Convergence of a Sequence $x_{n+1}=\frac{1}{2}(x_n+\frac{3}{x_n})$

Question

Prove the sequence given by $x_1= 1$ and $x_{n+1}= \frac{1}{2}(x_n+\frac{3}{x_n}$) for $n \geq$ 1 is convergent to $\sqrt3$?

Struggling to come up with any real idea of where to begin.

Answer 1

You can exploit the $x_n$, $\frac{1}{x_n}$ pairs to your advantage. Also, we remember that an absolutely decreasing sequence with a lower bound converges to that lower bound.

Use AM-GM: $$ x_{n+1} = \frac{x_n+\frac{3}{x_n}}{2} \ge \sqrt{x_n \times \frac{3}{x_n}}=\sqrt{3} $$ So all $x_n$ for $n>1$ are larger than or equal to $\sqrt{3}$. This is the lower bound and now we show that the sequence is absolutely decreasing. $$ \frac{x_{n+1}}{x_n}=\frac{\frac{1}{2} (x_n+\frac{3}{x_n})}{x_n}=\frac{1}{2}+\frac{1}{2}\frac{3}{x_n^2} $$ But $x_n \ge\sqrt3$ so for $n>1$ we have $$ \frac{x_{n+1}}{x_n} \le \frac{1}{2}+\frac{1}{2}\frac{3}{\sqrt{3}^2}=1 $$ So, $x_n$ is either constant (which it is not) or absolutely decreasing (which it is).

As a result, $x_n$ is absolutely decreasing with a lower bound (aka limit) of $\sqrt{3}$.

Just to be thorough, if it happens that for some $n$, $x_{n+1}=x_n$, then you can solve for $x_n$ which gives you, as expected, $x_n=\sqrt{3}$ and this would be an equilibrium point for the system generating the sequence i.e., $x_n$ will not change after that point.

Answer 2

Rewriting$$x_{n+1}=\frac{1}{2}\left(x_n+\frac{3}{x_n}\right)=x_n - x_n+\frac{1}{2}\left(x_n+\frac{3}{x_n}\right)=x_n-\frac{x_n^2-3}{2x_n}$$ This is Newton iterative scheme for solving $x^2-3=0$