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I am give $a_{n+1}=\sqrt{a_{n}+12}$ and $a_{n}∈[-12, 4]$. I need to prove $0≤a_{n}≤4$ for all $n≥2$. I have that $a_{2}∈[0,4]$ so it works for the first case and $a_{3}∈[\sqrt{12},4]$ so it holds for that case as well, but how do I use induction to show it holds for all cases?

user228802
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  • You can check this post for suggestions on how to write induction proofs : http://math.stackexchange.com/questions/1253956/proof-writing-how-to-write-a-clear-induction-proof – Alexandre Halm May 03 '15 at 18:37

2 Answers2

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Assume $a_n \leq 4 \Rightarrow a_{n+1} = \sqrt{a_n+12} \leq \sqrt{4+12} = 4$, and this completes the induction process.

DeepSea
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You make an induction hypothesis. Suppose that $a_n \in [0,4]$ for all $n \in \{2,3,\ldots, k\}$. Then we know $$12 \leq a_k+12 \leq 16$$ hence $$\sqrt{12} \leq \sqrt{a_k+12} \leq 4$$ and by definition $a_{k+1} = \sqrt{a_k+12}$. So the result holds for all $n \in \mathbb{N}$.

graydad
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