Convergence of $a_{n+2} = \sqrt{7-\sqrt{7+a_n}}$, with $a_1=\sqrt{7}$, $a_2 = \sqrt{7-\sqrt{7}}$?

Question

How to prove the convergence of the real sequence $\{a_n\}$, which is defined by $a_{n+2} = \sqrt{7-\sqrt{7+a_n}}$, with $a_1=\sqrt{7}$, $a_2 = \sqrt{7-\sqrt{7}}$? Furthermore, how to verifty that 2 is the limit?

Answer 1

Easy to see that $0 \le a_n \le \sqrt{7}$. Then $$|a_{n+2}-a_{m+2}|=\frac{|\sqrt{7+a_n}-\sqrt{7+a_m}|}{\sqrt{7-\sqrt{7+a_n}}+\sqrt{7-\sqrt{7+a_m}}} \le \frac{|\sqrt{7+a_n}-\sqrt{7+a_m}|}{\sqrt{7-\sqrt{7+\sqrt{7}}}+\sqrt{7-\sqrt{7+\sqrt{7}}}} < $$ $$ <|\sqrt{7+a_n}-\sqrt{7+a_m}|=\frac{|a_n-a_m|}{ \sqrt{7+a_n}+\sqrt{7+a_m}} \le \frac{|a_n-a_m|}{ \sqrt{7}+\sqrt{7}} <\frac{|a_n-a_m|}{5}$$ Since $|a_{n+2}-a_{m+2}|<\frac{|a_n-a_m|}{5}$ then $|a_{n+2N}-a_{m+2N}|<\frac{|a_n-a_m|}{5^N} \le \frac{\sqrt{7}}{5^N}$. Since $|a_{n+2N}-a_{m+2N}|< \frac{\sqrt{7}}{5^N}$ then sequence $a_n$ is Cauchy sequence. Then by Cauchy Convergence Criterion (http://mathonline.wikidot.com/the-cauchy-convergence-criterion) the sequence $a_n$ is convergent.