Is sequence $7a_{n+1} = a_n^2+3$ bounded, increasing? And find its limit as $n\to+\infty$ Consider sequence $a_n$ defined by 
$$a_1=2 $$
$$7a_{n+1} = a_n^2+3, n\ge2$$


*

*Show that $\frac{1}{2}\lt a_n \lt 3$ 

*Prove that it is increasing.

*Find its limit as $n\to+\infty$.
I know how to prove the first part by induction.
However, I don't know how to prove the second part.
And if the sequence is both bounded and increasing, which is monotonic, it is going to converge, right? But I don't think it is going to converge at all.
 A: First, for any value of $a_n$, $a_{n+1}>3/7$.  So, the sequence is bounded below by $3/7$.  Second, if $a_n<3$, then $a_{n+1}<12/7<3$.
Therefore, if the initial value $a_1<3$, then $3/7<a_n<12/7<3$ for $n\ge2$.
Next, let's examine the first difference $a_{n+1}-a_n$.  We have
$$\begin{align}
a_{n+1}-a_n&=\frac{a_n^2+3}{7}-a_n\\\\
&=\frac{a_n^2+7a_n+3}{7}\\\\
&=\frac17 \left(a_n-\frac{7+\sqrt{37}}{2}\right)\left(a_n-\frac{7-\sqrt{37}}{2}\right)\\\\
\end{align}$$
Therefore, $a_{n+1} < a_n$ whenever $0<\frac{7-\sqrt{37}}{2}<a_n<\frac{7+\sqrt{37}}{2}<7$.  
Inasmuch as $a_1=2$ $a_n$ is a monotonically decreasing sequence that is bounded below by $3/7$.  Then, by the Monotone Convergence Theorem, $a_n$ converges.
Finally, letting $\lim_{n\to \infty}a_n=L$, we can write
$$\begin{align}
\lim_{n\to \infty}a_{n+1}&=\lim_{n\to \infty}\frac{a_n^2+3}{7}\\\\
L&=\frac{L^2+3}{7}\\\\
L^2-7L+3&=0\\\\
L&=\bbox[5px,border:2px solid #C0A000]{\frac{7-\sqrt{37}}{2}}
\end{align}$$
where we ruled out the solution $L=\frac{7+\sqrt{37}}{2}>13/2$.
A: It is converging and the limit is going to be the smaller solution of $x^2-7x+3=0$, which is ${7-\sqrt{37}\over2}$.
Also it is strictly decreasing. 
Proof: 
Base case when $n=1,2$ is trivial and $a_n>a_{n+1}\implies {{a_n}^2+3\over7}>{{a_{n+1}}^2+3\over7}\implies a_{n+1}>a_{n+2}$
