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The sequence is: $x_{n+1}=\frac{1}{2}(x_{n}+\frac{a}{x_n})$ for$ , n \in \mathbb{N}_{0}$, $a>0$ and $x_{0}=a$

Hint: Show at first that $x^{2}_{n+1} - a \ge 0$ and than take $x_{n+1}-x_{n}$

I tried this way: $\frac{a}{x_0}$ is not negativ, so the sum of $x_{0}+\frac{a}{x_0}$ and $\frac{1}{2}(x_{0}+\frac{a}{x_0})$ is also not negativ.

I guess that I should prove it for $x_{n}$ that it's not negativ with induction.

Herrpeter
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    https://math.stackexchange.com/questions/1977236/for-a0-and-x-0-ge-a-prove-that-the-sequence-defined-as-x-n1-x-na-x-n?rq=1 –  Jun 11 '20 at 15:15
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    Showing that the sequence is Cauchy shouldn't be too hard. Then, since $\mathbb{R}$ is complete metric, the sequence converges. – K.defaoite Jun 11 '20 at 17:46

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Well, $$x_{n+1}=\frac{1}{2}\underbrace{\left(x_{n}+\frac{a}{x_n}\right)}_\text{$AM \ge GM$} \ge \sqrt{a} \implies x_{n+1}^2 \ge a$$ $$\iff x_{n+1}^2-a \ge0 \iff a-x_{k}^2 \le 0$$ for all $k \in \mathbb{N}$

I used here the $AM \ge GM$ inequality: $\frac{x+y}{2} \ge\sqrt{xy}$ for positive $x$ and $y$, and with our sequence being positive, we have nothing to worry about using it or dividing by one of its terms. $$x_{n+1}-x_n=\frac{1}{2}\left(\frac{a-x_n^2}{x_n}\right) \le0$$ $$\iff x_{n+1} \le x_n$$ for all $n \in \mathbb{N}$ which directly means the sequence is decreasing and thus converging. But $x_{n+1} \ge \sqrt{a}$ and thus $$x_1 \ge \dots \ge x_n \ge \dots \ge \sqrt{a}$$ Thus the limit is $\sqrt{a}$

Anas A. Ibrahim
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    Provided $a\ge\sqrt a$... –  Jun 11 '20 at 15:30
  • My last comment: $\frac58\le\frac14$ ($a=\frac12, n=0$) does not hold. The flaw is in the second paragraph. –  Jun 11 '20 at 19:20
  • @YvesDaoust I realised the mistake and fixed it. It turns out that I assumed $x_{n+1}^2 \ge a$ means that all $x_n^2 \ge a$ which is false because $x_0^2 \ge a$ is not true for $0<a<1$ – Anas A. Ibrahim Jun 11 '20 at 19:25