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A sequence $\{ {a_n}\}_{n=1}^\infty$ is defined recursively by

$a_1=3$ and $a_{n+1}=\sqrt{3a_n-2}$ for $n=1,2,3,...$

Show by induction on n that $\{ {a_n}\}_{n=1}^\infty$ is a decreasing sequence.


So far I have this:

Base case: For n=1 we have $a_2= \sqrt7 <3.$ So the proposition holds for $n=1$.

Inductive step: Assume there exists $k \in \mathbb{N}$ such that $a_{k+1}<a_k$.

Then for $n=k+1$ we have

\begin{align} a_{k+2} & = \sqrt{3a_{k+1}-2} \\ & = \sqrt{3(\sqrt{3a_{k}-2})-2}\\ \end{align}

I'm not sure how to get to $a_{k+1}<a_k$ from here. Any help would be appreciated. Thanks

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Hint:$a_{n+2}=\sqrt{3a_{n+1}-2}\lt\sqrt{3a_n-2}=a_{n+1}$

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Hint: $\displaystyle\require{cancel}\; a_{n+1} - a_n = \sqrt{3a_n-2} - \sqrt{3a_{n-1}-2} = \frac{(3a_n-\cancel{2}) - (3a_{n-1}-\cancel{2})}{\sqrt{3a_n-2} + \sqrt{3a_{n-1}-2}} = \ldots$

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HINT One way to do it is to analyze when $\sqrt{3x-2}<x.$ This turns out to be $x>2.$ So since $a_1>2$ you need only shown that $a_n>2\implies a_{n+1}>2$

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