IAL 2022 Jan Q6 | 國際數學站

Given that $A > B > 0$ , by letting $x = \arctan A$ and $y = \arctan B$

(a) prove that

\begin{align*} \arctan A - \arctan B =\,& \arctan \left( \frac{A - B}{1 + AB} \right)\\[2mm] \end{align*}

(3)

(b) Show that when $A = r + 2$ and $B = r$

\begin{align*} \frac{A - B}{1 + AB} =\,& \frac{2}{(1 + r)^2}\\[2mm] \end{align*}

(2)

\begin{align*} \sum_{r=1}^n \arctan \left( \frac{2}{(1 + r)^2} \right) =\,& \arctan(n + p) + \arctan(n + q) - \arctan 2 - \frac{\pi}{4}\\[2mm] \end{align*}

where $p$ and $q$ are integers to be determined.

(4)

(d) Hence, making your reasoning clear, determine

\begin{align*} \sum_{r=1}^\infty \arctan \left( \frac{2}{(1 + r)^2} \right)\\[2mm] \end{align*}

giving the answer in the form $k\pi - \arctan 2$ , where $k$ is a constant.

(2)