# Show that $x_{n+1} = \frac{2+x_n^2}{2x_n}$ is a decreasing sequence.

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?

• If $x_1 \lt 0$, show $-\sqrt{2} \lt x_n \lt 0$ by induction. If $x_1 \gt 0$, show $\sqrt{2} \gt x_n \gt 0$ by induction. – Ewan Delanoy Nov 14 '18 at 15:53
• It is possible to get a closed form for this. Just define $y_n = \frac{x_n - \sqrt{2}}{x_n+\sqrt{2}}$. The recurrence relation for $(y_n)$ is very simple: $y_{n+1} = y_n^2 \implies y_n = y_1^{2^{n-1}}$. – achille hui Nov 14 '18 at 16:08
• @achillehui How did you arrive at such definition? – roman Nov 14 '18 at 16:10
• @roman similar question has been asked on math.SE many times. When you are here long enough, you will pick up this trick. – achille hui Nov 14 '18 at 16:12

## 3 Answers

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}

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 $$x_n \geq\sqrt 2$$ 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&>2\sqrt2 x_n\\ x_n^2-2\sqrt2 x_n+2&=(x_n-\sqrt2)^2\\ &>0 \end{align} Which is true for any $$x_n\neq \sqrt2$$

• Without loss of generality, take $x_1=2$? What? $x_1$ is defined, and is equal to $4$... – 5xum Nov 14 '18 at 15:58
• This answer does not address the main issue of the OP, which is that the first inequality OP wrote is incorrect... – 5xum Nov 14 '18 at 16:03