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 $\sqrt a$.
Asked
Active
Viewed 223 times
0
-
Well, the question starts by asking you to show that the sequence is decreasing; so $x_{n + 1} - x_n < \sqrt{a}$ is useless, because you actually need to show that $x_{n + 1} - x_n \le 0$. Then you'll probably invoke a theorem about monotone convergence. – Nov 19 '18 at 16:30
-
I answered here to this question sometimes ago, in a different context (it was required that the approximating succession ${x_n}_{n\in\mathbb{N}}$ was made of only rational numbers, but the method of proof is basically the same). – Daniele Tampieri Nov 19 '18 at 16:31
1 Answers
2
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.
Arthur
- 199,419