Im asked to show the sequence $a_{n+1}=\sqrt{3+2a_{n}}$ where $a_{1}=0, a_{2}=1$ is increasing and bounded and therefore convergent.
I don't even know how to start the proof. Im sure it increases because its essentially adding a really small number each time and the number gets very small since its being square rooted a bunch of times but i'm having a real issue showing that it actually is increasing for all n.
$$
a_{2}=\sqrt{3} \\
a_{3}=\sqrt{3+2\sqrt{3}} \\
a_{4}=\sqrt{3+2\sqrt{3+2\sqrt{3}}}
$$
Everything I try to do leads nowhere and makes no sense. I've seen some proofs that go about by squaring both sides then solving the resulting quadratic equation assuming the sequence has a limit but im not sure how that shows that its increasing. If anyone has any tips that might push my in the right direction it would be greatly appreciated.
Best Answer
Note that $a_n \ge 0$ and
$$a_{n+1}^2 - a_n^2 = 3 + 2a_n - 3 -2a_{n-1} = 2(a_n - a_{n-1})$$
thus one can argue by induction that $a_n$ is increasing.
Although you didn't ask, you can also prove by induction that $a_n \le 3$.