For a positive integer N let SN=Σn=1^Ntan ^-1 ((2)/(n^2) and let S=lim N→∞SN.

Note $\tan^{-1}\frac{2}{n^2}=\tan^{-1}(n+1)-\tan^{-1}(n-1)$ because $\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}=\frac{2}{n^2}$ (the principal-value difference stays in $(0,\pi/2)$). The sum telescopes: $S_N=\tan^{-1}(N+1)+\tan^{-1}N-\tan^{-1}1-\tan^{-1}0$. For $N=2$, $S_2=\tan^{-1}3+\tan^{-1}2-\frac\pi4$, and $\tan^{-1}2+\tan^{-1}3=\frac{3\pi}{4}$ (the sum has tangent $-1$ and lies in $(\frac\pi2,\pi)$), so $S_2=\frac\pi2$ and (A) is true. For $N=3$, $S_3=\tan^{-1}4+\tan^{-1}3-\frac\pi4$ lies between $\frac\pi2$ and $\pi$ and $\tan S_3=-\frac92$: (B) true. As $N\to\infty$, $S=\frac\pi2+\frac\pi2-\frac\pi4=\frac{3\pi}{4}$: (C) true. Since the terms are positive, $S_N$ increases to $\frac{3\pi}4<\pi$, so $S$ cannot exceed $\pi$: (D) false. So the correct options are A, B and C only, each following from the telescoped closed form of the partial sums.