To show that a sequence is bounded above by $3$. Let $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2 x_n}$.
I would like to know if there is a way to see that this sequence is bounded above by 3 without induction.
I know how to use induction to show that it is in fact bounded by 2. I would like to not use induction though. Thanks for your time.
 A: OK. Let's give it a try. $x_{n} = \sqrt{2x_{n-1}}=2^{\frac{1}{2}}\cdot \sqrt{\sqrt{2x_{n-2}}}=2^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}\cdot \sqrt[4]{x_{n-2}}= 2^{\frac{1}{2}+\frac{1}{4}}\cdot \sqrt[4]{\sqrt{2x_{n-3}}}= 2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}\cdot x_{n-3}^{\frac{1}{8}}= 2^{\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{n-1}}}\cdot x_1^{\frac{1}{2^{n-1}}}= 2^{\frac{1}{2}+\frac{1}{4}+\cdots + \frac{1}{2^n}}= 2^{1-\frac{1}{2^n}}< 2 < 3.$
A: hint
$$x_{n+1}-3=\sqrt{2x_n}-3$$
$$=\frac{2x_n-9}{\sqrt{2x_n}+3}$$
$$=\frac{2(x_n-3)-3}{\sqrt{2x_n}+3}$$
$$<\frac{2(x_n-3)}{3}$$
$$<(\frac 23)^{n}(x_1-3)<0$$
A: Induction is pretty ... pervasive.  Even a simple comment like for $2^x$ is increasing because $2^{x+1} = 2\times 2^x > 2^x$ is using induction.
I could quibble and say let $x_n \ge 3$.  Then $x_n = \sqrt{2 x_{n-1}}$ and
$x_{n-1} = (\frac {x_n}2)^2 \ge (\frac 32)^2 > 3$.
So the is no least $n$ where $x_n \ge 3$ violating the well ordering principal.  So that's a contradiction and there is no $x_n \ge 3$.
What? that's not induction that's the well-ordering principal!
.... but the well-ordering principal is equivalent to induction so it is induction.
I'm not saying it's not provable with induction but I bet no-one here will be able to.
I trust Lee Mosher will attempt and succeed in finding where induction is hiding under the table.
A: For each $n \in \mathbb Z^+$, let $$b_n = -1 + \log_2 x_n.$$  Then $b_1 = -\frac{1}{2}$ and $$b_{n+1} = \frac{b_n}{2}.$$  So if $b_n < 0$, it follows that $b_{n+1} < 0$.  And here is the "induction under the table step"--therefore, $b_n < 0$ for all positive integers $n$.  Consequently, $\log_2 x_n < 1$, or $x_n < 2 < 3$ for all $n \in \mathbb Z^+$.  This is about as naked as we can make the issue, since the Peano axioms have induction built in.
