Show that the sequence $x_{n+1} = \frac12(x_n+ \sqrt{x_n})$ converges and determine the limit.

Question

Let $x_1=3$ and $x_{n+1} = \frac12(x_n+ \sqrt{x_n})$. Show that the sequence converges and determine the limit.

How should I formulate this in order to use $\varepsilon - \delta$? I assume that's what they would want me to use? I found $$x_2 = \frac12(3+\sqrt{3})= \frac{3+\sqrt{3}}{2},$$ but this doesn't seem to lead to anywhere.

Answer 1

Hint: Prove by induction: for all $n$, $x_n \ge 1$ and $x_{n+1} \le x_n$. Conclude that $x_n$ converges to $L \ge 1$. Find $L$.

Answer 2

Let $x_1=3$ and $x_{n+1} = \frac12(x_n+ \sqrt{x_n})$. Show that the sequence converges and determine the limit.

From the graph, $\displaystyle\lim_{n\to\infty}x_n=1$ if one starts from $x=3$ because choosing $x=1$ makes one stick at $x=1$ in the graph.

Answer 3

Another hint. Look at the function $f(x)=\frac{x+\sqrt{x}}{2}$ where $x_{n+1}=f(x_n)$. Find that $f'(x)=\frac{1}{2}+\frac{1}{4\sqrt{x}}>0, x>0$. Conclude that $f(x)$ is ascending for $x>0$. Check that $x_1>x_2$, then $x_2=f(x_1)\geq f(x_2)=x_3$, leading to $x_1>x_2\geq x_3$. And so on, by induction $(x_n)_{n\in\mathbb{N}}$ is monotone descending and bounded (say by $0$), thus the limit exists. It should be one of the solutions of $L=\frac{L+\sqrt{L}}{2}$.