# Proving a sequence has a lower bound

1. Dec 30, 2016

### Z90E532

1. The problem statement, all variables and given/known data
Given the sequence $\frac{1}{2}(x_n + \frac{2}{x_n})= x_{n+1}$, where $x_1 =1$:

Prove that $x_n$ is never less than $\sqrt{2}$, then use this to prove that $x_n - x_{n+1} \ge 0$ and conclude $\lim x_n = \sqrt{2}$.

2. Relevant equations

3. The attempt at a solution
Graphing the equation $\frac{1}{2}(x_n + \frac{2}{x_n})= y$, it's easy to see that that it is greater than $\sqrt{2}$ for all positive values of $x$. I'm just not sure how to do this algebraically.

Even if we take it that the lowest bound is $\sqrt{2}$, we can do $$x_n - x_{n+1} = x_n - \frac{x_n}{2}-\frac{1}{x_n} = \frac{x_n}{2} - \frac{1}{x_n} \ge 0$$ and if the minimum $x_n = \sqrt{2}$, we can see that it will always be $\ge 0$. I'm not sure how to do this without hand-waiving the first part, and I also don't know how to get from this to finding the limit of the sequence.

2. Dec 30, 2016

### Delta²

To find the limit is easy, if the limit exists then $\lim{x_n}=\lim{x_{n+1}}=L$ , use that and take the limit on both sides of the defining recurrence equation to make (using of the properties of limit for addition and division) a quadratic equation for L that is $\frac{1}{2}(L+\frac{2}{L})=L$.

To prove that all the terms (except the first) are bigger than $\sqrt2$ maybe induction on n could work.

3. Dec 30, 2016

### pasmith

Presumably this is for $n \geq 2$, since $x_1 = 1 < \sqrt{2}$.

Show that $$x_{n+1}^2 = 2 + \frac{(x_n^2 - 2)^2}{4x_n^2}.$$

The condition $x_n - x_{n+1} \geq 0$ is saying that the sequence is monotonic decreasing. A sequence which is monotonic decreasing and bounded below converges.

4. Dec 30, 2016

### Ray Vickson

Look at the arithmetic-geometric inequality; see
http://jwilson.coe.uga.edu/emt725/AMGM/AMGM.html