Let $x_n$ be defined as:
$$
\begin{cases}
x_{n+1} = \frac{2+x_n^2}{2x_n} \\
n\in \mathbb N \\
x_1 = 4
\end{cases}
$$
Show that $x_n$ is a decreasing sequence.
I'm having a hard time with the sequence above. I've started with assuming that $x_{n+1} < x_n$. Now having that in mind we may inspect the following inequality:
$$
x < \frac{2+x^2}{2x} \iff 2x^2 < 2+x^2 \iff x^2 < 2
$$
The inequality doesn't show what's needed but $\sqrt2$ seems to be a point to which the sequence converges. I've also tried calculations with various initial conditions for $x_1$ and it looks like for all $x_1 > 0$ the sequence converges to $\sqrt2$ while for $x_1 < 0$ it converges to $-\sqrt2$.
Finding a closed form seems to not be an options since this recurrence is non-linear and i don't think it has a closed form.
What would be a formal way to show that $x_n$ is decreasing?
Best Answer
Note that $x_{n+1} = \frac{1}{x_n}+\frac{x_n}{2}.$ Then for $x_n>\sqrt{2}$
$$\frac{x_{n+1}}{x_n} = \frac{1}{x_n^2}+\frac{1}{2}<1.$$
When you fix the direction of your inequality, you'll have shown that $x_n >\sqrt{2}$ for all $n$. So the inequality above shows $x_{n+1}<x_n.$
Or I guess you could let $f(x) = \frac{1}{x}+\frac{x}{2}$ and use calculus to show that $f(x)$ is increasing for $x>\sqrt{2}$ and conclude that if $x_n>\sqrt{2}$ then so must $x_{n+1}>\sqrt{2}.$