# 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. Commented Nov 14, 2018 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: $y_{n+1} = y_n^2 \implies y_n = y_1^{2^{n-1}}$. Commented Nov 14, 2018 at 16:08
• @achillehui How did you arrive at such definition? Commented Nov 14, 2018 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. Commented Nov 14, 2018 at 16:12

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 $$n$$.

I would like to outline a general approach to this type of questions. Namely, it is worth noting that in general whenever you have a sequence defined by an iterative process: $$x_{n+1} = f(x_{n})$$ (with $$f$$ continuous), then proving that f is increasing actually guarantees that $$sign(x_{n+1}-x_{n}) = sign(x_{2}-x_{1})$$.

This can be checked by direct substitution within the corresponding inequality, and by observing that iterates of increasing functions are also increasing.

In the situation described in the OP we can easily check that: (i) $$f^{'}(x)=1-2/x^{2}$$ and it is positive on the appropriate domain where the sequence is defined; (ii) that we obviously have by construction $$x_{2}>x_{1}$$ (simply rewrite $$f(x)=x+2/x$$). Hence the answer to the question as stated in the previous posts.

NB: it is important to clarify that in general even when $$f^{'}>0$$ the sequence defined by the iteration $$x_{n+1} = f(x_{n})$$ is by no means necessarily increasing, as it could well be decreasing if we had a situation where $$x_{2}.

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
Commented Nov 14, 2018 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
Commented Nov 14, 2018 at 16:03