The sequence $\left(s_n\right)_{n=1}^{\infty}$
is defined recursively as follows: let $s_1 = 1$, and then $s_{n+1} = \sqrt{1+2s_n}$
for $n \geq 1$. (So $s_1 = 1, s_2 = \sqrt{3}, s_3 = \sqrt{1+2\sqrt{3}}$, etc.).
Use a proof by induction to prove that $3$ is an upper bound for $\left(s_n\right)_{n=1}^{\infty}$.
Proof:
Base case: $n = 1$
$s_1 = 1 \leq 3$
Base case holds,
Induction Hypothesis:
Suppose that $s_{n+1} = \sqrt{1+2s_n} \leq 3$ holds for some $n$
Then for $n+1$:
$s_{n+2} = s_{n+1} + s_n$
$s_{n+2} = \sqrt{1+2s_n} + s_n$
Now I'm not sure what to do. Any help will be appreciated.
Best Answer
Suppose that $s_{n} = \sqrt{1+2s_{n-1}} \leq 3$ holds then we have $s_{n+1} = \sqrt{1+2s_n} \leq \sqrt{1+2*3} \leq 3$ holds for all $n$ by induction.