Inequality proof with sequences - showing $x_n

Question

So I was doing part 2 of this question and I wanted to know if my approach is correct.

$x_{n+1} - x_n = x_n^2 + 1/4 - x_n$

Now since it is a sequence of positive terms $x_n$ > 0

Therefore $x_n^2 + 1/4 > 1/4 > x_n$

Hence $x_n^2 + 1/4 - x_n > 0$

$x_{n+1} - x_n > 0$

$x_n < x_{n+1}$

The same idea from my answer to your other question works here as well: $;x_{n+1}-x_n=x_n^2-x_{n-1}^2,$. — dxiv, Aug 13 '18 at 07:03
In case it helps, here is another question about the same recurrence: $a_{n+1}=\frac{1}{4}+a_n^2$ is converging and have a limit. Drawing cobweb plot might help to get some intuition in problems like this, you can see some examples here and here. In this case, the function is $f(x)=x^2+\frac14$. — Martin Sleziak, Aug 21 '18 at 02:22

score 1 · Answer 1 · answered Aug 13 '18 at 06:20

1

(i).

Because by assumption of the induction $$x_{n+1}=x_n^2+\frac{1}{4}<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$$

$(ii).$

$$x_{n+1}-x_n=x_n^2-x_n+\frac{1}{4}=\left(x_n-\frac{1}{2}\right)^2>0.$$

answered Aug 13 '18 at 06:20

Michael Rozenberg

1 Answers1