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The question is: Let a sequence (list) of numbers be defined by $a_1=5$, $a_{n+1}=\sqrt{a_n+30}$. Prove that $a_n < 6$ for all $n\in \mathbb{N}$.

Here's my attempt at the solution. Am I missing any neccessary steps? Thank you.

For all $n\in \mathbb{N}$, let $P(n)$: $a_n<6$.

Base Case: When $n=1$, $P(1)$: $a_1<6\implies 5<6$. Therefore, the base case is true.

Inductive Assumption: Let $n\in \mathbb{N}$ be generic and assume $P(n)$ is true, i.e $a_n<6$.

Induction Step: Prove $P(n+1)$ is true. $a_n<6\implies a_n+30<36\implies \sqrt{a_n+30}<\sqrt{36}\implies \sqrt{a_n+30}<6\implies a_{n+1}<6$.

Therefore, $P(n+1)$ is true. By induction, $P(n)$ is true. $\blacksquare$

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  • $\begingroup$ Yes, looks correct. $\endgroup$ – rtybase Feb 22 '18 at 20:48
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    $\begingroup$ It's correct (besides the wrong usage of the arrow in the base case). $\endgroup$ – egreg Feb 22 '18 at 20:48
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$$a_{n+1}-6=\sqrt {a_n+30}-\sqrt {6+30} $$ $$=\frac {a_n-6}{a_{n+1}+6} $$

we just need to observe that $$\forall n>0 \;\;a_n\ge 0.$$

if $a_n <6$ then clearly $a_{n+1}<6$.

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