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Let $x_0>0$, and $x_{n+1}=\frac{1}{2}(x_{n}+\frac{4}{x_n})$. Show that $x=\lim_{n\rightarrow \infty} x_{n} $ exists.

My attempt: Let $x_0 \geq 2$, and $x_{n+1}=\frac{1}{2}(x_{n}+\frac{4}{x_n})$.

  1. $\{x_n\}$ is bounded.

We know that $x_{n+1}=\frac{1}{2}(x_{n}+\frac{4}{x_n})\geq (x_{n} \times\frac{4}{x_n})^{1/2}=2$.

This implies that $x_{n}\geq 2$ for each $n$.

  1. $\{x_n\}$ is monotonically decreasing:

Since, $x_n \geq 2$ for each $n$. Then, $x_n^2 \geq 4$. This implies $x_n \geq \frac{4}{x_n}$.

$x_{n+1}=\frac{1}{2}(x_{n}+\frac{4}{x_n})\leq \frac{1}{2}(x_n+x_n)$.

This implies, $x_{n+1}<x_n$ for each $n$.

By monotone convergence theorem, $\{x_n\}$ is decreasing and bounded. This implies $\{x_n\}$ converges.

I proved this question for $x_0\geq 2$, but in question $x_0>0$. Is my idea correct for this solution, and how do I fix $x_0>0$problem.

May I need to show for $x_0<2$, the sequence $\{x_n\}$ is monotonically increasing and it converges to 2.

Anyone please give their views on this point?

User124356
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2 Answers2

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Making the problem more general $$x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_n}\right)$$ rewrite it as $$x_{n+1}=x_n-\frac{x_n^2-a}{2x_n}$$ and recognize that this is the Newton iterative scheme for finding the zero of $f(x)=x^2-a$.

Claude Leibovici
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Let me give you an alternative solution for the boundedness and monotonicity of the sequence:
Note that any term of this sequence is positive and finite (even thought limit may not), i.e., $0\lt x_n\lt\infty.$ Therefore there is unique angle $\theta\in(0,\pi/2)$ such that $x_n=2\tan\theta$ (for a fixed $n$). Then $$x_{n+1}=\tan\theta+\cot\theta=2\csc 2\theta\ge2$$ for all $n\ge0.$ And therefore $$x_{n+1}-x_n=\dfrac12\left(\dfrac{4}{x_n}-x_n\right)=\dfrac{(2-x_n)(2+x_n)}{2x_n}\le0$$ for all $n\ge 1.$

Bumblebee
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