{1}{5}a^2+1$$
which gives $a=\frac{5-\sqrt{5}}{2}$.
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