I was wondering if the following can be proved with the definition of limits:
$$a_0=1$$ $$a_{n+1}=a_n-\frac{(a_n)^2-5}{2a_n}$$
The thing to prove is $$\lim_{n\to\infty}a_n=\sqrt5$$
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It might behoove you to rewrite the recursion term as $$a_{n+1}=\frac{a_n^2+5}{2a_n}.$$ – Cameron Buie Dec 20 '14 at 22:33
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Do you have to use the raw definition of a limit to prove this? If not, there are better methods to use. – JimmyK4542 Dec 20 '14 at 22:33
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1I disagree with the suggested rewriting. The form you have it in is immediately recognized as the Newton–Raphson method on the equation $x^2 = 5$. Wikipedia has an analysis of the convergence of this method. – A.E Dec 20 '14 at 22:39
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I know there are better methods, but I wanted to see if anyone could do this with the raw definition. I think that would be an interesting challenge – Alice Ryhl Dec 20 '14 at 22:55
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$$a_{n+1}-\sqrt{5} = \frac{(a_n-\sqrt{5})^2}{2a_n}\tag{1}$$
and we may prove by induction that $3=a_1>a_2>a_3>\ldots\geq \sqrt{5}$. Since $\left|a_2-\sqrt{5}\right|<\frac{1}{10}$,
by $(1)$ we have
$$ \left| a_{n+2}-\sqrt{5} \right| <\frac{1}{10^{2^n}},\tag{2} $$
that implies $\lim_{n\to +\infty} a_n=\sqrt{5}$ as wanted.
Jack D'Aurizio
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