Show that the sequence $a_{n+1}=\sqrt{2+a_n}$ is convergent.

Question

Let $a_{n+1}=\sqrt{2+a_n}, a_1=\sqrt{3}$. Show that it is convergent.

I know that this is a classic nested square root sequence but how do I prove it's convergence? To know it's limit I can just take the limit on both sides and find $L$, or rewrite $a_n=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}$

Answer 1

Define $$b_n=a_n-2$$therefore $$b_{n+1}=a_{n+1}-2=\sqrt{a_n+2}-2={a_n-2\over 2+\sqrt{a_n+2}}={b_n\over 2+\sqrt {a_n+2}}$$therefore $$|b_{n+1}|=|{b_n\over 2+\sqrt {a_n+2}}|\le |{b_n \over 2}|\le \cdots \le {|b_1|\over 2^n}$$therefore $b_n\to 0$ when $n\to \infty$ which means that $a_n\to 2$