From the recursive definition of a(n), above, show that \(a(n+1)^2-2 =-(a(n)^2-2)/(1+a(n)^2)\), and so \(| a(n+1)^2- 2|=|a(n)^2-2|\times1/(1+a(n)^2)\) for all \(n\geq 1\)

Prove that \(1/(1+a(n)^2)\leq 1/2\) for all \(n\geq 1\).

\[a(n+1)^2 -2 = (\frac{1+1}{(1+a(n)))^2} - 2\] = \[\frac{2 + a(n))^2}{1 + a(n))^2} -2 = \frac{2 - a(n)^2}{1 + a(n))^2}\] = \[-\frac{a(n)^2 -2}{1+ a(n)^2}\]

Use the two facts above to show that \(|a(n)^2-2|\leq \frac{1}{2^{n-1}}\), and so \(a(n)^2-2\to 0\) as \(n\to \infty\).

\(\vert a(n)^2 -2 \vert \leq \frac{1}{4^{n-1}}\) is true for n = 1.

Suppose that for some \(n \geq 1\) we have \(\vert a(n)^{2} - 2 \vert \leq \frac{1}{4^{n-1}}\) Then \[\vert a(n+1)^2 -2 \vert = \frac{\vert a(n)^2 -2 \vert}{(1+a(n)^2)} \leq \frac{1}{4^{n-1}} * \frac{1}{4} = \frac{1}{4^n}\] So by induction, \[\vert a(n)^2 -2 \vert \leq \frac{1}{4^{n-1}}\] for all \(n \geq 1\).

Factorize \(a(n)^2-2\) as a product of two linear factors with real number coefficients.

\[a(n)^2 -2 = (a(n) - \sqrt{2}(a(n) + \sqrt{2})\]

Use that facts that \(a(n)>0\) for all n and \(a(n)^2 \to 0\) to conclude that a(n) converges to \(\sqrt{2} \in \mathbb{R}\).

\[\vert a(n) - \sqrt{2} \vert = \frac{\vert a(n)^2 - 2 \vert}{(a(n) + \sqrt{2} )} < \vert a(n)^2 -2 \vert \leq \frac{1}{4^{n-1}}\]

so \(a(n) - \sqrt{2}\) converges to 0.