<s>[INST] 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.
 [/INST] 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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