Csdrhrt

Let $x_n$ be defined as:
$$
begin{cases}
x_{n+1} = frac{2+x_n^2}{2x_n} \
nin mathbb N \
x_1 = 4
end{cases}
$$
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?

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 $x_{n+1}>sqrt{2}.$

The condition, $x<frac{2+x^2}{2x}$, is the condition that would have to be true for the sequence to be increasing (since the condition says "the $n+1$-th element is larger than the $n$-th).

The actual condition has the inequality reversed, and you can prove that this holds by

first, through induction, proving that $x_n geqsqrt 2$ for all $n$.

then, proving your actual result.

One can solve
begin{align}
x_{n+1}=frac{x_n^2+2}{2x_n}&>sqrt2\
x_n^2+2&>2sqrt2 x_n\
x_n^2-2sqrt2 x_n+2&=(x_n-sqrt2)^2\
&>0
end{align}
Which is true for any $x_nneq sqrt2$