How can I prove that this sequence is monotonic?

Question

I have a sequence $(u_n)$ that is defined as:

$u_0 = 2$,

$u_{n+1} =\frac{u_n}{2} + \frac{1}{u_n}$

I have tried to prove that it is monotonic using induction but I wasn't able to succeed.

How can I prove it easily ?

Thank you

Answer 1

$$ u_{n+1}-u_n=\frac{u_n}{2}+\frac{1}{u_n}-u_n=\frac{1}{u_n}-\frac{u_n}{2} =\frac{2-u_n^2}{2u_n} $$ So the sign is determined once we know whether $u_n^2>2$ or $u_n^2<2$.

For $n=0$, we have $u_0>\sqrt{2}$, so we can try proving $u_n>\sqrt{2}$, for every $n$.

Suppose it does for $n$. Then $$ u_{n+1}^2-2=\left(\frac{u_n}{2}+\frac{1}{u_n}\right)^2-2= \frac{u_n^2}{4}+1+\frac{1}{u_n^2}-2= $$ Can you finish?

Answer 2

Let us show first that

$$u_n \ge \sqrt{2}$$

This can be done by induction: $u_1 \ge 2$, and if $u_m \ge \sqrt{2}$ then $$u_{m+1} = \dfrac{u_m}{2}+\dfrac{1}{u_m} \geq \sqrt{2}$$

The last inequality holds because of the following reason. Consider the function

$$f(x) = \dfrac{x}{2} + \dfrac{1}{x}$$

Its derivative:

$$f'(x) = \dfrac{1}{2} - \dfrac{1}{x^2}$$

Here we can note that if $x \ge \sqrt{2}$, then $f'(x) > 0$, which means that $f(x)$ is increasing in $[\sqrt{2}; +\infty)$. So, $f(x) \ge f(\sqrt{2}) = \dfrac{\sqrt{2}}{2} + \dfrac{1}{\sqrt{2}} = \sqrt{2}$ for $x \ge \sqrt{2}$.

So, we have just proven that $u_{m+1} = f(u_m)\ge \sqrt{2}$ (inductive step: if we assume that $u_m \ge \sqrt{2}$ then $u_{m+1} \ge \sqrt{2}$ as well).

Now note that if $u_m \ge \sqrt{2}$ then

$$\dfrac{u_m}{2} \ge \dfrac{1}{u_m} \Rightarrow u_m \ge \dfrac{u_m}{2} +\dfrac{1}{u_m} = u_{m+1}$$

This implies that $u_{m}$ is decreasing monotonically.