Bounded recursive sequence - Proof by induction

Question

Given the sequence $(x_n)$ defined by

\begin{cases} x_1 &= 1\\ x_{n+1} &= \frac{1}{2}\left(x_n + \frac{2}{x_n}\right), \end{cases} prove that $1 \leq x_n \leq \frac{3}{2}, \forall n \in N$.

I verified the base case for $n=1$ and $n=2$. Assumed that the boundaries hold for all $k \leq n$. Used the induction hypothesis to show that $1 \leq x_n \leq \frac{3}{2} \Rightarrow \frac{1}{2} \leq \frac{x_n}{2} \leq \frac{3}{4} $ and $1 \leq x_n \leq \frac{3}{2} \Rightarrow \frac{2}{3} \leq \frac{1}{x_n} \leq 1$.

Adding term by term I got $1 \leq \frac{7}{6} \leq \frac{1}{2}(x_n + \frac{2}{x_n}) \leq \frac{7}{4} $.

Is my reasoning correct? How could I show that $\frac{3}{2}$ is also an upper boundary, since it is smaller than $\frac{7}{4}$?

PS: I found similar questions for which $x_n \leq \sqrt{2}$.

Answer 1

For induction step we have to prove $1\leqslant x\leqslant\frac32\Rightarrow 2\leqslant x+\frac2x\leqslant3$ that is $x^2-3x+2\leqslant0$ and $x^2-2x+2\geqslant0$ for $1\leqslant x\leqslant\frac32$

Can you finish?

Answer 2

Let $a \gt 0$.

Exercise 1: $\left(\frac{a}{2} + \frac{1}{a}\right)^2 = 2 \Leftarrow\Rightarrow a = \sqrt 2$.

Exercise 2: $\left(\frac{a}{2} + \frac{1}{a}\right)^2 \ge 2$.

Hint: Apply the quadratic formula to a quartic polynomial.

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Using the exercises, if $n \gt 1$ then $x_n \gt \sqrt 2 \gt 1$, and we get 'one half' (and something stronger) of what the OP was looking for.

Since $x_2 = \frac{3}{2}$, $x_2 \le \frac{3}{2}$.

Assume for some fixed $k \gt 2$ that $x_k \leq \frac{3}{2}$.
We will show that $x_{k+1} \leq \frac{3}{2}$ is also true, allowing us to employ induction and find the bounds for the OP's sequence.

In general, if $\sqrt 2 \leq a \leq \frac{3}{2}$ then

$$\tag 1 \frac{a}{2} \leq \frac{3}{4}$$

and

$$\tag 2 \frac{1}{a} \leq \frac{1}{\sqrt 2}$$

So

$$\tag 3 \frac{1}{2}\left(a + \frac{2}{a}\right) = \frac{a}{2} + \frac{1}{a} \le \frac{3}{4} + \frac{1}{\sqrt 2} \lt \frac{3}{2}$$

Applying induction we conclude that for $n \gt 1$,

$$ \sqrt 2 \le x_n \le \frac{3}{2}$$

and we can also state that for $n \ge 1$ that

$$ 1 \le x_n \le \frac{3}{2}$$

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With the above work completed and working out some terms of the sequence $x_n$, an observant analyst would be naturally drawn to the question

If $\delta \gt 0$ and $a = \sqrt 2 + \delta$, can we assert that $\frac{a}{2} + \frac{1}{a}$ is less than $a$?

The answer is yes and is demonstrated using simple algebra.