In this question, the following theorem may be used without proof.
Let \(u_1, u_2, \ldots\) be a sequence of real numbers. If the sequence is
- bounded above, so \(u_n \leqslant b\) for all \(n\), where \(b\) is some fixed number
- and increasing, so \(u_n \leqslant u_{n+1}\) for all \(n\)
then there is a number \(L \leqslant b\) such that \(u_n \to L\) as \(n \to \infty\).
For positive real numbers \(x\) and \(y\), define \(\mathrm{a}(x,y) = \frac{1}{2}(x+y)\) and \(\mathrm{g}(x,y) = \sqrt{xy}\).
Let \(x_0\) and \(y_0\) be two positive real numbers with \(y_0 < x_0\) and define, for \(n \geqslant 0\)
\[ x_{n+1} = \mathrm{a}(x_n, y_n)\,, \]
\[ y_{n+1} = \mathrm{g}(x_n, y_n)\,. \]
- By considering \((\sqrt{x_n} - \sqrt{y_n})^2\), show that \(y_{n+1} < x_{n+1}\), for \(n \geqslant 0\). Show further that, for \(n \geqslant 0\)
- \(x_{n+1} < x_n\)
- \(y_n < y_{n+1}\).
Deduce that there is a value \(M\) such that \(y_n \to M\) as \(n \to \infty\).
Show that \(0 < x_{n+1} - y_{n+1} < \frac{1}{2}(x_n - y_n)\) and hence that \(x_n - y_n \to 0\) as \(n \to \infty\).
Explain why \(x_n\) also tends to \(M\) as \(n \to \infty\). - Let
\[ \mathrm{I}(p,q) = \int_0^{\infty} \frac{1}{\sqrt{(p^2 + x^2)(q^2 + x^2)}}\,\mathrm{d}x, \]
where \(p\) and \(q\) are positive real numbers with \(q < p\).
Show, using the substitution \(t = \frac{1}{2}\!\left(x - \dfrac{pq}{x}\right)\) in the integral
\[ \int_{-\infty}^{\infty} \frac{1}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}}\,\mathrm{d}t, \]
that
\[ \mathrm{I}(p,q) = \mathrm{I}\!\left(\mathrm{a}(p,q),\, \mathrm{g}(p,q)\right). \]
Hence evaluate \(\mathrm{I}(x_0, y_0)\) in terms of \(M\).