# Recursive sequence induction problem

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. $\blacksquare$

• Yes, looks correct. Feb 22 '18 at 20:48
• It's correct (besides the wrong usage of the arrow in the base case). Feb 22 '18 at 20:48

$$a_{n+1}-6=\sqrt {a_n+30}-\sqrt {6+30}$$ $$=\frac {a_n-6}{a_{n+1}+6}$$
we just need to observe that $$\forall n>0 \;\;a_n\ge 0.$$
if $a_n <6$ then clearly $a_{n+1}<6$.