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.
$\endgroup$ 13 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$Hint:$s_{n+1}-s_n=\sqrt{1+2s_n}-\sqrt{1+2s_{n-1}}$.
$\endgroup$ $\begingroup$You went a little off track in the induction step. You want to show that $s_{n+1}\le s_{(n+1)+1}$--that is, $$\sqrt{1+2s_n}\le\sqrt{1+2s_{n+1}}.$$ To do so, we will use the assumption that $s_n\le s_{n+1},$ which readily implies that $$1+2s_n\le1+2s_{n+1}.\tag{$\star$}$$ You'll want to prove that the square root function is monotonically increasing on its domain, from which one can show that (1) each point of the given sequence is positive, and (2) finish the induction step.
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