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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.

omololo
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    “A proof that a statement is true for all integers must - at some point or another - employ mathematical induction.” – From https://math.stackexchange.com/q/120674/42969. – Martin R Mar 03 '21 at 20:57
  • Thanks! That is an interesting question. I'm wondering however, if weakening the requirement of my upper bound from 2 to 3 (or 5 or 10...) can allow one to do away with induction. – omololo Mar 03 '21 at 21:04
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    I very much doubt it because the sequence is recursively defined. – Martin R Mar 03 '21 at 21:10
  • This problem appears in a calc 2 textbook that doesn't assume any proof writing skills and hence my optimism. Thank you for the comment. – omololo Mar 03 '21 at 21:36
  • Define $f(t)=\frac1{1-t}$ on $[0,1)$ and otherwise recursively $f(t+1)=\sqrt{2f(t)}$ for $t\ge 1$. Note that $f(n)=x_n$. If you can prove that $x_n<M$ for all $n\in\Bbb N$, then you can certainly also prove that $f(t)<M$ for all $t\in[1,\infty)$ – Hagen von Eitzen Mar 03 '21 at 21:36

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OK. Let's give it a try. $x_{n} = \sqrt{2x_{n-1}}=2^{\frac{1}{2}}\cdot \sqrt{\sqrt{2x_{n-2}}}=2^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}\cdot \sqrt[4]{x_{n-2}}= 2^{\frac{1}{2}+\frac{1}{4}}\cdot \sqrt[4]{\sqrt{2x_{n-3}}}= 2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}\cdot x_{n-3}^{\frac{1}{8}}= 2^{\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{n-1}}}\cdot x_1^{\frac{1}{2^{n-1}}}= 2^{\frac{1}{2}+\frac{1}{4}+\cdots + \frac{1}{2^n}}= 2^{1-\frac{1}{2^n}}< 2 < 3.$

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    A nice example of hiding the induction under the table: How does one prove that $\frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^n} = 1 - \frac{1}{2^n}$? – Lee Mosher Mar 03 '21 at 22:44
  • @LeeMosher https://math.stackexchange.com/questions/3365993/counting-proof-for-sum-k-0n-1-2k-2n-1/3366148#3366148 – RobPratt Mar 03 '21 at 23:00
  • Hmm.... that proof assumes that if a round robin event of $k$ rounds with $2^k$ teams where half the teams are eliminated each round is possible and will always work out evenly. I'm not sure that can be assumed without induction or the well ordering principal. Maybe... – fleablood Mar 03 '21 at 23:21
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hint

$$x_{n+1}-3=\sqrt{2x_n}-3$$ $$=\frac{2x_n-9}{\sqrt{2x_n}+3}$$

$$=\frac{2(x_n-3)-3}{\sqrt{2x_n}+3}$$

$$<\frac{2(x_n-3)}{3}$$ $$<(\frac 23)^{n}(x_1-3)<0$$

hamam_Abdallah
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    Another nice example of hiding the induction under the table: how does one prove that $\frac{2(x_n-3)}{3} < \left(\frac{2}{3}\right)^n (x_1-3)$? – Lee Mosher Mar 03 '21 at 22:45
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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.

fleablood
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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.

heropup
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