6. Given that \(A > B > 0\), by letting \(x = \arctan A\) and \(y = \arctan B\)
- prove that
$$\arctan A - \arctan B = \arctan \left( \frac { A - B } { 1 + A B } \right)$$
- Show that when \(A = r + 2\) and \(B = r\)
$$\frac { A - B } { 1 + A B } = \frac { 2 } { ( 1 + r ) ^ { 2 } }$$
- Hence, using the method of differences, show that
$$\sum _ { r = 1 } ^ { n } \arctan \frac { 2 } { ( 1 + r ) ^ { 2 } } = \arctan ( n + p ) + \arctan ( n + q ) - \arctan 2 - \frac { \pi } { 4 }$$
where \(p\) and \(q\) are integers to be determined.
- Hence, making your reasoning clear, determine
$$\sum _ { r = 1 } ^ { \infty } \arctan \left( \frac { 2 } { ( 1 + r ) ^ { 2 } } \right)$$
giving the answer in the form \(k \pi - \arctan 2\), where \(k\) is a constant.