The sum is equal to

MATHEMATICS · Single correct · AP · Calculus BC · Infinite Sequences and Series · Infinite Sequences and Series

Problem

The sum ∑n=1∞tan⁡−1(4nn4−2n2+2)\sum_{n=1}^{\infty}\tan^{-1}\left(\frac{4n}{n^4-2n^2+2}\right) is equal to (1) tan⁡−112+tan⁡−123\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{2}{3} (2) 4tan⁡−114\tan^{-1}1 (3) π2\frac{\pi}{2} (4) sec⁡−1(−2)\sec^{-1}(-\sqrt{2})

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