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Test $x_1=0, x_{n+1}=\frac{9+(x_n)^2}{6}$ for convergence and find its limit.


Note that $x_1=0, x_2=\frac{9+0²}{6}=\frac{9}{6}, x_3=\frac{9+9/6}{6}=\frac{17}{9}\dots$ with $x_1 < x_2 < x_3$. $(*)$

My plan for this is to show 1. that this recursively defined sequence is increasing by showing $a_{n+1}\geq a_{n}$ through induction and 2. to show that it is bounded above by $L=3$.


1.

Our base case is trivial by $(*)$. Our hypothesis is that $x_{n+1}\geq x_{n}$ and we want to show that $x_{n+2}\geq x_{n+1}$.

Inductive step: \begin{align*} x_{n+2}=\frac{9+(x_{n+1})^2}{6}&=\frac{9}{6}+\frac{(\frac{9+(x_n)^2}{6})^2}{6}\\&=\frac{9}{6}+\frac{1}{6^3}(9+(x_n)^2)^2\\ &\overset{(**)}\geq\frac{9}{6}+\frac{1}{6^3}36(x_n)^2\\&=\frac{9}{6}+\frac{(x_n)^2}{6}=\frac{9+(x_n)^2}{6}=x_{n+1}. \end{align*}

$(**)$ is true since \begin{align*} (9+(x_n)^2)^2\geq36(x_n)^2&\Leftrightarrow 9+(x_n)^2\geq6x_n\\&\Leftrightarrow (x_n)^2-6x_n+9\geq 0 \\&\Leftrightarrow (x_n -3)^2\geq 0. \end{align*}

2.

If we assume our limit $L$ exists, it must satisfy the equation $L=\frac{9+L^2}{6}\Leftrightarrow L^2+9-6L=0$. This is the case for $L=3$ as $9+9-18=0$ and as such $x_n$ is bounded by $0\leq x_n\leq 3$ with $\lim_{n\rightarrow\infty}x_n=3$.


I especially struggled with the inductive step. Would anybody know an easier and quicker way to show this? Also, do I have to argue the second step differently, by deducing $L=3$ through the quadratic formula first, or is it okay the way I did it? Thank you very much.

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Your conclusion that $(x_n)$ is bounded above by $3$ is unwarranted. Instead, I would do this by induction. Supposing that $x_n \le 3$, we have:

$$x_{n+1} \le \frac{9 + 3^2}{6} = 3$$

Now we conclude that $(x_n)$ is convergent and compute its limit as you have done.

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