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

Question

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.

Answer 1

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 $a_n>a_{n+1}\implies {{a_n}^2+3\over7}>{{a_{n+1}}^2+3\over7}\implies a_{n+1}>a_{n+2}$

Answer 2

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 $L=\frac{7+\sqrt{37}}{2}>13/2$.