How to find the green area
My attempt
Each area is $\pi r_n^2=\dfrac{4\pi}{\left(n^2+2\right)^2}$. We can approximate $\sum_1^\infty \dfrac{4\pi}{\left(n^2+2\right)^2}$ to be $\int_1^\infty \dfrac{4\pi}{\left(n^2+2\right)^2}dn$. Let us evaluate the indefinite integral $\int\dfrac{4\pi}{\left(n^2+2\right)^2}=4\pi\int\dfrac1{\left(n^2+2\right)^2}dn$ first.
We use a trig sub - let $n=\sqrt2\tan\theta\implies dn=\sqrt2\sec^2\theta~d\theta$. We end up with: \begin{align*} \int\dfrac1{\left(2\tan^2\theta+2\right)^2}~dn&=\int\dfrac1{4\left(\sec^2\theta\right)^2}~dn \\ &=\dfrac14\int\dfrac1{\sec^4\theta}\sqrt2\sec^2\theta~d\theta \\ &=\dfrac1{2\sqrt2}\int\dfrac1{\sec^2\theta}~d\theta \\ &=\dfrac1{2\sqrt2}\int\cos^2\theta~d\theta \\ &=\dfrac1{4\sqrt2}\int1+\cos(2\theta)~d\theta \\ &=\dfrac1{4\sqrt2}\left(\theta+\dfrac{\sin(2\theta)}2\right) \\ &=\dfrac1{8\sqrt2}(2\theta+\sin(2\theta)) \\ \theta&=\tan^{-1}\left(\dfrac n{\sqrt2}\right) \\ I&=\dfrac1{8\sqrt2}\left(2\tan^{-1}\left(\dfrac n{\sqrt2}\right)+\sin\left(2\tan^{-1}\left(\dfrac n{\sqrt2}\right)\right)\right). \end{align*}
