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Given that

$x_1=1$ and $\forall x \in \mathbb N :$ $x_{n+1} =$ $\sqrt{3+2x_n}$

Use the principle of mathematical induction to prove, for all $n \in \mathbb N $, that $x_n < x_{n+1}$

So far I have done the base case and a little of the induction hypothesis, but I do not know how to proceed.

Base case: $n=1$

Then $x_1 = 1$ and $x_2 = \sqrt{3+2(1)} = \sqrt{5} $. Clearly, $1<\sqrt{5}$ , so this base step works.

Now, for my induction hypothesis I have $\forall n \in \mathbb N, x_n < x_{n+1}$ and I want to show that $x_{n+1} < x_{n+2}$

I start with: $x_n < x_{n+1}$ $\to$ $x_n < \sqrt{3+2x_n}$

Then $x_n < \sqrt{3+2x_n}$ $\to$ $(x_n)^2 < 3+2x_n$

So, $\cfrac{(x_n)^2}{2} -3 < x_n$

Here is where I came to a complete stop. Does anyone have suggestions on what I did wrong, or how to go from here? Also, sorry for any format issues. I am getting used to mathjax.

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  • $\begingroup$ Do you mean $x_{n+1}$ or $x_n+1$? $\endgroup$
    – Aditya
    Apr 10 '20 at 4:59
  • $\begingroup$ "$q\lt q+1$" should not need induction. Surely you mean $x_n\lt x_{n+1}$ which is written x_n\lt x_{n+1}...? $\endgroup$
    – abiessu
    Apr 10 '20 at 5:00
  • $\begingroup$ Yes I mean x_n\lt x_{n+1} $\endgroup$
    – GF101
    Apr 10 '20 at 5:01
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$Hint:$ Find some constant $a\in \mathbb R$ such that $x_n \lt a$ $\forall n \in \mathbb N.$

And using above fact complete your induction step by considering $x_{n+1}^2-x_n^2$

$a=3$ $$x_n<3 \Rightarrow (3+2x_n)<9 \Rightarrow x_{n+1}<3$$ $$x_{n+1}^2-x_n^2=3+2x_n-x_n^2=(3-x_n)(1+x_n)>0$$

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You know that $x_n \lt x_{n+1}$

$\implies x_n \lt \sqrt{3+2x_n}$

$\implies 2x_n \lt 2 \sqrt{3+2x_n}$

$\implies 3+ 2x_n \lt 3+ 2 \sqrt{3+2x_n}$

$\implies \sqrt{3+ 2x_n} \lt \sqrt{3+ 2 \sqrt{3+2x_n}}$

$\implies x_{n+1} \lt \sqrt{3+ 2 x_{n+1}}$

$\implies x_{n+1} \lt x_{n+2}$

QED.

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