Showing that a recursive sequence is monotonous by using induction

Question

sorry that I can't post my question here by using MathJax. So here is a photo of my question

Answer 1

The sequence is bounded by below:

$$x_{n}=\frac{x_{n-1}}{2}+\frac{1}{x_{n-1}}$$

By AM-GM we have (using $x_0=2$),

$$x_{n} \geq 2\sqrt{\frac{x_{n-1}}{2}\frac{1}{x_{n-1}}}=\sqrt{2}$$

Now we show that it is monotonically decreasing. Note:

$$x_{n+1}=\frac{x_{n}}{2}+\frac{1}{x_{n}}$$

$$x_{n+1}=\frac{x_n^2+2}{2x_n}$$

So that,

$$x_{n+1}-x_{n}=\frac{x_n^2+2}{2x_n}-\frac{2x_n^2}{2x_n}$$

$$=\frac{2-x_n^2}{2x_n}$$

But $x_n \geq \sqrt{2}$ for all $n \geq 0$. This gives $-x_{n}^2 \leq -2$. So we have that $2-x_{n}^2 \leq 0$ and because $x_{n} > 0$ for all $n \geq 0$ we may write,

$$x_{n+1}-x_{n} \leq 0$$

$$x_{n+1} \leq x_{n}$$

So we can conclude the limit is $\sqrt{2}$ because if $L=\lim_{n \to \infty} x_{n}=\lim_{n \to \infty} x_{n-1}$ then we have,

$$L=\frac{L}{2}+\frac{1}{L}$$

One solution we may eliminate because our sequence is strictly positive.

Answer 2

It is easier in such cases to look into the differences $x_{n+1}-x_n$ and prove they are positive or negative than proving the inequality above directly.

ADDITIONAL HINT:

Prove (by induction) that $x_n\geq\sqrt{2}$ for any $n$.

Answer 3

I know your point is to prove the sequence is decreasing by induction which is quite a standard way, though I would like to mention that here we could have a shorter approach on proving that the sequence is decreasing. So basically, it is quite easy to prove inductively that every element $x_i$ of the sequence is positive and that $x_{n}>\sqrt{2}$ for every $n$. To prove that it is decreasing we use a quick trick : $x_{n}>\sqrt{2} \ \Rightarrow \ x_{n}^2>2 \ \Rightarrow \frac{x_{n}}{2}>\frac{1}{x_{n}}$ and it follows that $x_{i}=\frac{x_{i-1}}{2}+\frac{1}{x_{i-1}}<\frac{x_{i-1}}{2}+\frac{x_{i-1}}{2}=x_{i-1} \ $ so we are done .