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?

  • $\begingroup$ 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. $\endgroup$ – Ewan Delanoy Nov 14 '18 at 15:53
  • $\begingroup$ 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: $y_{n+1} = y_n^2 \implies y_n = y_1^{2^{n-1}}$. $\endgroup$ – achille hui Nov 14 '18 at 16:08
  • $\begingroup$ @achillehui How did you arrive at such definition? $\endgroup$ – roman Nov 14 '18 at 16:10
  • $\begingroup$ @roman similar question has been asked on math.SE many times. When you are here long enough, you will pick up this trick. $\endgroup$ – achille hui Nov 14 '18 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 $x_{n+1}<x_n.$

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 $x_n \geq\sqrt 2$ 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 $x_n\neq \sqrt2$

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

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