Fix $a > 0$ and let $x_1 > \sqrt a$

Question

My current thought is to take $x_{n+1}-x_n$ and to prove the difference is smaller than $\sqrt a$. But it won't help me to prove $x_n$ converges to $\sqrt a$.

Answer 1

Assuming $x_1>\sqrt\alpha$, show that $x_{n+1}>\sqrt\alpha$. Then use that to show $x_{n+1}-x_n<0$, which proves the sequence is monotonically decreasing.

Therefore (it is monotonically decreasing and bounded below) the sequence has a limit $L$. The limit must satisfy: $$ L=\frac12\left(L+\frac\alpha L\right) $$ which means you can figure out exactly what it is.