If we want to use the Monotone Convergence Theorem to show that the sequence ${x_n}$ is convergent, we must show that:
- ${x_n}$ is either monotone increasing and bounded above.
- ${x_n}$ is either monotone decreasing and bounded below.
This is what I have so far: $x_n - x_{n-1} = \frac{1}{2}(x_{n-1} + \frac{a}{x_{n-1}}) - x_{n-1} = \frac{a - {x_{n-1}}^2}{2x_{n-1}}$.
We can see from the above expression that the sequence is increasing when $x_{n-1} < \sqrt{2}$ and the sequence is decreasing when $x_{n-1} > \sqrt{2}$. I have a feeling that the sequence is bounded by $\sqrt{2}$ and we will probably need to use subsequence? How should I continue?
Thank you so much!
Let’s talk about the simplest case of the question like $$x_n=\frac{1}{2}\left(x_{n-1}+\frac{a}{x_{n-1}}\right) \textrm{ for } n\in \mathbb{N} $$ where $x_0 $ and $ a$ are greater than $0$. Then we are going to prove that the sequence $a_n$ is increasing and bounded below by $\sqrt a$.
Starting with a positive $x_0$, we have a positive sequence $x_n$. Then we can use H.M.$\ge $G.M. to get its lower bound as $$ x_n=\frac{1}{2}\left(x_{n-1}+\frac{a}{x_{n-1}}\right) \geqslant \sqrt{x_{n-1} \cdot \frac{a}{x_{n-1}}}=\sqrt{a} $$ Then use the idea of OP, we have $$ \begin{aligned} x_n-x_{n-1} & =\frac{1}{2}\left(x_{n-1}+\frac{a}{x_{n-1}}\right)-x_{n-1} . \\ & =\frac{1}{2 x_{n-1}}\left(x_{n-1}^2+a-2 x ^2_{n-1}\right) \\ & =\frac{1}{2 x_{n-1}}\left(a-x ^2_{n-1}\right) \\ & \leqslant 0 \end{aligned} $$ Therefore $x_n$ is monotonic decreasing and bounded below and hence is convergent.