Can't do this induction proof with sqrt(4) repeating

Question

Ok so I realize that the last term will be sqrt(6) but I just don't know how to manipulate this expressions to make it provable by induction. I tried rewriting it using exponents but had no luck with that

Answer 1

The sequence we are dealing with is recursively defined as $a_n=\sqrt{4+a_{n-1}}$. We need to prove that $a_n<3$ for all $n$.

Base case: $\sqrt{4} = 2 < 3$

Inductive step: We assume that for $n=k$ we have $a_k<3$ and then attempt to show that it follows the inequality holds for $n=k+1$ as well: $$a_k<3 \implies a_k+4 < 7 \implies a_k + 4< 9 \implies \sqrt{a_k+4} < 3 \implies a_{k+1} < 3$$