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Let we have the following sequence $$x_1=1$$ $$x_{n+1}=\frac{x_n^2+a}{2x_n}$$ such that $$a>0$$ how can I prove it is a convergent sequence and its limit is $\sqrt{a}$

Mhmd Diyene
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    Do you know monotone convergence theorem? –  Sep 21 '15 at 03:01
  • To show that it's limit is $\sqrt a$, take the limit of both sides of the equation. Solve the resulting equation. This does not show convergence, however. – Paul Sinclair Sep 21 '15 at 03:05

2 Answers2

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Calculate $ x_{n+1} - \sqrt{a} $,

$$ x_{n+1} - \sqrt{a} = \frac{(x_n - \sqrt{a})^2}{2x_n},$$

and

$$ x_2 = \frac{a + 1}{2} \geqslant \sqrt{a}.$$ Thus by induction we can prove that $\forall n \geqslant 2,\,\, x_n \geqslant \sqrt{a}$.

Now that

$$ x_{n+1} - \sqrt{a} = \frac{x_n - \sqrt{a}}{2x_n} \cdot (x_n - \sqrt{a}) = \frac12\left(1 - \frac{\sqrt{a}}{x_n}\right) \cdot (x_n - \sqrt{a})$$

and we know that

$$ 0 < \frac{\sqrt{a}}{x_n} \leqslant 1.$$

So

$$ 0 \leqslant \frac12\left(1 - \frac{\sqrt{a}}{x_n}\right) < \frac12$$

and the sequence $\{x_n - \sqrt{a}\}$ converges to $0$ because

$$ \forall n > 2, \,\, 0 \leqslant x_n - \sqrt{a} < \frac{1}{2^{n-2}} (x_2 - \sqrt{a}).$$

corindo
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First suppose that the sequence is convergent say $x_n \to x$. Then apply the limit me both sides of the above equality. Using the limit of subsequence $x_ {n + 1}$ is also x, we have $x=\dfrac{x^2+a}{2x}$ and thus $2x^2=x^2+a$ what imply $x=\sqrt{a}$. Now we proof that $x_n$ is convergent, you can user the method of successive approximations.

Alcides de carvalho jr
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