The sequence $\left(s_n\right)_{n=1}^{\infty}$ is defined recursively as follows: let $s_1 = 1$ and $s_{n+1} = \sqrt{1+2s_n}$ for $n \geq 1$ (So $s_1 = 1, s_2 = \sqrt{3}, s_3 = \sqrt{1+\sqrt{3}}$, etc…)
Show that the sequence $\left(s_n\right)_{n=1}^{\infty}$ is monotonically increasing.
Workings:
Proof:
Base Case: n = 1
$s_1 = 1$
$s_2 = \sqrt{3}$
$1 \leq \sqrt{3}$
Base case holds
Induction Hypothesis
Suppose that $s_n \leq s_{n+1}$ holds for some $n$.
Then for $n+1$
$s_{n+1} = \sqrt{1+2s_n}$
Now I'm not too sure on what to do. Any help will be appreciated.
Best Answer
We prove by induction.
Clearly $\sqrt{3} > 1$. Assume $s_n \geq s_{n-1}$.
Then $s_{n+1} = \sqrt{1 + 2 s_n} \geq \sqrt{1 + 2 s_{n-1}} = s_n$ using the inductive hypothesis.