I'm stuck on how to show this sequence monotonically decreasing: $U_{n+1}=U_{n}^{2}+\frac18$ and $U_0=\frac12$

Question

as infos I have : $U_{0}=\frac{1}{2}$

$U_{n+1}=U_{n}^{2}+\frac{1}{8}$

I have already proved that $U_{n}$ is positive as the exercise requested but I still don't know how to use it with $U_{n+1}$ to show that $U_{n}$ is decreasing monotonically . Thanks for your attention.

Answer 1

Proceed by induction.

First verify that $u_1 < u_0$.

Next suppose $u_{n}<u_{n-1}$ for some positive integer n.

As you've already shown, the terms of the sequence are positive, hence $$u_n<u_{n-1}$$ $$\Rightarrow {u_n}^2 < {u_{n-1}}^2$$ $$\Rightarrow {u_n}^2 + \frac{1}{8} < {u_{n-1}}^2 + \frac{1}{8}$$ $$\Rightarrow u_{n+1} < u_n$$ which completes the induction.

Answer 2

We can prove it by induction .

$U_1=\frac{1}{4}+\frac{1}{8}<\frac{1}{2}=U_{0}$

Let $n>1$, we assume that $U_{n-1}>U_{n}$, let s prove that $U_{n}>U_{n+1}$.

Given that $U_{n-1}>U_{n}$, therefore $U_{n-1}^2>U_{n}^2$

$$U_{n+1}=U_n^2+\frac{1}{8}<U_{n-1}^2+\frac{1}{8}=U_n$$

Answer 3

It's easy to observe that $U_n<\frac{1}{2}$ for $n\in{\mathbb{N}}$

Let $u=\frac{1}{2}-\frac{\sqrt{2}}{4}$ and we get $0<U_{n+1}-u=(U_n-u)(U_n+u)<(1-\frac{\sqrt{2}}{4})(U_{n}-u)<U_n-u$