Prove a limit exists, and find the answer

Question

Suppose $x_1 = 1$ and that $2x_{n+1} = x_n + \frac{3}{x_n}$ for $n \geq 1$. Prove that $\lim_{n\to\infty} x_n$ exists, and find the limit.

Would anyone mind giving me help here, the way we prove these exist confuses me. Thanks for any help!!

Answer 1

Assume that the sequence has a limit and then it is simple to determine what that limiting value would be (taking the limit of and applying typical limit theorems to the definition of the sequence). Then you just need to show that the sequence is always moving closer to that limit point.

Answer 2

Can you show that:

$\lim_{n \to \infty} x_n = \sqrt{3}$

[If you follow the definition then the next step:

Prove that for all $\epsilon > 0$, there exists N such that if $n \geq N$, then $|x_n - \sqrt{3}| < \epsilon$

Assuming that $\epsilon$ is given, can you find $N$?]

A better approach is to show that the sequence converges monotonically and is bounded and hence has a limit. Then show that the limit is $\sqrt{3}$

1. Let $x_n > \sqrt{3}$, then it follows $\sqrt{3} < x_{n+1} < x_n$

(i) $x_{n+1} - x_n = \frac{1}{2}(\frac{3}{x_n} - x_n) = \frac{3-x_n^2}{2x_n} < 0$

(ii) $x_{n+1}^2 - 3 = \frac{1}{4}(x_n^2 + 6 + \frac{9}{x_n^2}) - 3 = \frac{1}{4}(x_n^2 - 6 + \frac{9}{x_n^2}) = \frac{1}{4}(x_n - \frac{3}{x_n})^2 = \frac{(x_n^2-3)^2}{4x_n^2} > 0$

2. A monotonic decreasing sequence bounded below is convergent. If $x_1 < \sqrt{3}$, we have already shown above that $x_2 > \sqrt{3}$. For $n > 2$, the sequence is monotonically decreasing and bounded below by $\sqrt{3}$

3. The limit $\lim_{n \to \infty} x_n$ satisfies $x = \frac{1}{2}(x + \frac{3}{x})$. We can solve it to see $x = \sqrt{3}$